Circles & Sectors

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More questions on this topic: IGCSE Shape & Space: Circles & Sectors (166) A2 Pure: Radians (88)

From a Foundation paper

Edexcel

Edexcel · Old spec

AQA

OCR

Foundation June 2025 Paper 2 Q22

EdexcelFoundationCurrent spec5 marks2D Area & Perimeter Circles & Sectors

22 Here is a plan of part of Macsen’s garden.

There is a circle inside a triangle.
The circle has a diameter of 7 m.

Triangle with base 16 m and height 14 m; a circle of diameter 7 m inside it; the region inside the triangle but outside the circle is shaded

Macsen will cover the shaded area with gravel.

Gravel is sold in bags.
Each bag of gravel covers an area of 12.5 m²

(a) Work out the number of bags of gravel Macsen will need. (4)

Macsen finds that each bag of gravel only covers an area of 11 m²

(b) How does this affect your answer to part (a)? (1)

Mark scheme (a)

Answer	Mark	Mark scheme
6	P1	for process to find a relevant area, eg \(16 \times 14 \div 2\ (= 112)\) or \(\pi \times \left(\dfrac{7}{2}\right)^2\ (= 38.4\ldots)\)
	P1	for process to find the shaded area, eg \(\text{``}112\text{''} - \text{``}38.4\ldots\text{''}\ (= 73.51\ldots)\) or \(\text{``}8.96\text{''} - \text{``}3.07\ldots\text{''}\ (= 5.88\ldots)\)
	P1	for a process to find the number of bags required for a full or partial area, eg \(\text{``}73.51\ldots\text{''} \div 12.5\ (= 5.88\ldots)\) or \(\text{``}112\text{''} \div 12.5\ (= 8.96)\) or \(\text{``}38.4\ldots\text{''} \div 12.5\ (= 3.07\ldots)\) or [area] \(\div\, 12.5\) or uses 12.5 in a build up method to exceed [area], eg \(12.5 \times 6\ (= 75)\) oe
	A1	cao

Additional guidance

May be implied by \(\dfrac{49}{4}\pi\)

[area] can be any area but cannot be a length

Mark scheme (b)

Answer	Mark	Mark scheme
Statement	C1	for a valid statement relating to effect on number of bags needed, eg Acceptable examples Will need more bags It will increase Will need an extra bag or will now need 7 bags He won’t have enough There is no change (ft their [area] but must be supported by calculation) Not acceptable examples Will cover less area Needs to change the number of bags needed There is no change (unsupported or incorrect ft their [area]) He may need more bags Calculation using 11 with no supporting statement

Higher June 2025 Paper 2 Q3

EdexcelHigherCurrent spec5 marks2D Area & Perimeter Circles & Sectors

3 Here is a plan of part of Macsen’s garden.

There is a circle inside a triangle.
The circle has a diameter of 7 m.

Macsen will cover the shaded area with gravel.

Gravel is sold in bags.
Each bag of gravel covers an area of 12.5 m²

(a) Work out the number of bags of gravel Macsen will need. (4)

Macsen finds that each bag of gravel only covers an area of 11 m²

(b) How does this affect your answer to part (a)? (1)

Mark scheme (a)

Answer	Mark	Mark scheme
6	P1	for a process to find a relevant area, eg \(16 \times 14 \div 2\ (= 112)\) or \(\pi \times \left(\dfrac{7}{2}\right)^2\ (= 38.4\ldots)\)
	P1	for a process to find the shaded area, eg \(\text{``}112\text{''} - \text{``}38.4\ldots\text{''}\ (= 73.51\ldots)\) or \(\text{``}8.96\text{''} - \text{``}3.07\ldots\text{''}\ (= 5.88\ldots)\)
	P1	for a complete process to find the number of bags required for a full or partial area, eg \(\text{``}73.51\ldots\text{''} \div 12.5\ (= 5.88\ldots)\) or \(\text{``}112\text{''} \div 12.5\ (= 8.96)\) or \(\text{``}38.4\ldots\text{''} \div 12.5\ (= 3.07\ldots)\) or [area] \(\div\, 12.5\) or uses 12.5 in a build up method to exceed [area], eg \(12.5 \times 6\ (= 75)\) oe
	A1	cao

Additional guidance

May be implied by \(\dfrac{49}{4}\pi\)

[area] can be any area but cannot be a length.

Mark scheme (b)

Answer	Mark	Mark scheme
Statement	C1	for a valid statement relating to effect on number of bags needed, eg Acceptable examples Will need more bags It will increase Will need an extra bag or will now need 7 bags He won’t have enough There is no change (ft their [area] but must be supported by calculation) Not acceptable examples Will cover less area Needs to change the number of bags needed There is no change (unsupported or incorrect ft their [area]) He may need more bags A calculation using 11 with no supporting statement

Higher November 2024 Paper 3 Q11

EdexcelHigherCurrent spec3 marksCircles & Sectors

11 \(OAB\) is a sector of a circle with centre \(O\) and radius 4.7 m.

Major sector OAB of a circle with centre O and radius OA = 4.7 m; the reflex angle AOB is inside the sector

The sector has a perimeter of 34.3 m.

Find the size of the reflex angle \(AOB\).
Give your answer correct to the nearest degree. (3)

Mark scheme

Answer	Mark	Mark scheme
304	P1	for start of process to find the arc length eg \(34.3 - 2 \times 4.7\ (= 24.9)\) or for forming a suitable equation eg \(\dfrac{\angle AOB}{360} \times \pi \times 2 \times 4.7 + 2 \times 4.7 = 34.3\)
	P1	for process to isolate the arc length in an equation eg \(\dfrac{\angle AOB}{360} \times \pi \times 2 \times 4.7 = \text{``}24.9\text{''}\) or for complete process to find \(AOB\) eg \(\angle AOB = \dfrac{\text{``}24.9\text{''} \times 360}{\pi \times 2 \times 4.7}\)
	A1	answer in the range 303 to 304

Additional guidance

Condone omission of the addition of 2 radii in the equation for this mark only
eg \(\dfrac{\angle AOB}{360} \times \pi \times 2 \times 4.7 = 34.3\)

If an answer is given in the range in working and then rounded incorrectly award full marks

Higher June 2024 Paper 1 Q15

EdexcelHigherCurrent spec4 marksCircles & Sectors

15 \(OAB\) is a sector of a circle with centre \(O\) and radius 6 cm.

The length of the arc \(AB\) is \(5\pi\) cm.

Work out, in terms of \(\pi\), the area of the sector.
Give your answer in its simplest form. (4)

Mark scheme

Answer	Mark	Mark scheme
\(15\pi\)	P2	for process to find size of the angle eg \(\dfrac{5\pi \times 360}{2 \times \pi \times 6}\) oe (= 150) or for working with proportion eg \(\dfrac{5\pi}{12\pi}\ \left(= \dfrac{5}{12}\right)\) or \(\dfrac{12\pi}{5\pi}\ \left(= \dfrac{12}{5}\right)\)
	(P1	for a first step eg \(\dfrac{x}{360} \times 2 \times \pi \times 6 = 5\pi\))
	P1	(dep on P2) for process to find the area eg \(\dfrac{\text{``}150\text{''}}{360} \times \pi \times 6^2\) or \(\dfrac{5\pi}{12\pi} \times \pi \times 6^2\)
	A1	cao

Foundation June 2024 Paper 3 Q10

EdexcelFoundationCurrent spec2 marksCircles & Sectors

10 Here is a circle.

(a) On the diagram above, draw a radius of the circle. (1)

Here is another circle.

A circle with its centre marked and a straight line joining two points on the circle, not passing through the centre

(b) Write down the mathematical name for the straight line inside this circle. (1)

Answer	Mark	Mark scheme
radius drawn	B1	for radius drawn

Answer	Mark	Mark scheme
chord	B1	cao

Foundation November 2023 Paper 3 Q29

EdexcelFoundationCurrent spec4 marksCircles & Sectors Pythagoras

29 \(A\), \(B\), \(C\) and \(D\) are points on a circle such that \(ABCD\) is a square.

Square ABCD drawn inside a circle with its four corners on the circle; the regions between the square and the circle are shaded

The square \(ABCD\) has sides of length 3.5 cm.

Calculate the circumference of the circle.
Give your answer correct to 1 decimal place.
You must show all your working. (4)

Mark scheme

Answer	Mark	Mark scheme
15.6	P1	for beginning process to use Pythagoras to find diameter or radius, eg \(3.5^2 + 3.5^2\ (= 24.5)\) or \(1.75^2 + 1.75^2\ (= 6.125)\)
	P1	for complete process to find diameter or radius, eg \(\sqrt{3.5^2 + 3.5^2}\) or \(\sqrt{24.5}\ (= 4.94..)\) or \(\sqrt{1.75^2 + 1.75^2}\) or \(\sqrt{6.125}\ (= 2.47\ldots)\)
	P1	for process to find circumference of circle, eg \(\pi \times \text{``}4.94\ldots\text{''}\ (= 15.55\ldots)\) or \(2 \times \pi \times \text{``}2.47\ldots\text{''}\ (= 15.55\ldots)\)
	A1	for answer in range 15 to 16

Additional guidance

Award P1 for a correct Pythagorean statement eg
\(3.5^2 + 3.5^2 = \text{diameter}^2\)

4.94.. or 2.47.. truncated or rounded can imply P2

Accept use of 3.14 or better for \(\pi\)
Accept use of truncated values for 4.94.. or 2.47..

If an answer is shown in the range in working and then incorrectly rounded award full marks. Award 0 marks for a correct answer without correct supportive working.

Higher November 2023 Paper 3 Q12

EdexcelHigherCurrent spec2 marksCircles & Sectors

12 The diagram shows a shaded sector \(POQ\) of a circle with centre \(O\) and radius 6.2 cm.

Shaded major sector POQ of a circle centre O, radius 6.2 cm, with the reflex angle x marked at O

The area of the shaded sector is 82.6 cm²

Calculate the size of angle \(x\).
Give your answer correct to 3 significant figures. (2)

Mark scheme

Answer	Mark	Mark scheme
246	P1	for start of process to find angle inside sector eg \(\dfrac{x}{360} \times \pi \times 6.2^2 = 82.6\) or \(\dfrac{82.6}{\pi \times 6.2^2} \times 360\) or for an answer in the range 113 to 114
246	A1	for answer in range 246 to 247

Additional guidance

The size of the angle in the unshaded sector is 114

If an answer is shown in the range in working and then incorrectly rounded award full marks.

Higher November 2023 Paper 1 Q10

EdexcelHigherCurrent spec3 marksCircles & Sectors

10 The circumference of a circle is 10 m.

Work out the area of the circle.
Give your answer in terms of \(\pi\). (3)

Mark scheme

Answer	Mark	Mark scheme
\(\dfrac{25}{\pi}\)	P1	for showing \(\pi d = 10\) or \(2\pi r = 10\) or \(r = \dfrac{10}{2\pi}\) or \(d = \dfrac{10}{\pi}\)
	P1	(dep P1) for using \(r = \dfrac{5}{\pi}\) or \(r = \dfrac{10}{2\pi}\) or \(d = \dfrac{10}{\pi}\) in \(\pi r^2\) eg \(\pi \times \left(\dfrac{10}{2\pi}\right)^2\)
	A1	oe

Additional guidance

Need not be simplified but needs to show a single value of \(\pi\)

Higher June 2023 Paper 3 Q14

EdexcelHigherCurrent spec4 marksAdvanced Trigonometry Circles & Sectors

14 \(OAB\) is a triangle.
\(OBC\) is a sector of a circle, centre \(O\).

Triangle OAB joined to sector OBC. AB is 9 cm, AO is 6 cm, angle BAO is 35 degrees and the sector angle BOC is 80 degrees

Calculate the area of \(OBC\).
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Mark scheme
19.9	P1	for substituting into the cosine rule to find \(OB\), eg \((OB^2) = 9^2 + 6^2 - 2 \times 9 \times 6 \times \cos 35\)
	P1	for using correct order of operations eg \(81 + 36 - 88.46\ldots\ (= 28.5\ldots)\)
	P1	for process to find the area of the sector, eg \(\dfrac{80}{360} \times \pi \times {\text{``}5.34\ldots\text{''}}^2\)
	A1	for answer in the range 19.8 to 20

Additional guidance

May be implied by \(OB = 5.34\ldots\)
Values may be rounded or truncated.

If a correct answer within the range is shown in working but incorrectly rounded award full marks.

Foundation June 2023 Paper 1 Q6

EdexcelFoundationCurrent spec1 markCircles & Sectors

6 Freddie adds labels to this diagram of a circle.

A circle with three labels: Circumference with an arrow to the edge of the circle, Centre with an arrow to the centre point, and Diameter with an arrow to a line drawn from the centre to the edge of the circle

Explain why one of the labels is wrong. (1)

Mark scheme

Answer	Mark	Mark scheme
Explanation	C1	for a correct explanation. Acceptable examples he has labelled the radius incorrectly the diameter (label) should read radius the diameter goes from one side (of the circle) to the other the radius is labelled diameter Not acceptable examples a label is wrong there is nothing wrong with his labels the radius is wrong the centre is wrong, it should be the radius he has incorrectly labelled the diameter

Additional guidance

Do not accept a statement that is ambiguous, where one aspect contradicts another, eg. “the radius goes from the centre whereas the diameter goes all round the circle”

Foundation November 2022 Paper 2 Q21

EdexcelFoundationCurrent spec5 marks2D Area & Perimeter Circles & Sectors

21 The diagram shows a plan of Jason’s garden.

\(ABCO\) and \(DEFO\) are rectangles.
\(CDO\) is a right-angled triangle.
\(AFO\) is a sector of a circle with centre \(O\) and angle \(AOF = 90^\circ\)

Garden plan: rectangle ABCO with AB = 11 m and BC = 7 m, rectangle DEFO with ED = 7 m and FE = 9 m, triangle CDO, and quarter circle AFO with centre O

Jason is going to cover his garden with grass seed.
Each bag of grass seed covers 14 m\(^2\) of garden.
Each bag of grass seed costs £10.95

Work out how much it will cost Jason to buy all the bags of grass seed he needs. (5)

Mark scheme

Answer	Mark	Mark scheme
186.15	P1	for correctly finding the area of at least three sections, eg 3 of \(11 \times 7\ (= 77)\), or \(9 \times 7\ (= 63)\), or \(\dfrac{1}{2} \times 11 \times 9\ (= 49.5)\), or \(\dfrac{1}{4} \times \pi \times 7^2\ (= 38.4845..)\)
	P1	for a method to find the number of bags required for one area or a combination of areas eg \(\text{``}77\text{''} \div 14\ (= 5.5)\) or \(\text{``}227.9845..\text{''} \div 14\ (= 16.2846…)\)
	P1	for method to work out the total area for all four sections eg \(\text{``}77\text{''} + \text{``}63\text{''} + \text{``}49.5\text{''} + \text{``}38.4845…\text{''}\ (= 227.9845…)\) or adding the exact number of bags per section for all four sections eg \(\text{``}5.5\text{''} + \text{``}4.5\text{''} + \text{``}3.53..\text{''} + \text{``}2.74..\text{''}\ (= 16.28...)\)
	P1	for method to find the cost, eg integer number of bags \(\times\ 10.95\)
	A1	cao

Additional guidance

Note a trapezium for the rectangle and triangle should be classed as two areas.
Accept figures rounded or truncated to 1 dp or better throughout.

The third P1 mark is dependent upon correct processes seen for all four sections.

integer number of bags must come from area ÷ 14 rounded up

Higher November 2022 Paper 1 Q15

EdexcelHigherCurrent spec3 marksCircles & Sectors

15 The diagram shows a sector of a circle of radius 18 cm.

Sector with two radii of 18 cm and angle x degrees at the centre

The length of the arc is \(4\pi\) cm.

Work out the value of \(x\). (3)

Mark scheme

Answer	Mark	Mark scheme
40	P1	for a start to the process, eg \(\dfrac{x}{360} \times 2 \times \pi \times 18\ (= 4\pi)\) oe or \(\dfrac{4\pi}{2 \times \pi \times 18}\ \left(= \dfrac{x}{360}\right)\) oe
	P1	for a complete process to find \(x\), eg \(\dfrac{4\pi}{36\pi} \times 360\) oe
	A1	cao

Additional guidance

Any arrangement equivalent to this equation acceptable

Higher November 2022 Paper 2 Q4

EdexcelHigherCurrent spec5 marks2D Area & Perimeter Circles & Sectors

4 The diagram shows a plan of Jason’s garden.

\(ABCO\) and \(DEFO\) are rectangles.
\(CDO\) is a right-angled triangle.
\(AFO\) is a sector of a circle with centre \(O\) and angle \(AOF = 90^\circ\)

Plan of the garden: rectangle ABCO with AB = 11 m and BC = 7 m, rectangle DEFO with FE = 9 m and ED = 7 m, right-angled triangle CDO, and quarter circle AFO with centre O

Jason is going to cover his garden with grass seed.
Each bag of grass seed covers 14 m² of garden.
Each bag of grass seed costs £10.95

Work out how much it will cost Jason to buy all the bags of grass seed he needs. (5)

Mark scheme

Answer	Mark	Mark scheme
186.15	P1	for correctly finding the area of at least three sections, eg 3 of \(11 \times 7\ (= 77)\), or \(9 \times 7\ (= 63)\), or \(\frac{1}{2} \times 11 \times 9\ (= 49.5)\), or \(\frac{1}{4} \times \pi \times 7^2\ (= 38.4845..)\)
	P1	for a method to find the number of bags required for one area or a combination of areas eg \(\text{``}77\text{''} \div 14\ (= 5.5)\) or \(\text{``}227.9845..\text{''} \div 14\ (= 16.2846\ldots)\)
	P1	for method to work out the total area for all four sections eg \(\text{``}77\text{''} + \text{``}63\text{''} + \text{``}49.5\text{''} + \text{``}38.4845\ldots\text{''}\ (= 227.9845\ldots)\) or adding the exact number of bags per section for all four sections eg \(\text{``}5.5\text{''} + \text{``}4.5\text{''} + \text{``}3.53..\text{''} + \text{``}2.74..\text{''}\ (= 16.28...)\)
	P1	for method to find the cost, eg integer number of bags \(\times\) 10.95
	A1	cao

Additional guidance

Note a trapezium for the rectangle and triangle should be classed as two areas.
Accept figures rounded or truncated to 1 dp or better throughout.

This mark is dependent upon correct processes seen for all four sections.

integer number of bags must come from area ÷ 14 rounded up

Higher June 2022 Paper 1 Q21

EdexcelHigherCurrent spec5 marksCircles & Sectors

21 The diagram shows three circles, each of radius 4 cm.

The centres of the circles are \(A\), \(B\) and \(C\) such that \(ABC\) is a straight line and \(AB = BC = 4\) cm.

Three circles of radius 4 cm with centres A, B and C on a horizontal line; the two regions of the middle circle outside the other two circles, at the top and bottom, are shaded

Work out the total area of the two shaded regions.
Give your answer in terms of \(\pi\) (5)

Mark scheme diagram: equilateral triangle with vertices A, B and a point where the circles meet; the segment between the triangle and the arc is hatched — Mark scheme

Mark scheme diagram: isosceles triangle with apex A and the two points where the circles meet; the segment between the chord and the arc is hatched — Mark scheme

Answer	Mark	Mark scheme
\(16\sqrt{3} - \dfrac{16\pi}{3}\)	P1	for identifying an angle of 60 or 120
P1	for process to find the area of a sector of angle 60 or 120 eg \(\pi 4^2 \times \dfrac{60}{360}\ \left(= \dfrac{8\pi}{3}\right)\) or \(\pi 4^2 \times \dfrac{120}{360}\ \left(= \dfrac{16\pi}{3}\right)\)
P1	for process to find the area of an equilateral triangle eg \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60\ (= 4\sqrt{3})\) or \(\dfrac{4 \times \sqrt{4^2 - 2^2}}{2}\ (= 2\sqrt{12} \text{ or } 4\sqrt{3})\) or the area of an isosceles triangle or area of a right-angled triangle eg \(\dfrac{1}{2} \times 4 \times 4 \times \sin 120\ (= 4\sqrt{3})\) or \(\dfrac{2 \times \sqrt{4^2 - 2^2}}{2}\ (= \sqrt{12} \text{ or } 2\sqrt{3})\)
P1	for using area of sector – area of triangle to find area of a segment eg \(\pi 4^2 \times \dfrac{60}{360} - \dfrac{1}{2} \times 4 \times 4 \times \sin 60\ \left(= \dfrac{8\pi}{3} - 4\sqrt{3}\right)\) or \(\pi 4^2 \times \dfrac{120}{360} - \dfrac{1}{2} \times 4 \times 4 \times \sin 120\ \left(= \dfrac{16\pi}{3} - 4\sqrt{3}\right)\)
A1	for \(16\pi - 4\left(\dfrac{16\pi}{6} - 4\sqrt{3} + \dfrac{16\pi}{6}\right)\) or \(16\sqrt{3} - \dfrac{16\pi}{3}\) oe

Higher June 2022 Paper 1 Q13

EdexcelHigherCurrent spec3 marksCircles & Sectors Pythagoras Show That & Proof

13 A right-angled triangle is formed by the diameters of three semicircular regions, A, B and C as shown in the diagram.

A right-angled triangle with a shaded semicircle on each side: A on the hypotenuse, B on the horizontal side and C on the vertical side

Show that

area of region A = area of region B + area of region C (3)

Mark scheme

Answer	Mark	Mark scheme
Relationship shown	M1	for use of Pythagoras’ theorem, eg \(d_A^{\,2} = d_B^{\,2} + d_C^{\,2}\) or \(a^2 = b^2 + c^2\) or \((2x)^2 = (2y)^2 + (2z)^2\) or \(a = \sqrt{b^2 + c^2}\) or uses a 3, 4, 5 triangle
	M1	for forming correct expressions for the areas of at least 2 of the 3 semicircles, eg at least two of \(\dfrac{1}{2}\pi\left(\dfrac{a}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{b}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{c}{2}\right)^2\) or at least two of \(\dfrac{1}{2}\pi x^2,\ \dfrac{1}{2}\pi y^2,\ \dfrac{1}{2}\pi z^2\) or at least two of \(\dfrac{1}{2}\pi\left(\dfrac{5}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{3}{2}\right)^2,\ \dfrac{1}{2}\pi\left(\dfrac{4}{2}\right)^2\)
	C1	for a fully correct and convincing chain of reasoning, eg showing that eg \(\dfrac{1}{2}\pi\left(\dfrac{a}{2}\right)^2 = \dfrac{1}{2}\pi\left(\dfrac{b}{2}\right)^2 + \dfrac{1}{2}\pi\left(\dfrac{c}{2}\right)^2\) can be reduced to \(a^2 = b^2 + c^2\) or that \((2x)^2 = (2y)^2 + (2z)^2\) is the same as \(\dfrac{1}{2}\pi x^2 = \dfrac{1}{2}\pi y^2 + \dfrac{1}{2}\pi z^2\)

Additional guidance

May be seen at any stage
Where \(d_A\), \(a\), \(2x\), etc are their diameters

Could be any Pythagorean triple

Where \(a\), \(b\), \(c\) are their diameters
Where \(2x\), \(2y\), \(2z\) are their diameters
Where 3, 4, 5 are their diameters

Foundation November 2021 Paper 3 Q27

EdexcelFoundationCurrent spec4 marksCircles & Sectors

27 The diagram shows a sector \(OPQR\) of a circle, centre \(O\) and radius 8 cm.

Quarter circle sector OPQR with a right angle at O, OP = 8 cm and OR = 8 cm. The segment between chord PR and arc PQR is shaded

\(OPR\) is a triangle.

Work out the area of the shaded segment \(PQR\).
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Mark scheme
18.3	P1	for finding the area of the triangle eg \(0.5 \times 8 \times 8\ (= 32)\)
	P1	for finding the area of the circle \(\pi \times 8 \times 8\ (= 201.06..)\)
	P1	for finding the area of the sector eg \(\dfrac{1}{4} \times \pi \times 8^2\) or \(\text{``}201.06..\text{''} \div 4\ (= 50.26...)\)
	A1	for an answer in the range 18.2 to 18.3

Additional guidance

Accept rounded or truncated figures

If the answer is given within the range but then rounded incorrectly award full marks.

Higher November 2021 Paper 3 Q7

EdexcelHigherCurrent spec4 marksCircles & Sectors

7 The diagram shows a sector \(OPQR\) of a circle, centre \(O\) and radius 8 cm.

Quarter circle sector OPQR with a right angle at O, OP = 8 cm and OR = 8 cm; the chord PR is drawn and the segment between PR and the arc PQR is shaded

\(OPR\) is a triangle.

Work out the area of the shaded segment \(PQR\).
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Mark scheme
18.3	P1	for finding the area of the triangle eg \(0.5 \times 8 \times 8\ (= 32)\)
	P1	for finding the area of the circle \(\pi \times 8 \times 8\ (= 201.06..)\)
	P1	for finding the area of the sector eg \(\dfrac{1}{4} \times \pi \times 8^2\) or \(\text{``}201.06..\text{''} \div 4\ (= 50.26...)\)
	A1	for an answer in the range 18.2 to 18.3

Additional guidance

Accept rounded or truncated figures

If the answer is given within the range but then rounded incorrectly award full marks.

Foundation November 2019 Paper 3 Q30

EdexcelFoundationCurrent spec3 marksCircles & Sectors Show That & Proof

30 A is in the shape of a quarter circle of radius 15 cm.
B is in the shape of a circle.

Shape A, a quarter circle with radius 15 cm, and shape B, a circle

The area of A is 9 times the area of B.

Show that the radius of B is 2.5 cm. (3)

Mark scheme

Answer	Mark	Mark scheme
Result shown	M1	for finding the area of A or the area of B, eg \((\pi \times 15^2) \div 4\ (=56.25\pi = 176.(7...)\) or \(177)\) or \(\pi \times 2.5^2\ (= 6.25\pi = 19.6(3...))\)
	M1	for finding the area of A and the area of B, eg \((\pi \times 15^2) \div 4\) or \(\text{``}6.25\pi\text{''} \times 9\ (=56.25\pi = 176.(7...)\) or \(177)\) AND \(\pi \times 2.5^2\) or \(\text{``}56.25\pi\text{''} \div 9\ (= 6.25\pi = 19.6(3...))\)
	C1	for conclusion eg, \(\sqrt{56.25\pi \div 9 \div \pi} = 2.5\) oe or \(\sqrt{\dfrac{6.25\pi \times 9 \times 4}{\pi}} = 15\) oe or \(56.25\pi \div 9 = 19.6(3...)\) and \(\pi \times 2.5^2 = 19.6(3...)\) oe or \(6.25\pi \times 9 = 176.(7...)\) or 177 and \((\pi \times 15^2) \div 4 = 176(.7..)\) or 177 oe or for \(((\pi \times 15^2) \div 4) \div (\pi \times 2.5^2) = 9\) oe

Additional guidance

May work without \(\pi\) or with an approximation of \(\pi\)
Values may be rounded or truncated

Higher November 2019 Paper 3 Q23

EdexcelHigherCurrent spec5 marksCircles & Sectors Volume & Surface Area

23 The diagram shows a sector \(OACB\) of a circle with centre \(O\).
The point \(C\) is the midpoint of the arc \(AB\).

The diagram also shows a hollow cone with vertex \(O\).
The cone is formed by joining \(OA\) and \(OB\).

Formula box: Volume of cone = one third pi r squared h; Curved surface area of cone = pi r l; with a cone labelled slant height l, vertical height h and radius r

A major sector OACB of a circle with centre O and C on the arc midway between A and B; beside it a cone with vertex O, with A and B together at one side of the base and C on the opposite side

The cone has volume 56.8 cm³ and height 3.6 cm.

Calculate the size of angle \(AOB\) of sector \(OACB\).
Give your answer correct to 3 significant figures.
You must show all your working. (5)

Mark scheme

Answer	Mark	Mark scheme
264	P1	correct substitution into the volume formula, eg \(56.8 = \dfrac{1}{3} \times \pi \times r^2 \times 3.6\)
	P1	completes process to find base radius or the value of \(r^2\), eg \(r = \sqrt{\dfrac{56.8 \times 3}{\pi \times 3.6}}\ (= 3.88158\ldots)\) or \(r^2 = \dfrac{56.8}{1.2\pi}\ (= 15.066)\)
	P1	Uses Pythagoras to find the sloping length, eg \(\sqrt{\text{``}3.88\ldots\text{''}^2 + 3.6^2}\ (= 5.29\ldots)\)
	P1	process to find an equation in \(AOB\), eg \(\pi \times \text{``}3.88\text{''} \times \text{``}5.29\text{''} = \dfrac{AOB}{360} \times \pi \times \text{``}5.29\text{''}^2\) or \(\dfrac{AOB}{360} \times \pi \times 2 \times \text{``}5.29\text{''} = 2 \times \pi \times \text{``}3.88\text{''}\) or \(\dfrac{AOB}{360} \times \text{``}5.29\text{''} = \text{``}3.88\text{''}\)
	A1	answer in the range 263.9 to 264.1

Additional guidance

\(AOB\) does not need to be the subject of the equation

Foundation November 2019 Paper 3 Q10

EdexcelFoundationCurrent spec2 marksCircles & Sectors

(a) On the diagram above, draw a diameter of the circle. (1)

(b) On the diagram below, draw a segment of the circle.
Shade the segment.

(1)

Answer	Mark	Mark scheme
Diameter drawn	B1	diameter drawn

Mark scheme (b)

Answer	Mark	Mark scheme
Segment shaded	B1	segment drawn unambiguously

Additional guidance

Line must go edge to edge (condone extending outside the circle). Freehand acceptable.
Can also draw a diameter here (as semi-circle).

Foundation June 2019 Paper 3 Q29

EdexcelFoundationCurrent spec5 marksCircles & Sectors Volume & Surface Area

29 Jeremy has to cover 3 tanks completely with paint.

A cylinder with diameter 1.6 m and height 1.8 m

Each tank is in the shape of a cylinder with a top and a bottom.
The tank has a diameter of 1.6 m and a height of 1.8 m.

Jeremy has 7 tins of paint.
Each tin of paint covers 5 m²

Has Jeremy got enough paint to cover completely the 3 tanks?
You must show how you get your answer. (5)

Mark scheme

Answer	Mark	Mark scheme
No Supported	P1	for finding the area of a circle eg \(\pi \times 0.8^2\ (= 2.01...)\)
	P1	for finding the curved surface area eg \(2\pi \times 0.8 \times 1.8\ (= 9.047\ldots)\)
	P1	for use of the coverage information with an area eg \(\text{``}2.01\ldots\text{''} \div 5\ (= 0.402\ldots)\) or \(\text{``}4.02\ldots\text{''} \div 5\ (= 0.804\ldots)\) or \(\text{``}9.047\ldots\text{''} \div 5\ (= 1.8095\ldots)\) or \(\text{``}11.058\text{''} \div 5\ (= 2.2116..)\) or \(\text{``}13.069\ldots\text{''} \div 5\ (= 2.6138\ldots)\) OR for process to find total coverage for comparison eg \(5 \times 7\ (= 35)\)
	P1	(dep P1) for finding total surface area for 3 tanks eg [total surface area] \({} \times 3\ (= 39.2\ldots)\) OR for complete process to find the number of tins needed for total area of 3 tanks eg \(\text{``}13.069\text{''}... \times 3 \div 5\ (= 7.84...)\) OR for complete process to find coverage needed from each tin eg \(\text{``}13.069\text{''}... \times 3 \div 7\ (= 5.6...)\)
	C1	for conclusion “No” supported by accurate figures eg 8 tins or 7.84 (\(\gt 7\)) or \(39.2 \gt 35\) or 5.6 (\(\gt 5\))

Additional guidance

For the first P1: Must be area of circle and not part of a volume, eg \(\pi r^2 h\)
May be seen as \(2\pi r^2\)

For the second P1: May be seen from \(2\pi rh\) or from \(\pi dh\)

For the third P1: Accept numbers without working written to no less than 2dp
Do not award if a volume has been used as part of the calculation.
An independent mark for \(5 \times 7\)

For the fourth P1: [total surface area] must come from the addition of two attempts at area, but not from volume.

For the C1: Clear statement that there is not enough paint supported by correct figures for comparison.
NB: \(2.6 \times 3 = 9\) tins needed is inaccurate
8 or 7.84 tins is sufficient without restating the 7, 5.6 m² is sufficient without restating the 5 but 39.2 and 35 are needed for comparison.
A statement of “No, 8 tins” alone gets 0 marks without supporting working.

Higher June 2019 Paper 2 Q12

EdexcelHigherCurrent spec4 marksCircles & Sectors

12 \(OAB\) is a sector of a circle with centre \(O\) and radius 7 cm.

The area of the sector is 40 cm²

Calculate the perimeter of the sector.
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Mark scheme
25.4	P2	for finding the size of the angle eg \(\dfrac{40 \times 360}{\pi \times 7^2}\) (= 93.5(4..)) or for working with proportion, eg \(\dfrac{40}{49\pi}\) (= 0.259(8…) or 0.26) or \(\dfrac{49\pi}{40}\) (= 3.84(8…) or 3.85)
	(P1	for finding the area of the circle eg \(\pi \times 7^2\) (= 153(.938..) or 154))
	P1	(dep on P2) for a process to find the arc length, eg \(\dfrac{\text{``}93.5(4...)\text{''}}{360} \times \pi \times 2 \times 7\) (= 11.4(28…)) or \(\dfrac{40}{49\pi} \times \pi \times 2 \times 7\) (= 11.4(28…)) or \(\pi \times 2 \times 7 \div \dfrac{49\pi}{40}\) (= 11.4(28…))
	A1	for answer in the range 25 to 25.44

Additional guidance

May be embedded

If an answer is shown in the range in working and then incorrectly rounded award full marks.
Accept \(\dfrac{178}{7}\)

Foundation November 2018 Paper 1 Q26

EdexcelFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors Show That & Proof

26 The diagram shows a square \(ABCD\) with sides of length 20 cm.
It also shows a semicircle and an arc of a circle.

Square ABCD with AB = 20 cm along the bottom. A semicircle on AB inside the square and a quarter-circle arc from A to C with centre B. The region between the arc and the semicircle is shaded

\(AB\) is the diameter of the semicircle.
\(AC\) is an arc of a circle with centre \(B\).

Show that \(\dfrac{\text{area of shaded region}}{\text{area of square}} = \dfrac{\pi}{8}\) (4)

Mark scheme

Answer	Mark	Mark scheme
shown	C1	for method to find area of semicircle, eg \(\pi \times 10^2 \div 2\ (= 50\pi)\)
	C1	for method to find area of quarter circle, for \(\pi \times 20^2 \div 4\ (= 100\pi)\)
	C1	for a complete method to find area shaded and area of square, eg \(\pi \times 20^2 \div 4 - \pi \times 10^2 \div 2\) and \(20 \times 20\)
	C1	fully correct working leading to \(\dfrac{\pi}{8}\)

Additional guidance

Can award first 3 marks if a value for \(\pi\) is used

Working out to find the area of the shaded region must be shown

Higher November 2018 Paper 3 Q16

EdexcelHigherCurrent spec5 marksAdvanced Trigonometry Circles & Sectors

16 Here is a shaded shape \(ABCD\).

Shaded shape ABCD made from triangle AOD and a sector of a circle with centre O

The shape is made from a triangle and a sector of a circle, centre \(O\) and radius 6 cm.
\(OCD\) is a straight line.

\(AD = 14\) cm
Angle \(AOD = 140^\circ\)
Angle \(OAD = 24^\circ\)

Calculate the perimeter of the shape.
Give your answer correct to 3 significant figures. (5)

Mark scheme

Answer	Mark	Mark scheme
39.9	P1	for finding the length of the minor or major arc eg \(\dfrac{220}{360} \times \pi \times 12\) (= 23(.03834..))
	P1	for substituting into the sine or cosine rule to find \(OD\) eg \(14 \div \sin 140 = OD \div \sin 24\) or (\(OD^2 =\)) \(6^2 + 14^2 - 2 \times 6 \times 14 \times \cos 24\) (=78.5....)
	P1	for a complete process to find the length \(OD\) eg \(14 \div \sin 140 \times \sin 24\) (=8.8(58778..))
	P1	for a complete process to find the perimeter eg \(\text{``}23(.03834..)\text{''} + 14 + \text{``}8.8(58778..)\text{''} - 6\)
	A1	for an answer in the range 39.8 to 40

Additional guidance

Allow appropriate rounding if calculation seen in parts

Must involve \(OD\) in the relationship but may be implied

May be seen in multiple calculations

If an answer in the range is seen in working and then incorrectly rounded award full marks.

Higher November 2018 Paper 2 Q13

EdexcelHigherCurrent spec3 marksAdvanced Trigonometry Circles & Sectors

13 The diagram shows a circle and an equilateral triangle.

A circle with its centre marked; a horizontal diameter is the base of an equilateral triangle whose top vertex is outside the circle

One side of the equilateral triangle is a diameter of the circle.
The circle has a circumference of 44 cm.

Work out the area of the triangle.
Give your answer correct to 3 significant figures. (3)

Mark scheme

Answer	Mark	Mark scheme
84.9	P1	shows a process to find the radius or diameter eg \(44 = 2 \times \pi \times r\) or \(r = \dfrac{22}{\pi}\) or \(d = \dfrac{44}{\pi}\) or \(r = 7.0028\) or \(d = 14.0056..\)
	P1	(dep on P1) complete method to find the area eg \(\frac{1}{2} \times \text{``}d\text{''}^2 \times \sin 60\) oe, \(\frac{1}{2} \times 14 \times \tan 60\) oe, \(\frac{1}{2} \times 14 \times \sqrt{14^2 - 7^2}\) oe
	A1	for answer in the range 84.8 to 85

Additional guidance

Allow \(r\) in the range 7 to 7.1 and \(d\) in the range 14 to 14.1
Could be shown on the diagram.

If the correct answer in the range is given in working and then rounded incorrectly award full marks.

Higher November 2018 Paper 1 Q7

EdexcelHigherCurrent spec4 marks2D Area & Perimeter Circles & Sectors Show That & Proof

7 The diagram shows a square \(ABCD\) with sides of length 20 cm.
It also shows a semicircle and an arc of a circle.

\(AB\) is the diameter of the semicircle.
\(AC\) is an arc of a circle with centre \(B\).

Show that \(\dfrac{\text{area of shaded region}}{\text{area of square}} = \dfrac{\pi}{8}\) (4)

Mark scheme

Answer	Mark	Mark scheme
shown	C1	for method to find area of semicircle, eg \(\pi \times 10^2 \div 2\ (= 50\pi)\)
	C1	for method to find area of quarter circle, for \(\pi \times 20^2 \div 4\ (= 100\pi)\)
	C1	for a complete method to find area shaded and area of square, eg \(\pi \times 20^2 \div 4 - \pi \times 10^2 \div 2\) and \(20 \times 20\)
	C1	fully correct working leading to \(\dfrac{\pi}{8}\)

Additional guidance

Can award first 3 marks if a value for \(\pi\) is used

Working out to find the area of the shaded region must be shown

Foundation June 2018 Paper 2 Q19

EdexcelFoundationCurrent spec5 marks2D Area & Perimeter Circles & Sectors

19 A farmer has a field in the shape of a semicircle of diameter 50 m.

A semicircle with its diameter of 50 m along the bottom

The farmer asks Jim to build a fence around the edge of the field.
Jim tells him how much it will cost.

Total cost = £29.86 per metre of fence plus £180 for each day’s work

Jim takes three days to build the fence.

Work out the total cost. (5)

Mark scheme

Answer	Mark	Mark scheme
4378.2(0)	P1	for a process to find the circumference of the circle or the semi circle, eg \(\pi \times 50\ (= 157.0796327)\) or \(0.5 \times \pi \times 50\ (= 78.53981634)\)
	P1	for a complete process to find the perimeter of the field, eg \((0.5 \times \pi \times 50) + 50\) (= 128.5...) OR for working with one cost eg \(\text{``}157.07...\text{''} \times 29.86\) (= 4690.11..) or \(\text{``}78.5...\text{''} \times 29.86\) (= 2345.198...) or \(50 \times 29.86\) (=1493) or \(3 \times 180\) (= 540)
	P1	For finding the costs of two different aspects eg 2 of \(\text{``}78.5...\text{''} \times 29.86\) (= 2345.1..) or \(50 \times 29.86\) (= 1493) or \(3 \times 180\) (= 540)
	P1	for a adding at least 2 costs eg \(\text{``}2345.1..\text{''} + \text{``}540\text{''}\) (=2885.1..) or \(\text{``}1493\text{''} + \text{``}540\text{''}\) (=2033) or \(\text{``}128.5...\text{''} \times 29.86\) (= 3838.2..)
	A1	for answer in the range 4377 – 4392

Additional guidance

Figures may be truncated or rounded

May use circle at this point, figures imply method
One cost is 1 length or labour
Figures may be truncated or rounded

Two different aspects means arc and straight edge or arc and labour or straight edge and labour
Condone circle and labour or circle and straight edge.

Finding the cost of the perimeter is two costs added and so implies the previous P1
The circle is not allowed to be counted as one of the two costs for this mark

Foundation June 2018 Paper 1 Q9

EdexcelFoundationCurrent spec2 marksCircles & Sectors

9 The centre of this circle is marked with a cross (×).

A circle with a straight line drawn from the centre cross to the edge of the circle

(a) Write down the mathematical name of the straight line shown in the circle. (1)

(b) Write down the mathematical name of the straight line that is touching the circle.

A circle with a straight line touching it at one point

(1)

Answer	Mark	Mark scheme
Radius	B1	cao

Answer	Mark	Mark scheme
Tangent	B1	cao

Foundation November 2017 Paper 1 Q26

EdexcelFoundationCurrent spec4 marksCircles & Sectors

26 The diagram shows a logo made from three circles.

Three circles with centre O. The inner circle has radius 4 cm, the shaded ring is 3 cm wide and the outer ring is 3 cm wide

Each circle has centre \(O\).

Daisy says that exactly \(\dfrac{1}{3}\) of the logo is shaded.

Is Daisy correct?
You must show all your working. (4)

Mark scheme

Answer	Mark	Notes
Daisy is wrong (supported)	P1	for process to find area of any relevant circle ie \(\pi \times 4^2\ (= 16\pi)\), \(\pi \times 7^2\ (= 49\pi)\), \(\pi \times 10^2\ (= 100\pi)\) or \(7^2\) and \(4^2\)
	P1	for completed method to find shaded area eg \(\text{``}\pi \times 7^2\text{''} - \text{``}\pi \times 4^2\text{''}\ (= 33\pi)\) or use of radii eg \(7^2 - 4^2\ (= 33)\)
	A1	for 2 comparable figures, eg \(33\pi\) and \(100\pi\) or 33 and 100 or 103 to 103.7 and 314 to 314.2 or 103 to 103.7 and 104.6 to 104.8
	C1	statement eg No because it should be \(\dfrac{33}{100}\) and their accurate figures Allow use of \(\pi = 3\) or better

Higher November 2017 Paper 2 Q20

EdexcelHigherCurrent spec5 marksBasic Trigonometry Circles & Sectors

Sector OAC with centre O, radius 10 m and angle AOC = 120 degrees. Tangents at A and C meet at B; the region between the tangents and the arc AC is shaded

\(OAC\) is a sector of a circle, centre \(O\), radius 10 m.

\(BA\) is the tangent to the circle at point \(A\).
\(BC\) is the tangent to the circle at point \(C\).

Angle \(AOC = 120^\circ\)

Calculate the area of the shaded region.
Give your answer correct to 3 significant figures. (5)

Mark scheme

Answer	Mark	Notes
68.5	B1	for angle \(OAB = 90^\circ\) or angle \(OCB = 90^\circ\), may be seen on diagram
	P1	for a process to find the length of \(AB\) or the length of \(CB\) \((= 10\sqrt{3}\) oe\()\) eg \(10 \times \tan 60^\circ\ (= 17.3\ldots)\) or the length of \(OB\ (= 20)\), eg \(10 \div \cos 60^\circ\)
	P1	for a process (dep previous P1) to find the area of the triangle \(OAB\) \((= 50\sqrt{3}\) oe\()\) or area of triangle \(OCB\) \((= 50\sqrt{3}\) oe\()\) or area of kite \(OABC\) \((= 100\sqrt{3}\) oe\()\)
	P1	for a process to find the area of the sector \(OAC\) e.g. \(\dfrac{1}{3} \times \pi \times 10^2\ (= 104.7\ldots)\), accept rounded or truncated to 3 significant figures or more
	A1	for 68.4 − 68.6

Foundation November 2017 Paper 2 Q18

EdexcelFoundationCurrent spec3 marksCircles & Sectors

18 Hasmeet walks once round a circle with diameter 80 metres.

Circle with 8 equally spaced points labelled A to H

There are 8 points equally spaced on the circumference of the circle.

(a) Find the distance Hasmeet walks between one point and the next point. (2)

Four of the points are moved, as shown in the diagram below.

The same circle with points A, C, E and G in the same places and points B, D, F and H moved so the points are no longer equally spaced

Hasmeet walks once round the circle again.

(b) Has the mean distance that Hasmeet walks between one point and the next point changed?
You must give a reason for your answer. (1)

Answer	Mark	Notes
31.4	P1	for working with circumference formula, eg \(\pi \times 80\ (= 251.(\ldots))\) oe
A1	for answer in the range 31.4 to 31.5 accept \(10\pi\)

Answer	Mark	Notes
No (supported)	C1	Mean distance stays the same with reason, eg total distance remains unchanged or same number of points

Higher November 2017 Paper 1 Q4

EdexcelHigherCurrent spec4 marksCircles & Sectors

4 The diagram shows a logo made from three circles.

Three circles with centre O. The inner circle has radius 4 cm; the shaded ring between the inner and middle circles is 3 cm wide; the unshaded outer ring is 3 cm wide

Each circle has centre \(O\).

Daisy says that exactly \(\dfrac{1}{3}\) of the logo is shaded.

Is Daisy correct?
You must show all your working. (4)

Mark scheme

Answer	Mark	Notes
Daisy is wrong (supported)	P1	for process to find area of any relevant circle ie \(\pi \times 4^2\ (=16\pi)\), \(\pi \times 7^2\ (=49\pi)\), \(\pi \times 10^2\ (=100\pi)\) or \(7^2\) and \(4^2\)
	P1	for completed method to find shaded area eg \(\text{``}\pi \times 7^2\text{''} - \text{``}\pi \times 4^2\text{''}\ (=33\pi)\) or use of radii eg \(7^2 - 4^2\ (=33)\)
	A1	for 2 comparable figures, eg \(33\pi\) and \(100\pi\) or 33 and 100 or 103 to 103.7 and 314 to 314.2 or 103 to 103.7 and 104.6 to 104.8
	C1	statement eg No because it should be \(\dfrac{33}{100}\) and their accurate figures Allow use of \(\pi = 3\) or better

Higher November 2017 Paper 2 Q3

EdexcelHigherCurrent spec3 marksCircles & Sectors

3 Hasmeet walks once round a circle with diameter 80 metres.

A circle with 8 equally spaced points A, B, C, D, E, F, G and H marked on its circumference

There are 8 points equally spaced on the circumference of the circle.

(a) Find the distance Hasmeet walks between one point and the next point. (2)

Four of the points are moved, as shown in the diagram below.

The same circle with points A, C, E and G in the same places and points B, D, F and H moved so that they are no longer equally spaced

Hasmeet walks once round the circle again.

(b) Has the mean distance that Hasmeet walks between one point and the next point changed?
You must give a reason for your answer. (1)

Answer	Mark	Notes
31.4	P1	for working with circumference formula, eg \(\pi \times 80\ (= 251.\ldots)\) oe
A1	for answer in the range 31.4 to 31.5 accept \(10\pi\)

Answer	Mark	Notes
No (supported)	C1	Mean distance stays the same with reason, eg total distance remains unchanged or same number of points

Higher June 2017 Paper 3 Q18

EdexcelHigherCurrent spec5 marksBasic Trigonometry Circles & Sectors

Circle with centre O and radius 5 cm through A, B and C; tangents DA and DC meet at D outside the circle, with OD drawn as a dashed line; B is on the far side of the circle from D

\(A\), \(B\) and \(C\) are points on a circle of radius 5 cm, centre \(O\).
\(DA\) and \(DC\) are tangents to the circle.
\(DO = 9\) cm

Work out the length of arc \(ABC\).
Give your answer correct to 3 significant figures. (5)

Mark scheme

Working	Answer	Mark	Notes
Note \(DOC = DOA\), \(ADO = CDO\)	21.6	P1	Recognises that \(OAD\) or \(OCD\) is 90° or right angle
		P1	for using trigonometry to set up an equation in \(DOA\) or \(ADO\) eg \(\cos DOA = \dfrac{5}{9}\)
		P1	for using inverse trigonometry to find \(DOA\) or \(ADO\) eg \(DOA = \cos^{-1} \dfrac{5}{9}\ (= 56.25\ldots)\)
		P1	for a complete process to find arc length \(ABC\) or \(AC\) eg \(\dfrac{360 - 2 \times \text{``}56.25..\text{''}}{360} \times 2 \times \pi \times 5\ (= 21.598..)\) or \(\dfrac{2 \times \text{``}56.25..\text{''}}{360} \times 2 \times \pi \times 5\ (= 9.8174..)\)
		A1	for answer in the range 21.5 to 21.65

Foundation June 2017 Paper 1 Q18

EdexcelFoundationCurrent spec5 marksCircles & Sectors

18 Balena has a garden in the shape of a circle of radius 10 m.
He is going to cover the garden with grass seed to make a lawn.

Grass seed is sold in boxes.
Each box of grass seed will cover 46 m² of garden.

Balena wants to cover all the garden with grass seed.

(a) Work out an estimate for the number of boxes of grass seed Balena needs.
You must show your working. (4)

(b) Is your estimate for part (a) an underestimate or an overestimate?
Give a reason for your answer. (1)

Mark scheme (a)

Answer	Mark	Notes
6 to 8	M1	evidence of recall of area formula with correct radius e.g. \(\pi \times 10^2\)
	M1	calculation to find number of boxes, (area) \(\div\) (coverage figure)
	M1	(indep) evidence of estimation, eg \(\pi\) in range 3 to 3.2, or coverage figure of 40, 42, 45, 48 or 50
	A1	(dep on M3) answer in the range 6 to 8

Answer	Mark	Notes
underestimate	C1	e.g. (ft from (a)) underestimate: true area greater so could need more boxes. Must relate to estimation, not rounding of answer.

Higher June 2017 Paper 2 Q17

EdexcelHigherCurrent spec5 marksAdvanced Trigonometry Circles & Sectors

Sector ONQ with centre O; A on ON and B on OQ form equilateral triangle AOB; the region of the sector outside the triangle is shaded

\(ONQ\) is a sector of a circle with centre \(O\) and radius 11 cm.

\(A\) is the point on \(ON\) and \(B\) is the point on \(OQ\) such that \(AOB\) is an equilateral triangle of side 7 cm.

Calculate the area of the shaded region as a percentage of the area of the sector \(ONQ\).
Give your answer correct to 1 decimal place. (5)

Mark scheme

Answer	Mark	Notes
66.5	B1	for recognising an angle of 60 at \(AOB\)
	P1	for a process to find the area of the sector, e.g. \(\dfrac{\text{``}60\text{''}}{360} \times \pi \times 11^2\) \(\left(= 63.3\ldots \text{ or } \dfrac{121\pi}{6}\right)\)
	P1	for a process to find the area of the triangle, e.g. \(\dfrac{1}{2} \times 7^2 \times \sin \text{``}60\text{''}\) \(\left(= 21.2\ldots \text{ or } \dfrac{49\sqrt{3}}{4}\right)\)
	P1	for a process to find the required percentage, eg. \(\dfrac{\text{``}63.3\ldots\text{''} - \text{``}21.2\ldots\text{''}}{\text{``}63.3\ldots\text{''}} \times 100\)
	A1	for answer in the range 66.5 to 66.6

Higher June 2017 Paper 3 Q8

EdexcelHigherCurrent spec4 marksCircles & Sectors Pythagoras

8 A square, with sides of length \(x\) cm, is inside a circle.
Each vertex of the square is on the circumference of the circle.

The area of the circle is 49 cm².

Work out the value of \(x\).
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Notes
5.59	M1	For use of \(\pi r^2 = 49\), where \(r\) is the radius or \(r = 3.9(49\ldots)\) or diameter \(= 7.8(9865\ldots)\)
	M1	For use of Pythagoras to set up an equation in \(x^2\) e.g. \(x^2 + x^2 = (d)^2\) or \(x^2 = r^2 + r^2\)	For use of trigonometry to set up an equation in \(x\) eg \(\sin 45 = x \div d\)
	M1	(dep on M2) Rearrange to \((x^2 =)\ 2 \times \text{``}3.949..\text{''}^2\)	Rearrange to \((x =)\ \text{``}7.898..\text{''} \times \sin 45\) oe
	A1	5.5 to 5.6

1MA0 Foundation June 2017 Paper 2 Q29

EdexcelFoundationOld spec5 marksCircles & Sectors

*29 The diagram shows a plan of Brian’s lawn.

Plan of a lawn made from two semicircles with centre O, diameters 9 m and 5 m, joined by two straight lines of 2 m

Diagram NOT accurately drawn

The edge of the lawn consists of two semicircles and two straight lines.
Each semicircle has centre O.
The diameters of the semicircles are 9 m and 5 m.

Brian is going to put lawn edging around the edge of the lawn.
Lawn edging is sold in 2.4 metre rolls.

Lawn edging
£3.99 per roll or 3 rolls for £10

Brian has £35

Has Brian got enough money to buy all the rolls of lawn edging he needs?
You must show all your working. (5)

Mark scheme

Answer	Mark	Notes
No (supported)	5	M1 for \(\pi \times 9 \div 2\) (= 14.137…) or \(\pi \times 5 \div 2\) (= 7.85…) or for \(\pi \times 9\) (= 28.27…) or \(\pi \times 5\) (= 15.7…) M1 for complete method to work out perimeter: \(2 + 2 + (\pi \times 9 \div 2) + (\pi \times 5 \div 2)\) (= 25.99...) M1 (dep M1) for method to find number of rolls required for their perimeter, eg “their total perimeter” \(\div\) 2.4 eg \(25.99.. \div 2.4\) (= 10.8), \(\text{``}47.98..\text{''} \div 2.4\) (= 19.9) or \(\text{``}43.98..\text{''} \div 2.4\) (= 18.3) M1 for method to work out cost eg \(3 \times 10 + 2 \times 3.99\) (= 37.98), or \(11 \times 3.99\) (= 43.89); 20 → 67.98, 19 → 63.99 or for method to find how many rolls can be bought for £35 (= 10) C1 for a conclusion supported by fully correct answers eg 37.98 (for comparing with 35) or 10 and 10.8 OR M1 for \(\pi \times 9 \div 2\) (= 14.137…) or \(\pi \times 5 \div 2\) (= 7.85…) or for \(\pi \times 9\) (= 28.27…) or \(\pi \times 5\) (= 15.7…) M1 for complete method to work out perimeter eg \(2 + 2 + (\pi \times 9 \div 2) + (\pi \times 5 \div 2)\) (= 25.99...) M1 for a method to find how many rolls can be bought for £35 (= 10) M1 for a method to work out the coverage of 10 rolls e.g. \(10 \times 2.4\) (= 24) C1 for a conclusion supported by fully correct answers eg 25.9(...) and 24

1MA0 Higher June 2017 Paper 2 Q11

EdexcelHigherOld spec5 marksCircles & Sectors

*11 The diagram shows a plan of Brian’s lawn.

Plan of a lawn made of a large semicircle and a smaller semicircle cut out, both with centre O; the small diameter is 5 m and there is 2 m of straight edge on each side

Diagram NOT accurately drawn

The edge of the lawn consists of two semicircles and two straight lines.
Each semicircle has centre \(O\).
The diameters of the semicircles are 9 m and 5 m.

Brian is going to put lawn edging around the edge of the lawn.
Lawn edging is sold in 2.4 metre rolls.

Lawn edging
£3.99 per roll or 3 rolls for £10

Brian has £35

Has Brian got enough money to buy all the rolls of lawn edging he needs?
You must show all your working. (5)

Mark scheme

Answer	Mark	Notes
No (supported)	5	M1 for \(\pi \times 9 \div 2\) (=14.137…) or \(\pi \times 5 \div 2\) (=7.85…) or for \(\pi \times 9\) (=28.27…) or \(\pi \times 5\) (=15.7…) M1 for complete method to work out perimeter eg \(2 + 2 + (\pi \times 9 \div 2) + (\pi \times 5 \div 2)\) (= 25.99...) M1 (dep M1) for method to find number of rolls required for their perimeter, eg “their total perimeter” \(\div\) 2.4 eg \(25.99.. \div 2.4\) (=10.8), \(\text{``}47.97..\text{''} \div 2.4\) (=19.9) or \(\text{``}43.47..\text{''} \div 2.4\) (=18.3) M1 for method to work out cost eg \(3 \times 10 + 2 \times 3.99\) (= 37.98) or \(11 \times 3.99\) (=43.89), 20 \(\to\) 67.98, 19 \(\to\) 63.00 or for method to find how many rolls can be bought for £35 (= 10) C1 for a conclusion supported by fully correct answers eg 37.98 (for comparing with 35) e.g 10.8 and 10 OR M1 for \(\pi \times 9 \div 2\) (=14.137…) or \(\pi \times 5 \div 2\) (=7.85…) or for \(\pi \times 9\) (=28.27…) or \(\pi \times 5\) (=15.7…) M1 for complete method to work out perimeter eg \(2 + 2 + (\pi \times 9 \div 2) + (\pi \times 5 \div 2)\) (= 25.99...) M1 for a method to find how many rolls can be bought for £35 (=10) M1 for a method to work out the coverage of 10 rolls e.g. \(10 \times 2.4\) (=24) C1 for a conclusion supported by fully correct answers eg 25.9(...) and 24

1MA0 Foundation June 2017 Paper 2 Q5

EdexcelFoundationOld spec3 marksCircles & Sectors

5 Here is a circle.

(a) Measure the diameter of the circle. (1)

(b) In the circle below, draw a sector.
Shade your sector.

(1)

Circle with a straight line joining two points on the circle

(1)

1MA0 Foundation November 2016 Paper 2 Q28

EdexcelFoundationOld spec4 marksCircles & Sectors Pythagoras

28 The diagram shows a square ABCD inside a circle.

Square ABCD inside a circle with diagonals of 6 cm from the centre to each vertex; regions between the square and circle shaded

Diagram NOT accurately drawn

The points A, B, C and D lie on the circle.
The radius of the circle is 6 cm.

Work out the total area of the shaded regions.
Give your answer correct to 3 significant figures. (4)

Mark scheme

Answer	Mark	Notes
41.1	4	M1 for method to work out the area of the circle or quarter circle or semi-circle eg \(\pi \times 6^2\) (= 113.(09..)) or \(\pi \times 6^2 \div 2\) (= 56.5(48..)) or \(\pi \times 6^2 \div 4\) (= 28.2(7…)) M1 for method to work out the area of the square eg (= 72) oe or a triangle eg \(\frac{1}{2} \times 6 \times 6\) (= 18) M1 for complete method to find shaded area. A1 for value in the range 41.04 - 41.112

1MA0 Higher November 2016 Paper 2 Q11

EdexcelHigherOld spec4 marksCircles & Sectors

11 The diagram shows a square \(ABCD\) inside a circle.

Square ABCD inside a circle with its diagonals drawn, each half-diagonal 6 cm; the regions between the square and the circle are shaded

Diagram NOT accurately drawn

The points \(A\), \(B\), \(C\) and \(D\) lie on the circle.
The radius of the circle is 6 cm.

Work out the total area of the shaded regions.
Give your answer correct to 3 significant figures. (4)

Mark scheme

Working	Answer	Mark	Notes
\(\pi \times 6^2 - 2 \times 6 \times 6\)	41.1	4	M1 for correct method to work out the area of the circle or quarter circle or semi-circle eg \(\pi \times 6^2\) (=113(.09..)); \(\pi \times 6^2 \div 2\) = 56.5(4..); \(\pi \times 6^2 \div 4\) = 28.2(7…)) M1 for method to work out the area of the square (=72) oe or a triangle eg \(\tfrac{1}{2} \times 6 \times 6\) (=18) M1 for complete method to find shaded area. A1 for value in the range 41.04 - 41.112

1MA0 Higher June 2016 Paper 2 Q22

EdexcelHigherOld spec3 marksCircles & Sectors

Sector OAB with sector OCD inside it, angle AOB = 75 degrees, OD = 6 cm, DB = 4 cm; the region between the two arcs is shaded

Diagram NOT accurately drawn

\(OAB\) is a sector of a circle, centre \(O\).
\(OCD\) is a sector of a circle, centre \(O\).
\(OCA\) and \(ODB\) are straight lines.

Angle \(AOB = 75^\circ\)
\(OD = 6\) cm
\(DB = 4\) cm

Calculate the perimeter of the shaded region.
Give your answer correct to 3 significant figures. (3)

Mark scheme

Answer	Mark	Notes
28.9	3	M2 for \(\dfrac{75}{360} \times 2 \times \pi \times 6\) oe \(+\ \dfrac{75}{360} \times 2 \times \pi \times 10\) oe (= 7.85... + 13.08... = 20.94..) (M1 for \(\dfrac{75}{360} \times 2 \times \pi \times 6\) oe or \(\dfrac{75}{360} \times 2 \times \pi \times 10\) oe) A1 for 28.9 to 28.95

1MA0 Higher June 2016 Paper 1 Q12

EdexcelHigherOld spec3 marksCircles & Sectors

12 The diagram shows a circle inside a square.

Square ABCD of side 10 cm with a circle touching all four sides; the four corner regions are shaded

Diagram NOT accurately drawn

\(ABCD\) is a square of side 10 cm.
Each side of the square is a tangent to the circle.

Work out the total area of the shaded regions in terms of \(\pi\).
Give your answer in its simplest form. (3)

Mark scheme

Answer	Mark	Notes
\(100 - 25\pi\)	3	M1 for \(\pi \times 5 \times 5\) or \(25\pi\) M1 for \((10 \times 10 - \text{``}\pi \times 5 \times 5\text{''})\) A1 for \(100 - 25\pi\) oe NB: ignore the inclusion of any units.

1MA0 Higher November 2015 Paper 2 Q20

EdexcelHigherOld spec4 marksCircles & Sectors

20 Here are two watch faces, A and B.

Two circular watch faces of radius 2 cm; face B has a 20 degree sector marked

Diagram NOT accurately drawn

Both watch faces are circular with radius 2 cm.

The materials used to make both watch faces have the same thickness.

A is made entirely of plastic.
B has a \(20^\circ\) sector of metal and a \(340^\circ\) sector of plastic.

The ratio of the cost per cm\(^2\) of the metal to the cost per cm\(^2\) of the plastic is 3 : 2

Work out the ratio of the cost of the materials for A to the cost of the materials for B.
Give your answer in its simplest form.

You must show all your working. (4)

Mark scheme

Working	Answer	Mark	Notes
\(\pi 2^2 \times 2 = 8\pi\) \(\pi 2^2 \times \tfrac{20}{360} \times 3 + \pi 2^2 \times \tfrac{340}{360} \times 2\) \(8\pi : \tfrac{74}{9}\pi\) 72 : 74 Or \(\tfrac{20}{360} \times 3 + \tfrac{340}{360} \times 2 = \tfrac{37}{18}\) \(2 : \tfrac{37}{18}\) 36 : 37	36 : 37	4	M1 method to find relative cost of design A e.g. \(\pi 2^2\), \(k \times \pi 2^2\) M1 for a complete method to find the relative cost of the watch face for design B e.g. \(\pi 2^2 \times \tfrac{20}{360} \times \tfrac{3}{5} + \pi 2^2 \times \tfrac{340}{360} \times \tfrac{2}{5}\) M1 (dep on M1, M1) for the cost of design A : cost of design B A1 cao Or M1 for method to find fraction of the sectors in design B \(\tfrac{20}{360}, \tfrac{340}{360}\) M1 for complete method to find the relative cost of the watch face for design B M1 (dep on M1,M1) for the cost of design A : cost of design B A1 cao

1MA0 Foundation June 2015 Paper 2 Q25

EdexcelFoundationOld spec4 marksCircles & Sectors

*25 The diagram shows the top of Levi’s birthday cake.

Diagram NOT accurately drawn

The top of the cake is in the shape of a circle.
The diameter of the circle is 7 inches.

A ribbon is going to be put around the side of the cake.
Ribbons are sold in 50 cm lengths.

1 inch is 2.54 cm.

Work out if one length of ribbon is long enough to go all the way around the cake.
You must show your working. (4)

Mark scheme

Answer	Mark	Notes
No supported by working	4	M1 for \(\pi \times 7\) (= 21.9 to 22) or \(\pi \times 7 \times 2.54\) (= 55.5 to 56) M1 (dep) for a complete method that could lead to two figures that are comparable e.g. \(\pi \times 7 \times 2.54\); \(\pi \times 7\) and \(50 \div 2.54\) A1 for correct comparable figures e.g. 55.5 to 56 (cm); 21.9 to 22 (in) and 19.6 to 19.7 (in) C1 (dep M2) for a correct conclusion based on their comparable figures OR M1 for \(50 \div \pi\) (= 15.9 to 15.92) or \(50 \div 2.54\pi\) (= 6.26 to 6.27) M1 (dep) for a complete method that could lead to two figures that are comparable e.g. \((50 \div \pi) \div 2.54\); \(50 \div \pi\) and \(7 \times 2.54\) A1 for correct comparable figures e.g. 6.26 to 6.27 (in); 15.9 to 15.92 (cm) and 17.7 to 17.8 (cm) C1 (dep M2) for a correct conclusion based on their comparable figures

1MA0 Higher June 2015 Paper 2 Q11

EdexcelHigherOld spec4 marksCircles & Sectors

*11 The diagram shows the top of Levi’s birthday cake.

Diagram NOT accurately drawn

The top of the cake is in the shape of a circle.
The diameter of the circle is 7 inches.

A ribbon is going to be put around the side of the cake.
Ribbons are sold in 50 cm lengths.

1 inch is 2.54 cm.

Work out if one length of ribbon is long enough to go all the way around the cake.

You must show your working. (4)

Mark scheme

Answer	Mark	Notes
No supported by working	4	M1 for \(\pi \times 7\) (= 21.9 to 22) or \(\pi \times 7 \times 2.54\) = (55.5 to 56) M1 (dep) for a complete method that could lead to two figures that are comparable eg \(\pi \times 7 \times 2.54\); \(\pi \times 7\) and \(50 \div 2.54\) A1 for correct comparable figures eg 55.5 to 56 (cm); 21.9 to 22 (in) and 19.6 to 19.7 (in) C1 (dep M2) for a correct conclusion based on their comparable figures OR M1 for eg \(50 \div \pi\) (= 15.9 to 15.92) or \(50 \div 2.54\pi\) (=6.26 to 6.27) M1 (dep) for a complete method that could lead to two figures that are comparable eg \((50 \div \pi) \div 2.54\); \(50 \div \pi\) and \(7 \times 2.54\) A1 for correct comparable figures eg 6.26 to 6.27 (in); 15.9 to 15.92 (cm) and 17.7 to 17.8 (cm) C1 (dep M2) for a correct conclusion based on their comparable figures

1MA0 Foundation November 2014 Paper 2 Q26

EdexcelFoundationOld spec4 marksCircles & Sectors

*26 Saphia is organising a conference.
People at the conference will sit at circular tables.

Diagram NOT accurately drawn

Each table has a diameter of 140 cm.
Each person needs 60 cm around the circumference of the table.

There are 12 of these tables in the conference room.
A total of 90 people will be at the conference.

Are there enough tables in the conference room? (4)

Mark scheme

Answer	Mark	Notes
No + reason	4	M1 for intention to find the circumference eg \(140 \times \pi\) (= 439.82...) A1 for circumference = 439 – 440 M1 (dep on previous M1) for a complete method shown that could arrive at two figures that are comparable, eg \(\text{``}C\text{''} \div 60 \times 12\) (= 87.96..), \(90 \div 12 \times 60\) (= 450), \(90 \times 60 \div \text{``}C\text{''}\) (= 12.27), \(\text{``}C\text{''} \div 90 \times 12\) (= 58.64..) C1 (dep on both M marks) for No and explanation that shows a correct comparison eg only 84 people could sit around the tables or that 13 tables are needed or that 480 cm is needed.

1MA0 Higher November 2014 Paper 2 Q11

EdexcelHigherOld spec4 marksCircles & Sectors

*11 Saphia is organising a conference.
People at the conference will sit at circular tables.

Diagram NOT accurately drawn

Each table has a diameter of 140 cm.
Each person needs 60 cm around the circumference of the table.

There are 12 of these tables in the conference room.
A total of 90 people will be at the conference.

Are there enough tables in the conference room? (4)

Mark scheme

Answer	Mark	Notes
No + reason	4	M1 for intention to find the circumference eg \(140 \times \pi\) (= 439.82...) A1 for circumference = 439 - 440 M1 (dep on M1) for a complete method shown that could arrive at two figures that are comparable eg \(\text{``}C\text{''} \div 60 \times 12\) (=87.96..), \(90 \div 12 \times 60\) (=450), \(90 \times 60 \div \text{``}C\text{''}\) (=12.27), \(\text{``}C\text{''} \div 90 \times 12\) (=58.64..) C1 (dep on both M marks) for No and explanation that shows a correct comparison eg only 84 people could sit around the tables or that 13 tables are needed or that 480 cm is needed.

1MA0 Foundation June 2014 Paper 2 Q29

EdexcelFoundationOld spec7 marksCircles & Sectors Pythagoras

29 Here is a right-angled triangle.

Right-angled triangle ABC with the right angle at C, AC = 32 cm and CB = 24 cm

Diagram NOT accurately drawn

(a) Work out the length of \(AB\). (3)

Inderpal is making two mirrors.

Mirror A: circle with diameter 60 cm; Mirror B: isosceles triangle with base 48 cm and height 32 cm

Diagram NOT accurately drawn

Mirror A is in the shape of a circle.
This mirror has a diameter of 60 cm.

Mirror B is in the shape of an isosceles triangle.
This mirror has base 48 cm and height 32 cm.

Inderpal buys metal strips to put around the edge of each mirror.
The metal strip is sold in lengths of one metre.
Each one metre length of metal strip costs £5.68

(b) Work out the total amount Inderpal pays.
You must show all your working. (4)

Mark scheme (b)

Answer	Mark	Notes
22.72	4	M1 for use of \(\pi \times 60\) oe M1 for method to calculate perimeter of triangle, eg 2 × ‘40’ + 48 (=128) M1(dep M2) for complete method to find total length of strip for both mirrors or to find the cost of strip for one mirror, eg 2 × £5.68 A1 for £22.72 from correct working

1MA0 Higher June 2014 Paper 2 Q4

EdexcelHigherOld spec2 marksCircles & Sectors

4 Here is a circle.

Diagram NOT accurately drawn

The diameter of the circle is 9 cm.

Work out the circumference of this circle.
Give your answer correct to 3 significant figures. (2)

1MA0 Foundation November 2013 Paper 2 Q27

EdexcelFoundationOld spec2 marksCircles & Sectors

27 A circle has a diameter of 140 cm.

Work out the circumference of the circle.
Give your answer correct to 3 significant figures. (2)

1MA0 Higher November 2013 Paper 2 Q12

EdexcelHigherOld spec2 marksCircles & Sectors

12 A circle has a diameter of 140 cm.

Work out the circumference of the circle.
Give your answer correct to 3 significant figures. (2)

1MA0 Foundation June 2013 Paper 2 Q6

EdexcelFoundationOld spec2 marksCircles & Sectors

6 The line OP is a radius of a circle, centre O.

(a) Draw accurately the circle.

(1)

(b) On the circle below, draw a chord.

(1)

1MA0 Higher March 2013 Paper 2 Q19

EdexcelHigherOld spec2 marksCircles & Sectors

Sector OAB with radius 15 cm and angle 30 degrees at O

Diagram NOT accurately drawn

\(OAB\) is a sector of a circle, centre \(O\).
The radius of the circle is 15 cm.
The angle of the sector is \(30^\circ\).

Calculate the area of sector \(OAB\).
Give your answer correct to 3 significant figures. (2)

1MA0 Higher March 2013 Paper 2 Q5

EdexcelHigherOld spec3 marksCircles & Sectors

Diagram NOT accurately drawn

Jon has a flower garden in the shape of a circle.
The diameter of the garden is 5 metres.

Jon wants to put fencing around the edge of the garden.
The fencing costs £1.80 per metre.

Work out the total cost of the fencing. (3)

Mark scheme

Working	Answer	Mark	Notes
\(\pi \times 5 \times 1.80\)	28.27	3	M1 for use of \(\pi \times x\) (with \(x = 5\) or \(x = 2.5\)) or \(2 \times \pi \times x\) (with \(x = 5\) or \(x = 2.5\)) M1 for \(\pi \times 5 \times 1.8(0)\) or \(2 \times \pi \times 2.5 \times 1.8(0)\) A1 for 28.26 or 28.27 or 28.28 or 28.3(0) or 28.8(0)

1MA0 Higher November 2012 Paper 1 Q12

EdexcelHigherOld spec3 marks2D Area & Perimeter Circles & Sectors

12 The diagram shows a circle drawn inside a square.

Circle of radius 6 cm inside a square of side 12 cm; the region in one corner between the circle and the square is shaded

Diagram NOT accurately drawn

The circle has a radius of 6 cm.
The square has a side of length 12 cm.

Work out the shaded area.
Give your answer in terms of \(\pi\). (3)

Mark scheme

Answer	Mark	Notes
\(36 - 9\pi\)	3	M1 for \(\pi \times 6 \times 6\) or \(36\pi\) seen value 113.03-113.2 M1 for (12×12 – ‘\(\pi \times 6 \times 6\)’)÷4 or value 7.7-7.8 A1 for \(36 - 9\pi\) oe OR M1 for \(\pi \times 6 \times 6 \div 4\) or \(9\pi\) seen or value 28.2-28.3 M1 for 6×6 – ‘\(\pi \times 6 \times 6 \div 4\)’ or value 7.7-7.8 A1 for \(36 - 9\pi\) oe NB: for M marks \(\pi\) may be given numerically.

1MA0 Foundation June 2012 Paper 2 Q27

EdexcelFoundationOld spec5 marks2D Area & Perimeter Circles & Sectors

*27 Mr Weaver’s garden is in the shape of a rectangle.

In the garden
there is a patio in the shape of a rectangle
and two ponds in the shape of circles with diameter 3.8 m.

The rest of the garden is grass.

Rectangular garden 17 m by 9.5 m with a 2.8 m wide patio along one end and two circular ponds of diameter 3.8 m in the grass

Diagram NOT accurately drawn

Mr Weaver is going to spread fertiliser over all the grass.
One box of fertiliser will cover 25 m\(^2\) of grass.

How many boxes of fertiliser does Mr Weaver need?
You must show your working. (5)

Mark scheme

Working	Answer	Mark	Notes
\((17 - 2.8) \times 9.5 = 134.9\) \(\pi \times (3.8 \div 2)^2 = 11.34..\) \(134.9 - 2 \times 11.34 = 112.21\) \(112.21 \div 25 = 4.488\)	5	5	M1 for \((17 - 2.8) \times 9.5\) (=134.9) or \(17 \times 9.5 - 2.8 \times 9.5\) (\(= 161.5 - 26.6 = 134.9\)) M1 for \(\pi \times (3.8 \div 2)^2\) (=11.33 – 11.35) M1(dep on M1) for ‘134.9’ \(- 2 \times\) ‘11.34’ A1 for 112 - 113 C1(dep on at least M1) for 'He needs 5 boxes' ft from candidate's calculation rounded up to the next integer.

1MA0 Higher June 2012 Paper 2 Q5

EdexcelHigherOld spec5 marks2D Area & Perimeter Circles & Sectors

*5 Mr Weaver’s garden is in the shape of a rectangle.

In the garden
there is a patio in the shape of a rectangle
and two ponds in the shape of circles with diameter 3.8 m.

The rest of the garden is grass.

Rectangular garden 17 m by 9.5 m with a rectangular patio 2.8 m wide along one end and two circular ponds of diameter 3.8 m; the rest is grass

Diagram NOT accurately drawn

Mr Weaver is going to spread fertiliser over all the grass.
One box of fertiliser will cover 25 m\(^2\) of grass.

How many boxes of fertiliser does Mr Weaver need?
You must show your working. (5)

Mark scheme

Working	Answer	Mark	Notes
(17 – 2.8) × 9.5 = 134.9 \(\pi \times (3.8 \div 2)^2 = 11.34...\) 134.9 – 2 × 11.34... = 112.21 112.21 ÷ 25 = 4.488	5	5	M1 for (17 – 2.8) × 9.5 (=134.9) or 17×9.5 – 2.8×9.5 ( = 161.5 – 26.6 = 134.9) M1 for \(\pi \times (3.8 \div 2)^2\) (= 11.33 – 11.35) M1 (dep on M1) for ‘134.9’ – 2 × ‘11.34’ A1 for 112 - 113 C1 (dep on at least M1) for ‘He needs 5 boxes’ ft from candidate’s calculation rounded up to the next integer

Foundation June 2025 Paper 2 Q25

AQAFoundationCurrent spec4 marksCircles & Sectors

25 Here is a circle, centre \(O\).

Circle centre O with a shaded sector of angle x at O

Not drawn accurately

(a) The circle has a circumference of 20 cm

Assume that angle \(x\) is \(90^\circ\)

Work out the shaded area.

Give your answer as a decimal. [3 marks]

(b) In fact, angle \(x\) is smaller than \(90^\circ\)

What does this mean about the shaded area?

Tick one box. [1 mark]

It is smaller than the answer to part (a)
It is the same as the answer to part (a)
It is larger than the answer to part (a)
It could be any of the above

Mark scheme (a)

Answer	Mark	Comments
\(20 \div \pi\) or [6.36, 6.37] or 6.4 or \(20 \div 2\pi\) or [3.18, 3.185] or 3.2	M1	oe
\(\pi \times (20 \div 2\pi)^2 \times \dfrac{90}{360}\) or \(\dfrac{100}{\pi} \times \dfrac{90}{360}\) or \(\dfrac{25}{\pi}\)	M1dep	oe full method
[7.93, 7.97]	A1

Additional guidance

Accept [3.14, 3.142] for \(\pi\) throughout

Ignore subsequent rounding after answer [7.93, 7.97] seen in working

Foundation June 2025 Paper 1 Q17

AQAFoundationCurrent spec1 markCircles & Sectors

17 A chord is drawn on a circle.

Which statement is correct?

Tick one box. [1 mark]

The chord must be longer than the radius.
The chord must be equal in length to the radius.
The chord must be shorter than the radius.
The chord could be longer than, equal in length to or shorter than the radius.

Foundation June 2025 Paper 1 Q11

AQAFoundationCurrent spec1 markCircles & Sectors Constructions

11 In this question use a pair of compasses.

The line shown is the diameter of a circle, centre X.

Draw the circle. [1 mark]

A horizontal line segment with its midpoint marked X

Higher June 2025 Paper 2 Q8

AQAHigherCurrent spec4 marksCircles & Sectors

8 Here is a circle, centre \(O\).

Not drawn accurately

(a) The circle has a circumference of 20 cm

Assume that angle \(x\) is \(90^\circ\)

Work out the shaded area.

Give your answer as a decimal. [3 marks]

(b) In fact, angle \(x\) is smaller than \(90^\circ\)

What does this mean about the shaded area?

Tick one box. [1 mark]

It is smaller than the answer to part (a)
It is the same as the answer to part (a)
It is larger than the answer to part (a)
It could be any of the above

Mark scheme (a)

Answer	Mark	Comments
\(20 \div \pi\) or [6.36, 6.37] or 6.4 or \(20 \div 2\pi\) or [3.18, 3.185] or 3.2	M1	oe
\(\pi \times (20 \div 2\pi)^2 \times \dfrac{90}{360}\) or \(\dfrac{100}{\pi} \times \dfrac{90}{360}\) or \(\dfrac{25}{\pi}\)	M1dep	oe full method
[7.93, 7.97]	A1

Additional guidance

Accept [3.14, 3.142] for \(\pi\) throughout

Ignore subsequent rounding after answer [7.93, 7.97] seen in working

Higher June 2025 Paper 3 Q3

AQAHigherCurrent spec3 marksCircles & Sectors

3 Here is a circle, centre \(O\)

Circle with centre O. C labels a line touching the circle at the top. D labels a shaded region between a vertical chord and the arc on the right. A labels a line from O straight down to the circle. B labels a shaded region bounded by two radii and an arc on the lower left

Match each letter to the correct word.

One has been done for you. [3 marks]

Boxes A, B, C and D on the left and boxes Arc, Chord, Diameter, Radius, Sector, Segment and Tangent on the right. A is already joined to Radius

Foundation November 2024 Paper 1 Q26

AQAFoundationCurrent spec4 marksCircles & Sectors

26 A large circle and a small circle are shown.

The radius of the large circle is 12 cm

radius of large circle : radius of small circle = 4 : 1

A small white circle inside a large circle. The region of the large circle outside the small circle is shaded

Not drawn accurately

Work out the shaded area.

Give your answer in terms of \(\pi\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(12 \div 4\) or 3	M1	oe may be on the diagram may be seen in a ratio
\(\pi \times\) their \(3 \times\) their 3 or \(9\pi\) or [28.2, 28.3]	M1dep	oe
\(\pi \times 12 \times 12\) or \(144\pi\) or [452.1, 452.45]	M1	oe
\(135\pi\)	A1	SC2 135

Additional guidance

Condone eg \(\pi 9\) for \(9\pi\)
Condone use of \(\dfrac{22}{7}\) or 3.1 or better for \(\pi\) up to M3
Answer 135 with \(135\pi\) in working Answer 135 without \(135\pi\) in working	M1M1M1A0 SC2
Answer [423.8, 424.3]	M1M1M1A0

Higher November 2024 Paper 1 Q7

AQAHigherCurrent spec4 marksCircles & Sectors

7 A large circle and a small circle are shown.

The radius of the large circle is 12 cm

radius of large circle : radius of small circle = 4 : 1

Not drawn accurately

Work out the shaded area.

Give your answer in terms of \(\pi\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(12 \div 4\) or 3	M1	oe may be on the diagram may be seen in a ratio
\(\pi \times\) their \(3 \times\) their 3 or \(9\pi\) or [28.2, 28.3]	M1dep	oe
\(\pi \times 12 \times 12\) or \(144\pi\) or [452.1, 452.45]	M1	oe
\(135\pi\)	A1	SC2 135

Additional guidance

Condone eg \(\pi 9\) for \(9\pi\)
Condone use of \(\dfrac{22}{7}\) or 3.1 or better for \(\pi\) up to M3
Answer 135 with \(135\pi\) in working Answer 135 without \(135\pi\) in working	M1M1M1A0 SC2
Answer [423.8, 424.3]	M1M1M1A0

Higher November 2024 Paper 2 Q1

AQAHigherCurrent spec3 marksCircles & Sectors

1 Choose a word from the list below to complete each sentence.

arc centre circumference diameter
radius sector segment tangent

(a) The length of the \(\ldots\ldots\ldots\ldots\ldots\) is double the length of the radius. [1 mark]

(b) A \(\ldots\ldots\ldots\ldots\ldots\) is a region created by drawing a chord through a circle. [1 mark]

Foundation June 2024 Paper 1 Q22

AQAFoundationCurrent spec4 marksCircles & Sectors Volume & Surface Area

22 Here is a cone.

Cone with vertical height 12 cm, slant height 13 cm and base radius 5 cm

(a)

Curved surface area of a cone \(= \pi r l\)
where \(r\) is the radius and \(l\) is the slant height

Beth tries to work out the curved surface area in terms of \(\pi\)

Curved surface area of the cone \(= \pi \times 5 \times 12\)
\(= 60\pi\) cm\(^2\)

What mistake has she made? [1 mark]

(b) Adam uses \(\pi = 3\) to estimate the area of the base of the cone.

Work out his estimate. [2 marks]

Is Beth’s estimate more than or less than Adam’s estimate?

Tick a box.

More than
Less than

Give a reason for your answer. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Correct statement	B1	eg she used the height instead of the slant height or she used the vertical height or she used 12 (instead of 13)

Additional guidance

Check diagram
For ‘vertical’ accept anything that implies she has used the wrong height
Condone ‘length’ to mean ‘height’ or ‘slant height’
12 or 13 circled on the diagram must be accompanied by a supporting statement
Indicates ‘12’ in the calculation	B1
She should have done \(\pi \times 5 \times 13\)	B1
It should be \(65\pi\)	B1
She used the wrong height / the (value of) \(l\) is wrong	B1
She hasn’t used the slant height (she used the (vertical) height)	B1
She hasn’t used the 13	B1
She hasn’t used the 13 and should be \(5 \times 12 \times 13 \times \pi\)	B0
The multiplication used the wrong number(s)	B0
She hasn’t used a value for \(\pi\)	B0
An incorrect statement with a correct statement eg she used 13 instead of 12 and didn’t square the radius	B0

Mark scheme (c)

Answer	Mark	Comments
‘More than’ indicated or implied by statement and valid reason	B1	eg valid reasons 3.14 is greater (than 3) Beth’s number is bigger (than Adam’s) (the correct answer is) 78.5 (with their answer to (b) less than 78.5)

Additional guidance

If calculations are used, the outcomes must be correct
Accept 78 or 79 for 78.5 unless from incorrect working
‘Less than’ indicated	B0
Do not penalise use of the same incorrect formula in (b) and (c) eg \(3 \times 10 = 30\) in (b) and \(3.14 \times 10 = 31.4\) in (c) with ‘More than’ ticked	B1
Ignore a non-contradictory reason with a correct reason eg 3.14 is bigger than 3 and nearer the true value of pi	B1
Acceptable reasons
Adam has rounded (pi) down / Adam only used 3	B1
There is an extra 0.14 to multiply by	B1
Her number has decimal places	B1
Her number is to more significant figures	B1
Non-acceptable reasons
3.14 will give a bigger answer / 3.14 is more accurate	B0

Foundation June 2024 Paper 2 Q16

AQAFoundationCurrent spec2 marksCircles & Sectors

16 The diagram shows a circle, centre \(O\), and three straight lines.

Circle centre O. Region A is shaded between a chord and the arc on the left. Region B is shaded between two radii and the arc on the right

Use one word to describe each shaded region.

Choose from

arc

chord

sector

segment

tangent

[2 marks]

Region A \(\ldots\ldots\ldots\)
Region B \(\ldots\ldots\ldots\)

Higher June 2024 Paper 1 Q12

AQAHigherCurrent spec3 marksCircles & Sectors

12 A sector has radius 12 cm and angle \(60^\circ\)

Not drawn accurately

Work out the length of the arc.

Give your answer in terms of \(\pi\) [3 marks]

Mark scheme

Answer	Mark	Comments
\(2 \times 12 \times \pi\) or \(24\pi\) or \(\dfrac{60}{360}\) or division by 6	M1	oe accept [3.14, 3.142] or \(\dfrac{22}{7}\) for \(\pi\) accept use of 0.17 or better for \(\dfrac{1}{6}\)
\(\dfrac{60}{360} \times 2 \times 12 \times \pi\)	M1dep	oe eg \(\dfrac{24\pi}{6}\)
\(4\pi\)	A1	condone \(\pi 4\)

Additional guidance

Answer \(24\pi\) from \(\pi \times 12^2 \times \dfrac{60}{360}\) scores M1 for \(\dfrac{60}{360}\)

M1M0A0

Higher June 2024 Paper 1 Q7

AQAHigherCurrent spec4 marksCircles & Sectors Volume & Surface Area

7 Here is a cone.

(a)

Curved surface area of a cone \(= \pi r l\)
where \(r\) is the radius and \(l\) is the slant height

Beth tries to work out the curved surface area in terms of \(\pi\)

Curved surface area of the cone \(= \pi \times 5 \times 12\)
\(= 60\pi\ \text{cm}^2\)

What mistake has she made? [1 mark]

(b) Adam uses \(\quad \pi = 3 \quad\) to estimate the area of the base of the cone.

Work out his estimate. [2 marks]

Is Beth’s estimate more than or less than Adam’s estimate?

Tick a box.

More than
Less than

Give a reason for your answer. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Correct statement	B1	eg she used the height instead of the slant height or she used the vertical height or she used 12 (instead of 13)

Additional guidance

Check diagram
For ‘vertical’ accept anything that implies she has used the wrong height
Condone ‘length’ to mean ‘height’ or ‘slant height’
12 or 13 circled on the diagram must be accompanied by a supporting statement
Indicates ‘12’ in the calculation	B1
She should have done \(\pi \times 5 \times 13\)	B1
It should be \(65\pi\)	B1
She used the wrong height / the (value of) \(l\) is wrong	B1
She hasn’t used the slant height (she used the (vertical) height)	B1
She hasn’t used the 13	B1
She hasn’t used the 13 and should be \(5 \times 12 \times 13 \times \pi\)	B0
The multiplication used the wrong number(s)	B0
She hasn’t used a value for \(\pi\)	B0
An incorrect statement with a correct statement eg she used 13 instead of 12 and didn’t square the radius	B0

Mark scheme (c)

Answer	Mark	Comments
‘More than’ indicated or implied by statement and valid reason	B1	eg valid reasons 3.14 is greater (than 3) Beth’s number is bigger (than Adam’s) (the correct answer is) 78.5 (with their answer to (b) less than 78.5)

Additional guidance

If calculations are used, the outcomes must be correct
Accept 78 or 79 for 78.5 unless from incorrect working
‘Less than’ indicated	B0
Do not penalise use of the same incorrect formula in (b) and (c) eg \(3 \times 10 = 30\) in (b) and \(3.14 \times 10 = 31.4\) in (c) with ‘More than’ ticked	B1
Ignore a non-contradictory reason with a correct reason eg 3.14 is bigger than 3 and nearer the true value of pi	B1
Acceptable reasons
Adam has rounded (pi) down / Adam only used 3	B1
There is an extra 0.14 to multiply by	B1
Her number has decimal places	B1
Her number is to more significant figures	B1
Non-acceptable reasons
3.14 will give a bigger answer / 3.14 is more accurate	B0

Higher June 2024 Paper 2 Q1

AQAHigherCurrent spec2 marksCircles & Sectors

1 The diagram shows a circle, centre \(O\), and three straight lines.

Circle centre O. Region A is shaded between a chord and the arc; region B is shaded between two radii and the arc

Use one word to describe each shaded region.

Choose from

arc chord sector segment tangent

[2 marks]

Foundation November 2023 Paper 1 Q27

AQAFoundationCurrent spec5 marksCircles & Sectors

27 Shape A is a circle with radius \(\dfrac{\sqrt{17}}{2}\) cm

Shape B is a sector of a circle with radius 5 cm

Shape A: a circle with radius √17/2 cm. Shape B: a sector with radius 5 cm and angle 60°

Not drawn accurately

Which shape has the greater area, A or B?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
\((\pi \times) \left(\dfrac{\sqrt{17}}{2}\right)^2\)	M1	oe condone missing brackets
\(\dfrac{17}{4}(\pi)\) or \(4\dfrac{1}{4}(\pi)\) or \(4.25(\pi)\)	A1	oe fraction, mixed number or decimal
\((\pi \times) 5^2\) or \((\pi \times) 25\) or \(\dfrac{60}{360}\) used	M1	oe
\(\dfrac{25}{6}(\pi)\) or \(4\dfrac{1}{6}(\pi)\) or \(4.1(6\ldots)(\pi)\) or \(4.17(\pi)\)	A1	oe fraction, mixed number or decimal
A with values in comparable form or A by \(\dfrac{1}{12}(\pi)\) or A by \(0.08(3\ldots)(\pi)\)	A1	eg values \(\dfrac{51}{12}(\pi)\) and \(\dfrac{50}{12}(\pi)\) \(4\dfrac{1}{4}(\pi)\) and \(4\dfrac{1}{6}(\pi)\) \(4.2(5)(\pi)\) and \(4.1(6\ldots)(\pi)\) \(4.2(5)(\pi)\) and \(4.17(\pi)\) accept ‘circle’ for A allow comparison of fraction or decimal parts only if integer parts shown as equal

Additional guidance

For the final mark, presence or absence of \(\pi\) must be the same for both values

Accept consistent use of a numerical value of \(\pi\) throughout.
The value can be 3 or 3.1 or 3.14 or 3.142 or better

Foundation November 2023 Paper 3 Q26

AQAFoundationCurrent spec5 marksCircles & Sectors Plans, Elevations & Nets

26 An open cardboard cylinder has radius 1.9 cm and height 10.2 cm

Cylinder with radius 1.9 cm and height 10.2 cm

(a) Harry assumes that the net of the cylinder is a rectangle with length \(l\) cm

Not drawn accurately

Rectangle with length l cm and height 10.2 cm

Work out the area of this rectangle. [3 marks]

In fact, the net is a parallelogram, not a rectangle.

Not drawn accurately

Parallelogram with top side l cm and perpendicular height 10.2 cm

(b) What does this mean about the area of the net?

Tick one box. [1 mark]

It is less than the area of the rectangle
It is equal to the area of the rectangle
It is more than the area of the rectangle

Tick one box. [1 mark]

It is less than the perimeter of the rectangle
It is equal to the perimeter of the rectangle
It is more than the perimeter of the rectangle

Mark scheme (a)

Answer	Mark	Comments
\(2 \times 1.9 \times \pi\) or \(3.8\pi\) or [11.9, 11.94]	M1	oe
\(2 \times 1.9 \times \pi \times 10.2\) or \(3.8\pi \times 10.2\) or their \([11.9, 11.94] \times 10.2\)	M1dep	oe
[121, 122]	A1	SC1 [115, 116]

Additional guidance

SC1 [115, 116] is from using the area of the circle

Higher November 2023 Paper 1 Q10

AQAHigherCurrent spec5 marksCircles & Sectors

10 Shape A is a circle with radius \(\dfrac{\sqrt{17}}{2}\) cm

Shape B is a sector of a circle with radius 5 cm

Not drawn accurately

Which shape has the greater area, A or B?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
\((\pi \times) \left(\dfrac{\sqrt{17}}{2}\right)^2\)	M1	oe condone missing brackets
\(\dfrac{17}{4}(\pi)\) or \(4\dfrac{1}{4}(\pi)\) or \(4.25(\pi)\)	A1	oe fraction, mixed number or decimal
\((\pi \times)\ 5^2\) or \((\pi \times)\ 25\) or \(\dfrac{60}{360}\) used	M1	oe
\(\dfrac{25}{6}(\pi)\) or \(4\dfrac{1}{6}(\pi)\) or \(4.1(6…)(\pi)\) or \(4.17(\pi)\)	A1	oe fraction, mixed number or decimal
A with values in comparable form or A by \(\dfrac{1}{12}(\pi)\) or A by \(0.08(3…)(\pi)\)	A1	eg values \(\dfrac{51}{12}(\pi)\) and \(\dfrac{50}{12}(\pi)\) \(4\dfrac{1}{4}(\pi)\) and \(4\dfrac{1}{6}(\pi)\) \(4.2(5)(\pi)\) and \(4.1(6…)(\pi)\) \(4.2(5)(\pi)\) and \(4.17(\pi)\) accept ‘circle’ for A allow comparison of fraction or decimal parts only if integer parts shown as equal

Additional guidance

For the final mark, presence or absence of \(\pi\) must be the same for both values

Accept consistent use of a numerical value of \(\pi\) throughout.
The value can be 3 or 3.1 or 3.14 or 3.142 or better

Foundation November 2023 Paper 3 Q3

AQAFoundationCurrent spec2 marksAngles in Polygons Circles & Sectors

(a) Write down the name of a triangle with three equal sides. [1 mark]

(b) Write down the name for the straight line inside this circle.

A circle with a straight line joining two points on its circumference, not passing through the centre

[1 mark]

Higher June 2023 Paper 1 Q26

AQAHigherCurrent spec4 marksCircles & Sectors Pythagoras

26 A circle, centre \(O\), has circumference \(20\pi\) cm

\(Q\) is a point on the circle.

\(OPQR\) is a square.

Circle centre O. Square OPQR with vertex O at the centre and vertex Q on the circle; diagonal OQ is dashed

Not drawn accurately

perimeter of the square : circumference of the circle \(= \sqrt{a} : \pi \quad\) where \(a\) is an integer.

Work out the value of \(a\).

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(20\pi \div 2\pi\) or 10	M1	oe may be seen on diagram implied by diameter = 20
\(x^2 + x^2 = (\text{their } 10)^2\) or \(2x^2 = 100\) or \(x^2 = 50\) or their \(10 \times \sin 45\) or their \(10 \times \cos 45\) or their \(10 \times \dfrac{1}{\sqrt{2}}\)	M1	oe any letter (condone \(a\)) their 10 is their length \(OQ\) (the radius of the circle)
\(\sqrt{\text{their } 10^2 \div 2}\) or \(\sqrt{50}\) or \(5\sqrt{2}\) or \(4 \times \sqrt{50}\) or \(4 \times\) their \(10 \times \sin 45\) or \(4 \times\) their \(10 \times \cos 45\) or \(40 \times \dfrac{1}{\sqrt{2}}\) or \(\dfrac{40\sqrt{2}}{2}\) or \(20\sqrt{2}\)	M1dep	oe value for the length of one side of the square or the perimeter of the square eg \(\dfrac{10}{\sqrt{2}}\) dep on previous mark
2 with full working seen for M3	A1
Alternative method 2
\(20\pi \div 2\pi\) or 10 or side length of square \(= 5\sqrt{a}\)	M1	oe may be seen on diagram implied by diameter = 20
(Perimeter of square \(= 20\sqrt{a}\) and) side length of square \(= 5\sqrt{a}\) and \(\left(5\sqrt{a}\right)^2 + \left(5\sqrt{a}\right)^2 = (\text{their } 10)^2\)	M1	oe their 10 is their length \(OQ\) (the radius of the circle) condone missing brackets if recovered
\(25a + 25a = (\text{their } 10)^2\) or \(50a = 100\)	M1dep	dep on M1M1
2 with full working seen for M3	A1

Additional guidance

2 with no working	M0M0M0A0
\(\sqrt{2}\) on answer line (may score method marks)	A0

Foundation November 2022 Paper 3 Q30

AQAFoundationCurrent spec4 marksCircles & Sectors

30 Part of a running track is the arc of a semicircle joined to a straight line.

The semicircle has diameter 45 metres.
The straight line has length 75 metres.

Track made of a semicircular arc of diameter 45 m joined to a straight line of length 75 m

Not drawn accurately

Abby runs once along this part of the track in 18 seconds.

Work out her average speed.

Give your answer to 2 significant figures. [4 marks]

Mark scheme

Answer	Mark	Comments
\(0.5 \times \pi \times 45\) or \(0.5 \times [141, 141.4]\) or [70.5, 70.7] or \(0.5 \times \pi \times 45 + 75\) or [145.5, 145.7]	M1	oe eg \(22.5\pi\)
\((0.5 \times \pi \times 45 + 75) \div 18\) or their \([145.5, 145.7] \div 18\)	M1	oe their [145.5, 145.7] can be any value
8.08(…) or 8.09(…)	A1	may be implied by 8.1
8.1	B1ft	ft any answer seen with greater than 2 sf SC2 3.9

Additional guidance

Up to M2 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts, B1ft may also be awarded
\(\dfrac{120}{18} = 6.67\) answer 6.7	M0M1A0B1ft
\(\dfrac{120}{18} = 6.7\)	M0M1A0B0ft
\(0.5 \times \pi \times 45\) and \(70.7 \div 18 = 3.93\) answer 3.9	M1M1A0B1ft
SC2 for an answer of 3.9 without working is when 75 is not used

Foundation November 2022 Paper 2 Q13

AQAFoundationCurrent spec3 marksCircles & Sectors

13 \(A\) and \(B\) are points on a circle.

\(C\) is the centre of the circle.

Circle with centre C; A and B on the circle; triangle ABC drawn

Not drawn accurately

Tick one box for each statement. [3 marks]

	Definitely true	Might be true	Cannot be true
Line \(AB\) is a tangent to the circle
\(AC\) is an arc of the circle
Triangle \(ABC\) is equilateral

Mark scheme

Answer	Mark	Comments
Line \(AB\) is a tangent to the circle: Cannot be true \(AC\) is an arc of the circle: Cannot be true Triangle \(ABC\) is equilateral: Might be true	B3	B1 for each any clear indication

Additional guidance

Only one cross in a row – mark the cross

A tick and cross(es) in a row – mark the tick

More than one tick in a row scores B0 for that row

Higher November 2022 Paper 3 Q11

AQAHigherCurrent spec4 marksCircles & Sectors

11 Part of a running track is the arc of a semicircle joined to a straight line.

The semicircle has diameter 45 metres.
The straight line has length 75 metres.

Arc of a semicircle with diameter 45 m joined at its lower end to a straight line of length 75 m, not drawn accurately

Abby runs once along this part of the track in 18 seconds.

Work out her average speed.

Give your answer to 2 significant figures. [4 marks]

Mark scheme

Answer	Mark	Comments
\(0.5 \times \pi \times 45\) or \(0.5 \times\) [141, 141.4] or [70.5, 70.7] or \(0.5 \times \pi \times 45 + 75\) or [145.5, 145.7]	M1	oe eg \(22.5\pi\)
\((0.5 \times \pi \times 45 + 75) \div 18\) or their [145.5, 145.7] \(\div\) 18	M1	oe their [145.5, 145.7] can be any value
8.08(…) or 8.09(…)	A1	may be implied by 8.1
8.1	B1ft	ft any answer seen with greater than 2 sf SC2 3.9

Additional guidance

Up to M2 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts, B1ft may also be awarded
\(\dfrac{120}{18} = 6.67\) answer 6.7	M0M1A0B1ft
\(\dfrac{120}{18} = 6.7\)	M0M1A0B0ft
\(0.5 \times \pi \times 45\) and \(70.7 \div 18 = 3.93\) answer 3.9	M1M1A0B1ft
SC2 for an answer of 3.9 without working is when 75 is not used

Foundation November 2022 Paper 1 Q2

AQAFoundationCurrent spec1 markCircles & Sectors

2 A circle has diameter 10 cm

Circle the radius. [1 mark]

5 cm
10 cm
20 cm
100 cm

Foundation June 2022 Paper 1 Q25

AQAFoundationCurrent spec4 marksCircles & Sectors

25 Here are two shapes, P and Q.

Shape P: three quarters of a circle, radius 20 cm. Shape Q: one third of a circle, radius 15 cm

Not drawn accurately

How many times bigger is the area of P than the area of Q?

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(20^2\) (\(\times\, \pi\)) or 400 (\(\times\, \pi\)) or \(15^2\) (\(\times\, \pi\)) or 225 (\(\times\, \pi\))	M1	oe
\(\dfrac{3}{4} \times 20^2\) (\(\times\, \pi\)) or 300 (\(\times\, \pi\)) or \(\dfrac{1}{3} \times 15^2\) (\(\times\, \pi\)) or 75 (\(\times\, \pi\))	M1dep	oe
\(\dfrac{3}{4} \times 20^2\) (\(\times\, \pi\)) or 300 (\(\times\, \pi\)) and \(\dfrac{1}{3} \times 15^2\) (\(\times\, \pi\)) or 75 (\(\times\, \pi\))	M1dep
300 (\(\times\, \pi\)) and 75 (\(\times\, \pi\)) and 4	A1	Accept \(P = 4Q\) for 4 SC2 40 (\(\times\, \pi\)) and 30 (\(\times\, \pi\)) and 30 (\(\times\, \pi\)) and 10 (\(\times\, \pi\)) and answer 3

Additional guidance

Answer 4 with no working	M0A0
Condone inconsistent use of \(\pi\) eg \(300\pi\) and 75 and 4	M3A1
Condone, for example, \(\pi 400\) for \(400\pi\)
Allow use of a numerical value for \(\pi\) for method marks and for the A mark with answer 4
Ignore units throughout

Foundation June 2022 Paper 1 Q17

AQAFoundationCurrent spec1 markCircles & Sectors

17 Part of this circle is shaded.

Circle with a chord near the top; the smaller region between the chord and the arc is shaded

Circle the name of the shaded part. [1 mark]

arc
sector
chord
segment

Higher June 2022 Paper 3 Q16

AQAHigherCurrent spec5 marksCircles & Sectors Volume & Surface Area

16 Here is a scale drawing of a reservoir.

Scale: 1 cm represents 500 m

Scale drawing of a reservoir with an irregular outline

Virat wants to estimate the volume of water in the reservoir.

He draws on the scale drawing a circle with radius 3 cm

The same reservoir outline with a circle drawn inside it; parts of the reservoir lie outside the circle

(a) Virat estimates the volume of the reservoir by assuming that

the reservoir is a cylinder whose cross section is the circle
the depth of the reservoir is 17 metres.

Work out Virat’s estimate in cubic metres. [3 marks]

(b) In fact,

the depth of the reservoir is 13.8 metres
the reservoir is not a cylinder (see diagram).

Which statement about the actual volume of the reservoir is correct?

Tick one box.

It is less than Virat’s estimate
It is greater than Virat’s estimate
It could be less than or greater than Virat’s estimate

Give a reason for your answer. [2 marks]

Mark scheme (a)

Answer	Mark	Comments
\(3 \times 500\) or 1500	M1	actual radius of circle in metres
(their \(1500)^2 \times \pi \times 17\) or \(38\,250\,000\pi\)	M1dep
[120 000 000, 120 200 000] or [\(1.2 \times 10^8\), \(1.202 \times 10^8\)]	A1	accept in words eg 120 million SC1 [480, 481] or [0.048, 0.0481]

Additional guidance

Do not award A mark if incorrect further work is seen

Mark scheme (b)

Answer	Mark	Comments
It could be less than or greater than Virat’s estimate (3rd box ticked) and statement that area is larger but depth is smaller	B2	B1 It is less than Virat’s estimate (1st box ticked) and statement that depth is smaller or It is greater than Virat’s estimate (2nd box ticked) and statement that area is larger or It could be less than or greater than Virat’s estimate (3rd box ticked) and statement that depth is smaller or It could be less than or greater than Virat’s estimate (3rd box ticked) and statement that area is larger

Additional guidance

For B2 their statement must refer to larger area and smaller depth
For B1 their statement must correctly refer to larger area or smaller depth for their box ticked
Examples of statements implying actual depth is smaller: height is less depth is lower it is shallower Virat’s estimate of the depth is bigger
Examples of statements implying actual area is larger: the width is bigger cross section is bigger shape is greater Virat’s estimate of the area is smaller
The reservoir could be bigger or smaller	B0
The reservoir is larger	B0
We do not know the depth	B0

Higher June 2022 Paper 1 Q11

AQAHigherCurrent spec4 marksCircles & Sectors

11 Here are two shapes, P and Q.

P: three quarters of a circle, radius 20 cm. Q: one third of a circle, radius 15 cm

Not drawn accurately

How many times bigger is the area of P than the area of Q?

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(20^2\) (\(\times \pi\)) or 400 (\(\times \pi\)) or \(15^2\) (\(\times \pi\)) or 225 (\(\times \pi\))	M1	oe
\(\dfrac{3}{4} \times 20^2\) (\(\times \pi\)) or 300 (\(\times \pi\)) or \(\dfrac{1}{3} \times 15^2\) (\(\times \pi\)) or 75 (\(\times \pi\))	M1dep	oe
\(\dfrac{3}{4} \times 20^2\) (\(\times \pi\)) or 300 (\(\times \pi\)) and \(\dfrac{1}{3} \times 15^2\) (\(\times \pi\)) or 75 (\(\times \pi\))	M1dep
300 (\(\times \pi\)) and 75 (\(\times \pi\)) and 4	A1	Accept P = 4Q for 4 SC2 40 (\(\times \pi\)) and 30 (\(\times \pi\)) and 30 (\(\times \pi\)) and 10 (\(\times \pi\)) and answer 3

Additional guidance

Answer 4 with no working	M0A0
Condone inconsistent use of \(\pi\) eg \(300\pi\) and 75 and 4	M3A1
Condone, for example, \(\pi 400\) for \(400\pi\)
Allow use of a numerical value for \(\pi\) for method marks and for the A mark with answer 4
Ignore units throughout

Foundation June 2022 Paper 2 Q4

AQAFoundationCurrent spec1 markCircles & Sectors

4 Which of these parts of a circle is a curve?

Circle your answer. [1 mark]

circumference
diameter
centre
radius

Higher November 2021 Paper 1 Q27

AQAHigherCurrent spec4 marksAdvanced Trigonometry Circles & Sectors

27 The vertices of a regular hexagon lie on a circle with centre \(O\) and radius 5 cm

Regular hexagon inside a circle centre O. Two radii of 5 cm go from O to adjacent vertices with a 60° angle between them. The segment between the circle and that side of the hexagon is shaded.

Not drawn accurately

Work out the shaded area.

Give your answer in the form \(\quad \dfrac{a\pi - b\sqrt{c}}{12} \quad\) where \(a\), \(b\) and \(c\) are integers. [4 marks]

Mark scheme

Answer	Mark	Comments
\(5^2 \times \pi\) (\(\div 6\)) or \(25\pi\) (\(\div 6\))	M1	oe allow 3.14 or better for \(\pi\) throughout
\(\dfrac{1}{2} \times 5 \times 5 \times \sin 60\) or \(\dfrac{1}{2} \times 5 \times 2.5 \tan 60\) or \(\dfrac{25}{2} \times \dfrac{\sqrt{3}}{2}\)	M1	oe correct method to work out the area of the triangle or the area of the hexagon implied by \(75 \sin 60\) or \(37.5 \tan 60\) or \(\dfrac{75\sqrt{3}}{2}\) oe
\(\dfrac{25\pi}{6} - \dfrac{25\sqrt{3}}{4}\)	A1	oe eg \(\dfrac{1}{6}\left(25\pi - \dfrac{75\sqrt{3}}{2}\right)\) implied by correct answer
\(\dfrac{50\pi - 75\sqrt{3}}{12}\)	A1	oe in correct form eg \(\dfrac{50\pi - 15\sqrt{75}}{12}\)

Additional guidance

Using Pythagoras to work out the perpendicular height of the triangle may lead to an area of \(\dfrac{5\sqrt{18.75}}{2}\) for the triangle or \(15\sqrt{18.75}\) for the area of the hexagon

2nd M1

Foundation November 2021 Paper 3 Q21

AQAFoundationCurrent spec1 markCircles & Sectors

21 What is the name of the longest possible chord in a circle?

Circle your answer. [1 mark]

tangent
circumference
radius
diameter

Foundation November 2021 Paper 3 Q13

AQAFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors

(a) Work out the area of this triangle. [2 marks]

Right-angled triangle with vertical side 9.8 cm and base 2.6 cm

Not drawn accurately

(b) A circle has a radius of 11.5 cm

Not drawn accurately

Work out the area of the circle. [2 marks]

Mark scheme (b)

Answer	Mark	Comments
\(\pi \times 11.5^2\)	M1	oe accept [3.14, 3.142] for \(\pi\)
[415, 416] or \(\dfrac{529}{4}\pi\) or \(132.25\pi\)	A1	oe

Additional guidance

Accept \(\dfrac{529}{4} \times \pi\) or \(132.25 \times \pi\) or \(\pi \times \dfrac{529}{4}\) or \(\pi \times 132.25\)	M1A1
Condone \(\pi\dfrac{529}{4}\) or \(\pi 132.25\)	M1A1

Higher November 2020 Paper 3 Q29

AQAHigherCurrent spec5 marksAdvanced Trigonometry Circles & Sectors

29 A shape is made by joining triangle ABC to a semicircle with diameter AC.

Triangle ABC with AB = 8 cm, BC = 10 cm and angle ABC = 114°. A semicircle is drawn on AC on the opposite side from B

Not drawn accurately

Work out the total area of the shape. [5 marks]

Mark scheme

Answer	Mark	Comments
\(0.5 \times 8 \times 10 \times \sin 114\) or [36.5, 36.542]	M1	oe
\(8^2 + 10^2 - 2 \times 8 \times 10 \times \cos 114\) or [229, 229.1]	M1	oe eg \(164 - 160 \times \cos 114\)
\(\sqrt{8^2 + 10^2 - 2 \times 8 \times 10 \times \cos 114}\) or [15.1, 15.14] or [7.55, 7.57]	M1dep	oe dep on 2nd M1
\(0.5 \times \pi \times (0.5 \times \text{their } [15.1, 15.14])^2\) or \(0.5 \times \pi \times \text{their } [7.55, 7.57]^2\) or [89.49, 90.03]	M1dep	dep on 2nd and 3rd M1
[125.99, 126.572]	A1

Additional guidance

Diameter must come from using the cosine rule

2nd mark is not dependent on the first

Foundation November 2020 Paper 1 Q26

AQAFoundationCurrent spec5 marksCircles & Sectors Show That & Proof

26 Two wire shapes make an earring.

The shapes are

a circle with radius 21 mm
and
a quarter circle.

A large circle with a small quarter circle attached at the bottom

Not drawn accurately

radius of circle : radius of quarter circle = 7 : 2

(a) Show that the radius of the quarter circle is 6 mm [1 mark]

(b) Work out the total length of the wire in the earring.

Give your answer in the form \(\quad a\pi + b \quad\) where \(a\) and \(b\) are integers. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\(21 \div 7 \times 2\ (= 6)\) or \(21 \div 3 = 7\) and \(6 \div 3 = 2\) or \(21 \div 7 = 3\) and \(6 \div 2 = 3\) or \(7 \times 3 = 21\) and \(2 \times 3 = 6\)	B1	oe eg \(6 \div 2 = 3\) and \(7 \times 3 = 21\)

Additional guidance

\(3 \times 2\ (= 6)\)	B0
7 : 2 (=) 21 : 6 with no other working	B0
7 : 2 (=) 21 : 6 with multiplication by 3 shown by arrow(s)	B1
7 : 2 (=) 14 : 4 (=) 21 : 6	B1
Do not condone incorrect representation of a division eg \(7 \div 21 = 3\)	B0
Do not condone incorrect mathematical representation eg \(21 \div 7 = 3 \times 2 = 6\)	B0
\(21 \div 6 = 3.5\), \(3.5 \times 2 = 7\)	B1
\(21 \times 2 = 42\), \(42 \div 7 = 6\)	B1

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1
\(2 \times \pi \times 21\) or \(\pi \times 42\) or \(42\pi\) or [131.88, 132]	M1	oe condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 6 \div 4\) or \(\pi \times 12 \div 4\) or \(3\pi\) or [9.4, 9.43]	M1	oe arc length of quarter circle condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 6 \div 4 + 2 \times 6\) or \(3\pi + 12\) or [21.4, 21.43]	M1dep	oe dep on 2nd M1 this does not imply M1M1M1
\(45\pi + 12\)	A1
Alternative method 2
\(2 \times \pi \times 21\) or \(\pi \times 42\) or \(42\pi\) or [131.88, 132]	M1	oe condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 21\) and \(2 \times \pi \times 6 \div 4\) or \(42\pi\) and \(3\pi\) or \(2 \times \pi \times 21 + 2 \times 6\) or \(42\pi + 12\) or [143.88, 144]	M1dep	oe eg \(42\pi\) and [9.4, 9.43] or [131.88, 132] and \(3\pi\)
\(2 \times \pi \times 21 + 2 \times \pi \times 6 \div 4\) or \(42\pi + 3\pi\) or \(45\pi\) or [141, 141.43] or [153, 153.43]	M1dep	oe eg \(42\pi +\) [9.4, 9.43] or [131.88, 132] \(+ 3\pi\)
\(45\pi + 12\)	A1

Additional guidance

Condone \(\;3(15\pi + 4)\)	M1M1M1A1
Condone, for example, \(\pi 42\) for up to M1M1M1
\(21\pi + 3\pi + 12\)	M0M1M1A0 on alt 1
\(441\pi + 3\pi + 12\)	M0M1M1A0 on alt 1
\(42\pi + 36\pi + 12\)	M1M1M0A0 on alt 2
\(441\pi + 36\pi + 12\)	M0M0M0A0
Using \(\pi r^2\) instead of \(2\pi r\) throughout	M0M0M0A0
\(45\pi + 12\) in working with incorrect further work, eg \(45\pi + 12 = 57\pi\)	M1M1M1A0

Foundation November 2020 Paper 2 Q14

AQAFoundationCurrent spec3 marksCircles & Sectors

14 This circle has centre \(C\).

\(W\), \(X\) and \(Y\) are points on the circle.
\(WY\) is a straight line.

Circle with centre C. Straight line WY passes through C. Chord WX, with the region between chord WX and the arc WX shaded

Tick one box for each statement. [3 marks]

	True	False
\(WY\) is a diameter.
\(WX\) is a radius.
The shaded section is a sector.
Arc \(XY\) is part of the circumference.

Mark scheme

Answer	Mark	Comments
True False False True	B3	B2 three correct B1 two correct allow any unambiguous indication

Additional guidance

A tick and a cross in the same row – mark the tick

Only a cross used in a row – regard cross as their selection for that row

Higher November 2020 Paper 1 Q12

AQAHigherCurrent spec5 marksCircles & Sectors Show That & Proof

12 Two wire shapes make an earring.

The shapes are

a circle with radius 21 mm
and
a quarter circle.

A large circle with a small quarter circle attached to the bottom of it

Not drawn accurately

radius of circle : radius of quarter circle \(= 7 : 2\)

(a) Show that the radius of the quarter circle is 6 mm [1 mark]

(b) Work out the total length of the wire in the earring.

Give your answer in the form \(\quad a\pi + b \quad\) where \(a\) and \(b\) are integers. [4 marks]

Mark scheme (a)

Answer	Mark	Comments
\(21 \div 7 \times 2\ (= 6)\) or \(21 \div 3 = 7\) and \(6 \div 3 = 2\) or \(21 \div 7 = 3\) and \(6 \div 2 = 3\) or \(7 \times 3 = 21\) and \(2 \times 3 = 6\)	B1	oe eg \(6 \div 2 = 3\) and \(7 \times 3 = 21\)

Additional guidance

\(3 \times 2\ (= 6)\)	B0
\(7 : 2\ (=)\ 21 : 6\) with no other working	B0
\(7 : 2\ (=)\ 21 : 6\) with multiplication by 3 shown by arrow(s)	B1
\(7 : 2\ (=)\ 14 : 4\ (=)\ 21 : 6\)	B1
Do not condone incorrect representation of a division eg \(7 \div 21 = 3\)	B0
Do not condone incorrect mathematical representation eg \(21 \div 7 = 3 \times 2 = 6\)	B0
\(21 \div 6 = 3.5\), \(3.5 \times 2 = 7\)	B1
\(21 \times 2 = 42\), \(42 \div 7 = 6\)	B1

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1
\(2 \times \pi \times 21\) or \(\pi \times 42\) or \(42\pi\) or [131.88, 132]	M1	oe condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 6 \div 4\) or \(\pi \times 12 \div 4\) or \(3\pi\) or [9.4, 9.43]	M1	oe arc length of quarter circle condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 6 \div 4 + 2 \times 6\) or \(3\pi + 12\) or [21.4, 21.43]	M1dep	oe dep on 2nd M1 this does not imply M1M1M1
\(45\pi + 12\)	A1
Alternative method 2
\(2 \times \pi \times 21\) or \(\pi \times 42\) or \(42\pi\) or [131.88, 132]	M1	oe condone [3.14, 3.142] for \(\pi\)
\(2 \times \pi \times 21\) and \(2 \times \pi \times 6 \div 4\) or \(42\pi\) and \(3\pi\) or \(2 \times \pi \times 21 + 2 \times 6\) or \(42\pi + 12\) or [143.88, 144]	M1dep	oe eg \(42\pi\) and [9.4, 9.43] or [131.88, 132] and \(3\pi\)
\(2 \times \pi \times 21 + 2 \times \pi \times 6 \div 4\) or \(42\pi + 3\pi\) or \(45\pi\) or [141, 141.43] or [153, 153.43]	M1dep	oe eg \(42\pi +\) [9.4, 9.43] or [131.88, 132] \(+ 3\pi\)
\(45\pi + 12\)	A1

Additional guidance

Condone \(\quad 3(15\pi + 4)\)	M1M1M1A1
Condone, for example, \(\pi 42\) for up to M1M1M1
\(21\pi + 3\pi + 12\)	M0M1M1A0 on alt 1
\(441\pi + 3\pi + 12\)	M0M1M1A0 on alt 1
\(42\pi + 36\pi + 12\)	M1M1M0A0 on alt 2
\(441\pi + 36\pi + 12\)	M0M0M0A0
Using \(\pi r^2\) instead of \(2\pi r\) throughout	M0M0M0A0
\(45\pi + 12\) in working with incorrect further work, eg \(45\pi + 12 = 57\pi\)	M1M1M1A0

Higher November 2019 Paper 1 Q26

AQAHigherCurrent spec4 marksCircles & Sectors

26 Here are a circle and a sector of the circle.

They each have radius \(r\).

A circle with radius r, and a sector of the same circle with two radii r and reflex angle x at the centre

Not drawn accurately

circumference of circle = perimeter of sector

Work out the size of angle \(x\).

Give your answer in terms of \(\pi\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: using the radius
\(2\pi r\)	M1
\(2\pi r \times \dfrac{x}{360}\)	M1dep	oe length of arc
\(2\pi r = 2\pi r \times \dfrac{x}{360} + 2r\) or \(\pi = \dfrac{\pi x}{360} + 1\) or \(2\pi = \dfrac{2\pi x}{360} + 2\)	M1dep	oe equation
\(\dfrac{360(\pi - 1)}{\pi}\) or \(\dfrac{360\pi - 360}{\pi}\) or \(360 - \dfrac{360}{\pi}\)	A1	oe expression in \(\pi\) with \(r\) cancelled throughout
Alternative method 2: using the diameter
\(\pi d\)	M1	oe
\(\pi d \times \dfrac{x}{360}\)	M1dep	oe length of arc
\(\pi d = \pi d \times \dfrac{x}{360} + d\) or \(\pi = \dfrac{\pi x}{360} + 1\)	M1dep	oe equation
\(\dfrac{360(\pi - 1)}{\pi}\) or \(\dfrac{360\pi - 360}{\pi}\) or \(360 - \dfrac{360}{\pi}\)	A1	oe expression in \(\pi\) with \(d\) cancelled throughout

Additional guidance

Ignore attempts to simplify, cancel or expand a correct expression

M1M1M1A1

Foundation November 2019 Paper 3 Q1

AQAFoundationCurrent spec1 markCircles & Sectors

1 On a circle, which of these is not a straight line?

Circle your answer. [1 mark]

circumference
radius
chord
diameter

Foundation June 2019 Paper 1 Q25

AQAFoundationCurrent spec4 marksCircles & Sectors

25 A shaded semicircle is inside a circle as shown.

Circle of radius 10 cm with a shaded semicircle of diameter 8 cm inside it

Not drawn accurately

The radius of the circle is 10 cm

The diameter of the semicircle is 8 cm

How many times bigger is the unshaded area than the shaded area? [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: areas
\(\pi \times 10^2\) or \(100\pi\)	M1	implied by [314, 314.2]
\(\pi \times (8 \div 2)^2\) or \(\pi \times 4^2\) or \(16\pi\) or \(\pi \times (8 \div 2)^2 \div 2\) or \(\pi \times 4^2 \div 2\) or \(16\pi \div 2\) or \(8\pi\)	M1	implied by [50.2, 50.3] or [25.12, 25.14] \(92\pi\) or \(84\pi\) or 92 : 8 or 8 : 92 or 84 : 16 or 16 : 84 implies M1M1
(their \(100(\pi)\) – their \(8(\pi)\)) \(\div\) their \(8(\pi)\) or \(92(\pi) \div 8(\pi)\) or their \(100(\pi) \div\) their \(8(\pi)\ (-\ 1)\) or \(12\dfrac{1}{2}\ (-\ 1)\) or \(12.5\ (-\ 1)\)	M1dep	dep on M2 absence of \(\pi\) must be consistent condone \(16(\pi)\) as their \(8(\pi)\) in first calculation only, ie condone (their \(100(\pi)\) – their \(16(\pi)\)) \(\div\) their \(16(\pi)\) or \(84(\pi) \div 16(\pi)\), but not their \(100(\pi) \div\) their \(16(\pi)\ (-\ 1)\)
\(11\dfrac{1}{2}\) or 11.5	A1	condone \(\dfrac{23}{2}\)
Alternative method 2: scale factor
\(\dfrac{10}{8 \div 2}\) or \(\dfrac{10}{4}\) or \(\dfrac{5}{2}\) or \(\dfrac{10 \times 2}{8}\) or \(\dfrac{20}{8}\) or 2.5	M1	oe scale factor of lengths eg \(\dfrac{2}{5}\) or 0.4 accept 2 : 5 or 5 : 2 oe ratio \(\pi\) may be present, but must be consistent in numerator and denominator
(their \(\dfrac{5}{2}\))\(^2\) or \(\dfrac{25}{4}\)	M1dep	oe scale factor of areas eg \(\dfrac{4}{25}\) accept 4 : 25 or 25 : 4 oe ratio
\(2 \times\) their \(\dfrac{25}{4}\ (-\ 1)\) or \(\dfrac{25}{2}\ (-\ 1)\) or \(12\dfrac{1}{2}\ (-\ 1)\) or \(12.5\ (-\ 1)\)	M1dep	oe eg \(2 \div\) their \(\dfrac{4}{25}\ (-\ 1)\)
\(11\dfrac{1}{2}\) or 11.5	A1	condone \(\dfrac{23}{2}\)

Additional guidance

Accept, for example, \(\pi 8\) or \(\pi \times 8\) or \(8 \times \pi\) for \(8\pi\)
An answer of \(11.5\pi\) with no incorrect working	M1M1M1A0
Consistent use of \(\pi d^2\) for the area of a circle gives the area of the circle as \(400\pi\), the area of the semicircle as \(32\pi\) and the area of the shaded part as \(368\pi\). This also gives the answer 11.5, but scores zero	M0M0M0A0
Irrespective of where their answer comes from and the presence of other measures such as circumference, students can gain the first two marks of alternative method 1 if it is clear that the methods or values given are for area eg 1 Big area \(= 100\pi\), little area \(= 8\pi\), big circumference \(= 20\pi\), little circumference \(= 4\pi\), \(20 \div 4 = 5\) eg 2 \(100\pi,\ 8\pi,\ 20\pi,\ 4\pi\)	M1M1M0A0 M0M0
Do not award the second mark if the value of \(8\pi\) comes from \(\pi d\) This is implied by, eg, ‘Area of circle \(= 20\pi\), area of semi-circle \(= 8\pi\)’	M?M0 M0M0
\(\dfrac{100(\pi) - 16(\pi)}{16(\pi)}\) (which may give an answer of 5.25)	M1M1M1A0
\(\dfrac{100(\pi)}{16(\pi)}\) (which may give an answer of 6.25)	M1M1M0A0

Higher June 2019 Paper 3 Q16

AQAHigherCurrent spec2 marksCircles & Sectors

16 Here are two sectors from different circles.

Sector A with angle x and radius 1.5r; Sector B with angle 2x and radius r

Not drawn accurately

Which sector has the bigger area?

Tick a box.

Sector A
Sector B

Show working to support your answer. [2 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{x}{360} \times \pi \times (1.5r)^2\) or \(\dfrac{1}{160}\pi x r^2\) or \(0.019…xr^2\) or \(\dfrac{2x}{360} \times \pi \times r^2\) or \(\dfrac{1}{180}\pi x r^2\) or \(0.017…xr^2\)	M1	oe eg (working in radians) \(\dfrac{1}{2} \times (1.5r)^2 \times x\) or \(\dfrac{1}{2} \times r^2 \times 2x\)
\(\dfrac{1}{160}\pi x r^2\) and \(\dfrac{1}{180}\pi x r^2\) and A or \(0.019…xr^2\) and \(0.017…xr^2\) and A	A1	oe eg (working in radians) \(\dfrac{9}{8}r^2 x\) and \(r^2 x\) and A

Additional guidance

Methods must be algebraic, containing \(x\), \(\pi\) and \(r\)

If a box is not ticked, must say ‘A’ without contradiction in working to award M1A1

To award A1 their areas must be in a comparable form eg
\(\dfrac{2.25}{360}\pi x r^2\) and \(\dfrac{2}{360}\pi x r^2\) and A ticked

Ignore further incorrect working after A1 scored

Higher June 2019 Paper 1 Q9

AQAHigherCurrent spec4 marksCircles & Sectors Similar Shapes

9 A shaded semicircle is inside a circle as shown.

A circle with radius 10 cm, containing a shaded semicircle of diameter 8 cm

Not drawn accurately

The radius of the circle is 10 cm

The diameter of the semicircle is 8 cm

How many times bigger is the unshaded area than the shaded area? [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: areas
\(\pi \times 10^2\) or \(100\pi\)	M1	implied by [314, 314.2]
\(\pi \times (8 \div 2)^2\) or \(\pi \times 4^2\) or \(16\pi\) or \(\pi \times (8 \div 2)^2 \div 2\) or \(\pi \times 4^2 \div 2\) or \(16\pi \div 2\) or \(8\pi\)	M1	implied by [50.2, 50.3] or [25.12, 25.14] \(92\pi\) or \(84\pi\) or 92 : 8 or 8 : 92 or 84 : 16 or 16 : 84 implies M1M1
(their \(100(\pi)\) – their \(8(\pi)\)) \(\div\) their \(8(\pi)\) or \(92(\pi) \div 8(\pi)\) or their \(100(\pi) \div\) their \(8(\pi)\) (– 1) or \(12\dfrac{1}{2}\) (– 1) or 12.5 (– 1)	M1dep	dep on M2 absence of \(\pi\) must be consistent condone \(16(\pi)\) as their \(8(\pi)\) in first calculation only, ie condone (their \(100(\pi)\) – their \(16(\pi)\)) \(\div\) their \(16(\pi)\) or \(84(\pi) \div 16(\pi)\), but not their \(100(\pi) \div\) their \(16(\pi)\) (– 1)
\(11\dfrac{1}{2}\) or 11.5	A1	condone \(\dfrac{23}{2}\)
Alternative method 2: scale factor
\(\dfrac{10}{8 \div 2}\) or \(\dfrac{10}{4}\) or \(\dfrac{5}{2}\) or \(\dfrac{10 \times 2}{8}\) or \(\dfrac{20}{8}\) or 2.5	M1	oe scale factor of lengths eg \(\dfrac{2}{5}\) or 0.4 accept 2 : 5 or 5 : 2 oe ratio \(\pi\) may be present, but must be consistent in numerator and denominator
(their \(\dfrac{5}{2}\))\(^2\) or \(\dfrac{25}{4}\)	M1dep	oe scale factor of areas eg \(\dfrac{4}{25}\) accept 4 : 25 or 25 : 4 oe ratio
\(2 \times\) their \(\dfrac{25}{4}\) (– 1) or \(\dfrac{25}{2}\) (– 1) or \(12\dfrac{1}{2}\) (– 1) or 12.5 (– 1)	M1dep	oe eg \(2 \div\) their \(\dfrac{4}{25}\) (– 1)
\(11\dfrac{1}{2}\) or 11.5	A1	condone \(\dfrac{23}{2}\)

Additional guidance

Accept, for example, \(\pi 8\) or \(\pi \times 8\) or \(8 \times \pi\) for \(8\pi\)
An answer of \(11.5\pi\) with no incorrect working	M1M1M1A0
Consistent use of \(\pi d^2\) for the area of a circle gives the area of the circle as \(400\pi\), the area of the semicircle as \(32\pi\) and the area of the shaded part as \(368\pi\). This also gives the answer 11.5, but scores zero	M0M0M0A0
Irrespective of where their answer comes from and the presence of other measures such as circumference, students can gain the first two marks of alternative method 1 if it is clear that the methods or values given are for area eg 1 Big area \(= 100\pi\), little area \(= 8\pi\), big circumference \(= 20\pi\), little circumference \(= 4\pi\), \(20 \div 4 = 5\) eg 2 \(100\pi\), \(8\pi\), \(20\pi\), \(4\pi\)	M1M1M0A0 M0M0
Do not award the second mark if the value of \(8\pi\) comes from \(\pi d\) This is implied by, eg, ‘Area of circle \(= 20\pi\), area of semi-circle \(= 8\pi\)’	M?M0 M0M0
\(\dfrac{100(\pi) - 16(\pi)}{16(\pi)}\) (which may give an answer of 5.25)	M1M1M1A0
\(\dfrac{100(\pi)}{16(\pi)}\) (which may give an answer of 6.25)	M1M1M0A0

Higher June 2019 Paper 1 Q3

AQAHigherCurrent spec1 markCircles & Sectors

3 Work out the arc length, in metres, of a semicircle of radius 6 metres.

Circle your answer. [1 mark]

\(3\pi\)
\(6\pi\)
\(12\pi\)
\(18\pi\)

Foundation November 2018 Paper 3 Q27

AQAFoundationCurrent spec5 marksCircles & Sectors

27 A motor racing circuit consists of

two parallel straight sections, each of length 0.75 km
a semicircle of diameter 0.9 km
three equal, smaller semicircles.

Circuit made of two parallel straights of 0.75 km joined on the right by a large semicircle of diameter 0.9 km and on the left by three equal smaller semicircles

Not drawn accurately

The length of a motor race must be greater than 305 km

What is the lowest number of full laps needed at this circuit?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
\(0.9 \times \pi \div 2\) or \(0.9\pi \div 2\) or \(0.45\pi\) or \(0.9 \times [3.14, 3.142] \div 2\) or \([2.82, 2.83] \div 2\) or \(2.8 \div 2\) or 1.4…	M1	Large semicircle
\(0.9 \div 3 \times \pi \div 2\) or \(0.3\pi \div 2\) or \(0.15\pi\) or \(0.9 \div 3 \times [3.14, 3.142] \div 2\) or \(0.94\ldots \div 2\) or 0.47…	M1	Small semicircle May be implied from using 1.4… for three small semicircles in next mark
their 1.4… + 3 × their 0.47… + 2 × 0.75 or \(0.9\pi + 2 \times 0.75\) or \(2 \times\) their 1.4… \(+ 2 \times 0.75\) or 4.3…	M1dep	oe dep on both marks
\(305 \div\) their 4.3… or [70.4, 70.94]	M1dep	dep on previous mark
71 with working	A1

Additional guidance

\(0.9\pi\) or 2.8… with no evidence of incorrect method	M1M1
\(0.45\pi \div 2\)	M0

Foundation November 2018 Paper 2 Q23

AQAFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors

23 Two identical quarter circles are cut from a rectangle as shown.

Rectangle of height 12 cm. Two identical quarter circles are cut from it, leaving a shaded region between their arcs

Not drawn accurately

Work out the shaded area. [4 marks]

Mark scheme

Answer	Mark	Comments
\(12 \times 6\) or 72	M1	oe area of rectangle
\(\pi \times 6^2\) or \(36\pi\) or [113, 113.112]	M1	oe may be implied eg \(\pi \times 6^2 \div 4\) or \(9\pi\) or [28.2, 28.3]
\(\pi \times 6^2 \div 2\) or \(18\pi\) or [56.4, 56.6]	M1dep	oe dep on 2nd M1
[15.4, 15.5] or \(72 - 18\pi\)	A1

Additional guidance

\(72 - 18\pi = 54\pi\)	M1M1M1A0
\(\pi \times 6^2 \div 2\) scores 2nd and 3rd M1
\(12 \times 6 = 72\) \(72 \div 2 = 36\) (unless identified as half of rectangle)	(1st) M0
\(\pi \times 6^2\) scores 2nd M1 even if subsequently used incorrectly eg \(\pi \times 6^2 = 36\pi\) \(36\pi \times 2 = 72\pi\)	(2nd) M1
Ignore units throughout

Foundation November 2018 Paper 2 Q17

AQAFoundationCurrent spec1 markCircles & Sectors

17 A circle has circumference \(C\) and diameter \(d\).

Circle with circumference labelled C and a diameter labelled d

\(C = kd\)

What value does the constant \(k\) represent? [1 mark]

Mark scheme

Answer	Mark	Comments
Pi or \(\pi\)	B1	accept a value in range [3.14, 3.142]

Additional guidance

Accept incorrect spelling if intention is clear eg accept pie
Answer \((C =)\ \pi d\)	B0
Answer \((C =)\ \pi d \quad (k =)\ \pi\)	B1

Higher November 2018 Paper 3 Q9

AQAHigherCurrent spec5 marks2D Area & Perimeter Circles & Sectors

9 A motor racing circuit consists of

two parallel straight sections, each of length 0.75 km
a semicircle of diameter 0.9 km
three equal, smaller semicircles.

Not drawn accurately

The circuit: two parallel straight sections of 0.75 km joined on the right by a semicircle of diameter 0.9 km and on the left by three equal smaller semicircles

The length of a motor race must be greater than 305 km

What is the lowest number of full laps needed at this circuit?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
\(0.9 \times \pi \div 2\) or \(0.9\pi \div 2\) or \(0.45\pi\) or \(0.9 \times [3.14, 3.142] \div 2\) or \([2.82, 2.83] \div 2\) or \(2.8 \div 2\) or 1.4…	M1	Large semicircle
\(0.9 \div 3 \times \pi \div 2\) or \(0.3\pi \div 2\) or \(0.15\pi\) or \(0.9 \div 3 \times [3.14, 3.142] \div 2\) or \(0.94… \div 2\) or 0.47…	M1	Small semicircle May be implied from using 1.4… for three small semicircles in next mark
\(\text{their } 1.4…\) \(+\; 3 \times \text{their } 0.47…\) \(+\; 2 \times 0.75\) or \(0.9\pi + 2 \times 0.75\) or \(2 \times \text{their } 1.4… + 2 \times 0.75\) or 4.3…	M1dep	oe dep on both marks
\(305 \div \text{their } 4.3…\) or [70.4, 70.94]	M1dep	dep on previous mark
71 with working	A1

Additional guidance

\(0.9\pi\) or 2.8… with no evidence of incorrect method	M1M1
\(0.45\pi \div 2\)	M0

Higher November 2018 Paper 2 Q8

AQAHigherCurrent spec4 marks2D Area & Perimeter Circles & Sectors

8 Two identical quarter circles are cut from a rectangle as shown.

Not drawn accurately

A tall rectangle of height 12 cm. Two quarter circles are cut from the right-hand corners, meeting at the midpoint of the right-hand side; the region left between them is shaded

Work out the shaded area. [4 marks]

Mark scheme

Answer	Mark	Comments
\(12 \times 6\) or 72	M1	oe area of rectangle
\(\pi \times 6^2\) or \(36\pi\) or [113, 113.112]	M1	oe may be implied eg \(\pi \times 6^2 \div 4\) or \(9\pi\) or [28.2, 28.3]
\(\pi \times 6^2 \div 2\) or \(18\pi\) or [56.4, 56.6]	M1dep	oe dep on 2nd M1
[15.4, 15.5] or \(72 - 18\pi\)	A1

Additional guidance

\(72 - 18\pi = 54\pi\)	M1M1M1A0
\(\pi \times 6^2 \div 2\) scores 2nd and 3rd M1
\(12 \times 6 = 72\) \(72 \div 2 = 36\) (unless identified as half of rectangle)	(1st) M0
\(\pi \times 6^2\) scores 2nd M1 even if subsequently used incorrectly eg \(\pi \times 6^2 = 36\pi\) \(36\pi \times 2 = 72\pi\)	(2nd) M1
Ignore units throughout

Higher November 2018 Paper 2 Q5

AQAHigherCurrent spec1 markCircles & Sectors

5 A circle has circumference \(C\) and diameter \(d\).

A circle with a diameter labelled d and the circumference labelled C

\(C = kd\)

What value does the constant \(k\) represent? [1 mark]

Mark scheme

Answer	Mark	Comments
Pi or \(\pi\)	B1	accept a value in range [3.14, 3.142]

Additional guidance

Accept incorrect spelling if intention is clear eg accept pie
Answer (\(C =\)) \(\pi d\)	B0
Answer (\(C =\)) \(\pi d\) (\(k =\)) \(\pi\)	B1

Foundation November 2018 Paper 1 Q4

AQAFoundationCurrent spec1 markCircles & Sectors

4 On a circle, which of these is always longer than the diameter?

Circle your answer. [1 mark]

chord
arc
radius
circumference

Foundation June 2018 Paper 2 Q21

AQAFoundationCurrent spec4 marksCircles & Sectors

(a) A circle has radius 4.2 cm

Work out the length of the circumference.

Give your answer to 1 decimal place. [3 marks]

(b) The circle below has centre O.

Draw a sector on the circle. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
\(2 \times 4.2 \times \pi\) or \(8.4\pi\)	M1	oe allow [3.14, 3.142] for \(\pi\)
[26.376, 26.393]	A1	may be implied by 26.4 as answer
26.4	B1ft	ft their value to at least 2 dp rounded correctly to 1 dp

Additional guidance

26.4	M1A1B1ft
26.3 only	M0A0B0ft
55.4 only	M0A0B0ft

Mark scheme (b)

Answer	Mark	Comments
Sector drawn correctly	B1	two radii joined at the centre of the circle

Additional guidance

Mark intention
Diameter drawn	B1
Any number of sectors (eg diameter and radius drawn)	B1
Ends of radii joined to form segment with whole sector shaded	B1
Ends of radii joined to form segment without whole sector shaded	B0

Higher June 2018 Paper 1 Q18

AQAHigherCurrent spec5 marksCircles & Sectors Prisms

18 Here is a tunnel for a toy train.

A tunnel shaped like half of a thick cylindrical tube, lying on its flat side

The diagram below shows the cross section of the tunnel.

Cross section: a semicircular ring. The outer arc AD has radius 10 cm and the inner arc BC has radius 7 cm, measured from the same centre. AB = 3 cm and CD = 3 cm along the base

Not drawn accurately

\(AD\) is a semicircular arc of radius 10 cm
\(BC\) is a semicircular arc of radius 7 cm
The length of the tunnel is 30 cm

Work out the total area of all six faces of the tunnel.

Give your answer in terms of \(\pi\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(\pi \times 10^2 - \pi \times 7^2\) or \(100\pi - 49\pi\) or \(51\pi\) or \(\dfrac{1}{2} \times \pi \times 10^2 - \dfrac{1}{2} \times \pi \times 7^2\) or \(\dfrac{1}{2} \times 100\pi - \dfrac{1}{2} \times 49\pi\) or \(\dfrac{1}{2} \times 51\pi\) or \(25.5\pi\)	M1	oe implied by \(102\pi\) method to work out front and/or back faces – must not be part of a method to work out volume (\(\times\) 30) may be taken to be full circles
\(2 \times \pi \times 10 \times 30\) or \(600\pi\) or \(\dfrac{1}{2} \times 2 \times \pi \times 10 \times 30\) or \(300\pi\) or \(2 \times \pi \times 7 \times 30\) or \(420\pi\) or \(\dfrac{1}{2} \times 2 \times \pi \times 7 \times 30\) or \(210\pi\) or \(1020\pi\) or \(510\pi\)	M1	oe method to work out outer and/or inner curved surfaces may be taken to be full circles \(1122\pi\) implies M1M1
\(\left(\dfrac{1}{2} \times \pi \times 10^2 - \dfrac{1}{2} \times \pi \times 7^2\right) \times 2\) \(+ \dfrac{1}{2} \times 2 \times \pi \times 10 \times 30\) \(+ \dfrac{1}{2} \times 2 \times \pi \times 7 \times 30\) or \(2 \times 25.5\pi + 300\pi + 210\pi\) or \(561\pi\)	M1dep	oe dep on M1M1 correct method to work out total of front, back, outer curved and inner curved surfaces
\(2 \times 30 \times 3\) or 180	M1	implied by an answer of \(n\pi + 180\) do not award if 180 is used as \(180\pi\)
\(561\pi + 180\)	A1

Additional guidance

\(150\pi\) and \(105\pi\) implies use of radius for curved surface areas	max M1M0M0M1A0
Condone use of [3.14, 3.142] for \(\pi\) up to M1M1M0M1A0

Higher June 2018 Paper 2 Q1

AQAHigherCurrent spec1 markCircles & Sectors

1 Here is a circle.

A circle with a chord near the bottom; the smaller region between the chord and the arc is shaded

Circle the word that describes the shaded part. [1 mark]

segment
chord
sector
arc

Foundation November 2017 Paper 1 Q27

AQAFoundationCurrent spec2 marksCircles & Sectors

27 Here is a quarter circle of radius 6 cm

A quarter circle with a right angle at the centre and radius 6 cm

Not drawn accurately

Work out the area of the quarter circle.

Give your answer in terms of \(\pi\). [2 marks]

Mark scheme

Answer	Mark	Comments
\(\pi \times 6 \times 6\) or \(36\pi\) or [113, 113.112] or \(9 \times\) [3.14, 3.142] or [28.26, 28.3]	M1	oe accept [3.14, 3.142] for \(\pi\)
\(9\pi\) or \(9 \times \pi\) or \(\pi 9\) or \(\pi \times 9\)	A1

Additional guidance

\(36\pi\) followed by an incorrect method eg \(36\pi \div 2 = 18\pi\) with answer \(18\pi\)	M1A0
Answer of \(9\pi\) from \(\pi \times 3^2\)	M0A0
\(9\pi\) and [28.26, 28.3] given on answer line	M1A0
\(\pi r^2\) stated but followed by 36 or 9	M0A0

Higher November 2017 Paper 2 Q22

AQAHigherCurrent spec5 marksCircles & Sectors

22 The cross section of an earring is a semicircle, centre \(C\), radius 25 mm

The earring is black and white.

The shaded area is black.

Semicircle AE with centre C on the straight edge and radius 25 mm (CA). A white sector BCD with centre C sits inside; the rest of the semicircle is shaded

Not drawn accurately

Sector \(BCD\) is white and has radius 12 mm

Sector BCD with centre C, radius 12 mm and angle 150 degrees

Not drawn accurately

Is more than 20% of the semicircle white?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
(\(\dfrac{1}{2} \times\)) \(\pi \times 25 \times 25\) or \(625\pi\) or \(312.5\pi\) or [1962.5, 1964] or [981, 982] or \(\pi \times 12 \times 12\) or \(144\pi\) or [452, 452.45]	M1	oe Area of circle or semicircle radius 25 or area of circle radius 12
\(\dfrac{150}{360}\) or \(\dfrac{5}{12}\) or 0.41(6…) or 0.417 or 0.42 or \(\dfrac{360}{150}\) or \(\dfrac{12}{5}\) or 2.4	M1	May be seen in two steps eg \(\times\, 150 \div 360\)
their \(\dfrac{150}{360} \times \pi \times 12 \times 12\) or \(\pi \times 12 \times 12 \div\) their \(\dfrac{360}{150}\) or \(60\pi\) or [188.4, 188.52]	M1dep	oe dep on M2 Area of sector
\(\dfrac{\text{their } [188.4, 188.52]}{\text{their } [981, 982]}\) (\(\times\, 100\)) or [0.19, 0.1922] or [19, 19.22]	M1dep	oe dep on M3 their [981, 982] must be the area of semicircle radius 25
[19, 19.22] and No or [0.19, 0.1922] and 0.2 and No	A1
Alternative method 2
(\(\dfrac{1}{2} \times\)) \(\pi \times 25 \times 25\) or \(625\pi\) or \(312.5\pi\) or [1962.5, 1964] or [981, 982] or \(\pi \times 12 \times 12\) or \(144\pi\) or [452, 452.45]	M1	oe Area of circle or semicircle radius 25 or area of circle radius 12
\(\dfrac{150}{360}\) or \(\dfrac{5}{12}\) or 0.41(6…) or 0.417 or 0.42 or \(\dfrac{360}{150}\) or \(\dfrac{12}{5}\) or 2.4	M1	May be seen in two steps eg \(\times\, 150 \div 360\)
their \(\dfrac{150}{360} \times \pi \times 12 \times 12\) or \(\pi \times 12 \times 12 \div\) their \(\dfrac{360}{150}\) or \(60\pi\) or [188.4, 188.52]	M1dep	oe dep on M2 Area of sector
their \([188.4, 188.52] \times 5\) or [942, 942.6]	M1dep	oe dep on M3
[942, 942.6] and [981, 982] and No	A1	oe eg \(300\pi\) and \(312.5\pi\) and No
Alternative method 3
(\(\dfrac{1}{2} \times\)) \(\pi \times 25 \times 25\) or \(625\pi\) or \(312.5\pi\) or [1962.5, 1964] or [981, 982] or \(\pi \times 12 \times 12\) or \(144\pi\) or [452, 452.45]	M1	oe Area of circle or semicircle radius 25 or area of circle radius 12
\(0.2 \times\) their [981, 982] or \(62.5\pi\) or [196.2, 196.4]	M1dep	oe dep on 1st M1 their [981, 982] must be the area of semicircle radius 25
\(\dfrac{150}{360}\) or \(\dfrac{5}{12}\) or 0.41(6…) or 0.417 or 0.42 or \(\dfrac{360}{150}\) or \(\dfrac{12}{5}\) or 2.4	M1	May be seen in two steps eg \(\times\, 150 \div 360\)
their \(\dfrac{150}{360} \times \pi \times 12 \times 12\) or \(\pi \times 12 \times 12 \div\) their \(\dfrac{360}{150}\) or \(60\pi\) or [188.4, 188.52]	M1dep	oe dep on 1st M1 and 3rd M1 Area of sector
[188.4, 188.52] and [196.2, 196.4] and No	A1	oe eg \(60\pi\) and \(62.5\pi\) and No

Additional guidance

Alt 3 20% of [981, 982] does not score 2nd M1 unless evaluated correctly

Foundation November 2017 Paper 2 Q16

AQAFoundationCurrent spec3 marksCircles & Sectors

16 A wheel is made of a circular rim and 8 spokes as shown.

Circle with 8 spokes from the centre to the rim, equally spaced

Not drawn accurately

The length of each spoke is 37 cm

Work out the total length of the rim and spokes. [3 marks]

Mark scheme

Answer	Mark	Comments
\(2 \times \pi \times 37\) or \(\pi \times 74\) or \(8 \times 37\) or 296	M1	Accept [3.14, 3.142] for \(\pi\)
[232, 233] or \(74\pi\)	A1	May be implied by eg \(74\pi + \ldots\)
[528, 529] or \(74\pi + 296\)	A1

Additional guidance

\(360 - 37 \times 8\)	M1A0A0
\(37 \times 8\) or 296 seen and then eg halved or doubled	M1

Higher November 2017 Paper 1 Q7

AQAHigherCurrent spec2 marksCircles & Sectors

7 Here is a quarter circle of radius 6 cm

Quarter circle with a right angle at the bottom left corner and horizontal radius 6 cm

Not drawn accurately

Work out the area of the quarter circle.

Give your answer in terms of \(\pi\). [2 marks]

Mark scheme

Answer	Mark	Comments
\(\pi \times 6 \times 6\) or \(36\pi\) or [113, 113.112] or \(9 \times [3.14, 3.142]\) or [28.26, 28.3]	M1	oe accept [3.14, 3.142] for \(\pi\)
\(9\pi\) or \(9 \times \pi\) or \(\pi 9\) or \(\pi \times 9\)	A1

Additional guidance

\(36\pi\) followed by an incorrect method eg \(36\pi \div 2 = 18\pi\) with answer \(18\pi\)	M1A0
Answer of \(9\pi\) from \(\pi \times 3^2\)	M0A0
\(9\pi\) and [28.26, 28.3] given on answer line	M1A0
\(\pi r^2\) stated but followed by 36 or 9	M0A0

Higher June 2017 Paper 2 Q25

AQAHigherCurrent spec5 marksAdvanced Trigonometry Circles & Sectors Show That & Proof

25 The diagram shows a logo.

ABE and DCE are congruent triangles.
BCE is a sector of a circle, centre E.

Not drawn accurately

Show that the area of the logo is 510 cm\(^2\) to 2 significant figures. [5 marks]

Mark scheme

Answer	Mark	Comments
\(15^2 + 26^2 - 2 \times 15 \times 26 \times \cos 38\) or [286, 286.4] or [16.9, 17]	M1	May be seen in a square root May be seen on diagram
\(\dfrac{108}{360}\) or 0.3 or \(\dfrac{360}{108}\) or 3.33(…)	M1	oe eg \(108 \div 360\) or 30% May be seen in two steps eg \(\times 108 \div 360\)
their \(\dfrac{108}{360} \times \pi \times\) [286, 286.4] or \(\pi \times\) their [286, 286.4] \(\div\) their \(\dfrac{360}{108}\) or [269, 272.4114]	M1dep	dep on 1st and 2nd M1 oe eg \(\dfrac{108}{360} \times \pi \times (\text{their } [16.9, 17])^2\)
\((2 \times)\ \dfrac{1}{2} \times 15 \times 26 \times \sin 38\) or [120, 120.1] or [240, 240.2]	M1	oe
[509, 512.6114] and 510	A1	Must see a value in range [509, 512.6114] and 510

Additional guidance

\(15 \times 26 \times \sin 38\) scores 4th M1 unless subsequently doubled
If (sector) 270 and (2 triangles) 240 followed by \(270 + 240 = 510\)	M4A1
Working back from 510. Apply scheme but maximum mark is M4A0
Assuming angle AEB = 72 and then using sine rule to work out BE does lead to area = 510 to 2sf but can score a maximum of M0M1M0M1depA0 \(BE = \dfrac{26}{\sin 72} \times \sin 38 = 16.8\) (or 17) \(\dfrac{108}{360} \times \pi \times 16.8^2 = 266 \qquad 2 \times \dfrac{1}{2} \times 15 \times 26 \times \sin 38 = 240.2\) \(506.2 \to 510\)	M0 M1M0depM1 A0
BE = [16.9, 17] seen with no working scores first M1 (and possibly all other marks)
\(BE = 35 \div 2 = 17.5 \to 17\) does not score first M1

Foundation June 2017 Paper 1 Q20

AQAFoundationCurrent spec3 marks2D Area & Perimeter Circles & Sectors

20 Here is a circle touching a square.

A circle inside a square, touching all four sides

Not drawn accurately

The area of the square is 64 cm\(^2\)

Work out the area of the circle.

Give your answer in terms of \(\pi\). [3 marks]

Mark scheme

Answer	Mark	Comments
\(\sqrt{64}\) or 8 or \(64 = 8 \times 8\)	M1	Implied by a diameter or side length of 8 stated or shown on the diagram, or radius of 4 stated or used or shown on the diagram
\(\pi \times (\text{their } 8 \div 2)^2\) or \(\pi \times 4^2\) or \(\pi 4^2\) or [50.24, 50.272]	M1dep	oe Allow [3.14, 3.142] for \(\pi\)
\(16\pi\)	A1	Condone \(16 \times \pi\) or \(\pi \times 16\) or \(\pi 16\)

Additional guidance

\(64 - 16\pi\)	M1M1A0
Beware of incorrect methods which lead to the correct answer eg \(r = 8,\ \ 2 \times \pi \times 8 = 16\pi\) \(\sqrt{64} = 8,\ \ 8^2 = 16,\ \ 16\pi\)	M0M0A0 M1M0A0

Foundation June 2017 Paper 3 Q13

AQAFoundationCurrent spec2 marksCircles & Sectors

13 A circle is drawn on a centimetre grid.

A 5 by 5 centimetre grid with a circle in the middle that fits inside the central 3 by 3 block of squares

(a) Draw a tangent to the circle. [1 mark]

(b) Grace works out that the area of the circle is more than 9 cm\(^2\)

Why must this be wrong? [1 mark]

Mark scheme (a)

Answer	Mark	Comments
Correct tangent drawn	B1

Additional guidance

Accept unruled line if intention is clear
Tangent must be drawn without clear space between line and circle
Ignore square drawn on grid lines from part (b)
Tangent may be drawn as part of a square	B1
Accept tangent which does not extend to both sides of circle	B1
Accept tangent drawn and ignore any radius or diameter drawn	B1
Do not accept tangent and chord drawn together	B0

Mark scheme (b)

Answer	Mark	Comments
Valid reason for the area of the circle or the square around the circle	B1

Additional guidance

The area of the circle stated to be [4.5, 6.2] without incorrect working	B1
Area of circle of radius 1.5 (cm) is 7(.06…) or 7.07 or 7.1	B1
The square around it is only 9 cm\(^2\) or 9 squares or \(3 \times 3\) square	B1
There aren’t 9 squares in the circle	B1
The circle fits into a 9 cm\(^2\) square or 9 squares or \(3 \times 3\) square	B1
It only covers about [4.5, 6.2] squares	B1
Circle does not (completely) cover nine separate boxes	B1
There is one whole square and 8 part squares in the circle	B1
Because all of the space for 9 is not used up	B1
Calculate radius = 1.6(9…) (cm) or 1.7 (cm) from area of circle 9 (cm\(^2\)) and states radius of circle drawn is smaller	B1
She uses 9 squares that are half in and half out of the circle, she needed to work it out only using the squares inside the circle	B0
Does not fill up the whole square (no reference to 9)	B0
Because the radius is not big enough for it to be 9	B0

Higher June 2017 Paper 1 Q12

AQAHigherCurrent spec3 marks2D Area & Perimeter Circles & Sectors

12 Here is a circle touching a square.

Not drawn accurately

The area of the square is 64 cm\(^2\)

Work out the area of the circle.

Give your answer in terms of \(\pi\). [3 marks]

Mark scheme

Answer	Mark	Comments
\(\sqrt{64}\) or 8 or \(64 = 8 \times 8\)	M1	Implied by a diameter or side length of 8 stated or shown on the diagram, or radius of 4 stated or used or shown on the diagram
\(\pi \times (\text{their } 8 \div 2)^2\) or \(\pi \times 4^2\) or \(\pi 4^2\) or [50.24, 50.272]	M1dep	oe Allow [3.14, 3.142] for \(\pi\)
\(16\pi\)	A1	Condone \(16 \times \pi\) or \(\pi \times 16\) or \(\pi 16\)

Additional guidance

\(64 - 16\pi\)	M1M1A0
Beware of incorrect methods which lead to the correct answer eg \(\;r = 8,\ 2 \times \pi \times 8 = 16\pi\) \(\sqrt{64} = 8,\ 8^2 = 16,\ 16\pi\)	M0M0A0 M1M0A0

Foundation June 2025 Paper 1 Q28

OCRFoundationCurrent spec6 marksCircles & Sectors Pythagoras

28 The shaded shape below is made from a right-angled triangle, ABC, and a semicircle. The length of AB is 20 cm and the diameter of the semicircle, BC, is of length 12 cm.

Shaded shape: right-angled triangle ABC with the right angle at B, AB = 20 cm vertical, BC = 12 cm horizontal (dashed), and a semicircle on BC below it

Not to scale

Calculate the perimeter of the shaded shape.
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
62.16 to 62.18 or 62.2 with correct working or \(4\sqrt{34} + 6\pi + 20\) oe with correct working	6		Correct working requires at least the first M1 AND the second M1
		M2 for \(\sqrt{20^2 + 12^2}\) or M1 for \(20^2 + 12^2\) maybe implied by 544	Award M and A marks even if not used to find final answer and not contradicted
		A1 for 23.3… or \(4\sqrt{34}\) or \(\sqrt{544}\) oe with correct working	Award M2 A1 if M1A1 seen
		AND M2 for \(\dfrac{12\pi}{2}\) maybe implied by \(6\pi\) or by 18.84 to 18.852 or M1 for \(12\pi\) maybe implied by 37.68 to 37.704	Award M marks even if not used to find final answer and not contradicted Their \(12\pi\) must not come from area calculation or from \(2 \times \pi \times 12 \div 2\)
		If 0,1 or 2 scored, instead award SC3 for answer 62.16 to 62.18 or 62.2 or \(4\sqrt{34} + 6\pi + 20\) or \(\sqrt{544} + 6\pi + 20\) with no working or insufficient working If 0 or 1 scored, instead award SC2 for answer 42.16 to 42.18 or 42.2 or \(4\sqrt{34} + 6\pi\) or \(\sqrt{544} + 6\pi\) with no working or insufficient working If 0 scored, instead award SC1 for 23.3… or \(4\sqrt{34}\) or \(\sqrt{544}\) or 43.3… or \(4\sqrt{34} + 20\) or \(\sqrt{544} + 20\) with no working or insufficient working	Could be in working eg \(23.3 + 6\pi\)

Foundation June 2025 Paper 2 Q26

OCRFoundationCurrent spec6 marksCircles & Sectors

26 The diagram shows a sector of a circle AOB and a circle with radius \(r\) cm.

Sector AOB with centre O, radius OB = 12 cm and angle AOB = 120 degrees, and a separate circle with radius r cm

Not to scale

The sector has radius 12 cm and angle AOB = \(120^\circ\).

The ratio of the area of the sector AOB to the area of the circle is 3 : 4.

Calculate the value of \(r\).
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
8 with correct working	6		Correct working requires evidence of at least M1 and M1dep If candidates use numerical values for \(\pi\), allow 3.1 or 3.14… or \(\dfrac{22}{7}\). If candidates use other values e.g. 3 they must state \(\pi = 3\)
		M2 for \(\dfrac{120}{360} \times \pi \times 12^2\) oe or M1 for \(\dfrac{120}{360}\) oe \(\times\ k\pi\) or \(\pi \times 12^2\) or \(k \times \pi \times 12^2\) A1 for \(48\pi\)	For M1 \(k\) must be numerical and > 0 M1 may be implied by \(144\pi\) Award M2 A1 if M1 A1 seen Award M and A marks for the area of sector even if not used to find final answer and not contradicted.
		AND
		M1dep for their \(48\pi \div 3 \times 4\) or for their \(48 \div 3 \times 4\)	Dependent on M1. Implied by \(64\pi\) implied by 64
		M1dep for \([\pi \times]\,r^2 = [\pi \times](\)their \(48 \div 3 \times 4)\)	e.g. \(\pi r^2 = 64\pi\) or \(r^2 = 64\) or \(r = \sqrt{64}\). Dependent on previous M1dep If candidates only use trials to find square root, they must obtain the correct answer.
		If 0 or 1 scored instead award SC2 for \(r = 8\) with no/insufficient working or if 0 scored award SC1 for area of sector = \(48\pi\)	Do not award SC2 if clearly from wrong working Do not award SC1 if clearly from wrong working

Foundation June 2025 Paper 3 Q21

OCRFoundationCurrent spec4 marksCircles & Sectors Volume & Surface Area

21 The diagram shows a sphere with radius \(r\) cm.

Sphere with its centre marked and the radius r cm drawn to the surface

The sphere has a circumference of 50 cm.

Work out the surface area of the sphere.

[The surface area, \(A\), of a sphere with radius \(r\) is \(A = 4\pi r^2\).] [4]

Mark scheme

Answer	Marks	Part marks and guidance
796 or 795.67 to 796.4	4	M2 for [\(r =\)] \(\dfrac{50}{2\pi}\) oe or B2 for 7.96 or 7.956 to 7.958 or M1 for \(2\pi r = 50\) oe	Accept \(\dfrac{2500}{\pi}\) or 253.3[...]\(\pi\) for 4 marks May be \(50 \div \pi \div 2\) or \(25 \div \pi\) Note \(\dfrac{50}{2\pi} = 7.956 \approx 8\) then \(4 \times \pi \times 8^2\) scores M2 M1 A0
		and M1 for \(4 \times \pi \times (\textit{their}\ r)^2\) oe or \(4 \times \pi \times \left(\dfrac{50}{2\pi}\right)^2\) oe	For M1 substitution of their r must be explicit Their r may be any value from 3 to 10 understood to be their radius \(\pi\) may be substituted with 3.142 or calculator value, allowing 3.14 and \(\dfrac{22}{7}\)

Higher June 2025 Paper 5 Q19

OCRHigherCurrent spec5 marks2D Area & Perimeter Circles & Sectors

19 The diagram shows a shape formed by a semicircle joined to a sector of a circle, BAC, along a common length AB.

Sector BAC with centre A, angle BAC = 40°, and a semicircle on diameter AB below AB

Not to scale

The area of the sector BAC is \(16\pi\) cm\(^2\).

Work out the total area of the shape.
Give your answer in terms of \(\pi\).

You must show your working. [5]

Mark scheme

Answer	Marks	Part marks and guidance
\(34\pi\) final answer with correct working	5	M2 for \(\text{AB}^2 = \frac{16 \times 360}{40}\) or better or M1 for \(\frac{40}{360}[\pi]\text{AB}^2 [= 16[\pi]]\) oe or area of full circle = \(\frac{360}{40} \times 16\ [\pi]\) or \(144[\pi]\)	Correct working requires at least M2A1 For M2 accept AB\(^2\) = 144 Accept other variable for AB e.g. \(r\) , \(d\)
		A1 for AB = 12 or better e.g. \(\frac{\text{AB}}{2} = 6\) M1 for \(\pi\left(\frac{\textit{their } \text{AB}}{2}\right)^2 \div 2\ [+\ 16\pi]\) oe	After M1, A1 implies M2 their AB must be clearly indicated in working or on diagram . Allow AB as the answer to their work with the sector BAC
		If 0 or 1 scored instead award SC2 for answer \(34\pi\) with no or insufficient working If 0 scored, award SC1 for AB = 12 with no or insufficient working	Do not award SC2 if clearly from wrong working Do not award SC1 if clearly from wrong working

Higher June 2025 Paper 4 Q3

OCRHigherCurrent spec4 marks2D Area & Perimeter Circles & Sectors

3 The diagram shows a shape formed from a rectangle measuring 8 cm by 17 cm with a semicircle removed from one side.

Rectangle 17 cm by 8 cm with a semicircle cut out of the top side; the top side has 4 cm to the left of the semicircle and 3 cm to the right

Not to scale

Calculate the perimeter of the shape. [4]

Mark scheme

Answer	Marks	Part marks and guidance
55.7 to 55.71 or \(40 + 5\pi\)	4	B1 for diameter = 10 cm or radius = 5 cm	B1 might be seen on diagram and can be implied by 5 or 10 with \(\pi\) e.g. \(5\pi\), 31.4 to 31.42, \(5^2 \times \pi\)
		M1 for \(\left[\frac{1}{2}\right] 2 \times \pi \times\) their5 oe	M1 implied by \(10\pi\), \(5\pi\), 31.4 to 31.42 or 15.7 to 15.71 Award M marks even if not used to find final answer as long as it is not contradicted. Note : their 5 in the M1 should be their radius of the circle
		M1 for \(8 + 17 + 8 + 3 + 4 +\) their 15.7 or better	their 15.7 should involve \(\pi\) and be an attempt at the circumference of the semicircle, not the area or whole circle

Foundation June 2025 Paper 3 Q1

OCRFoundationCurrent spec3 marksCircles & Sectors

1 The diagram shows a circle, centre O, and a straight line.

Circle with centre O marked and a straight line joining two points on the circle, not through the centre

(a) Write down the mathematical name for

(i) the straight line [1]

(ii) the perimeter of the circle. [1]

(b) On the diagram, draw a diameter. [1]

Mark scheme (b)

Answer	Marks	Part marks and guidance
Straight line drawn joining two points on circumference and through centre	1		Accept good freehand but no daylight with centre blob and finishing at circumference \(\pm\) 2mm by eye.

Higher November 2024 Paper 5 Q15

OCRHigherCurrent spec8 marksCircles & Sectors Pythagoras

(a) The diagram shows a rectangle, OPQR, and a circle, centre O, which passes through Q.
OR = 6 cm and RQ = 8 cm.

Circle centre O with rectangle OPQR inside it; Q lies on the circle. OR = 6 cm and RQ = 8 cm

Not to scale

Find the circumference of the circle.
Give your answer in terms of \(\pi\). [4]

(b) AOB is a sector of a circle, centre O and radius 8 cm.
Angle AOB = \(x^\circ\).
The arc, AB, has length \(2\pi\) cm.

Sector AOB with radius OB = 8 cm, angle x° at O and arc AB of length 2π cm

Not to scale

Find the area of the sector.
Give your answer in terms of \(\pi\). [4]

Mark scheme (a)

Answer	Marks	Part marks and guidance
\(20\pi\) final answer	4	M2 for \(\sqrt{8^2 + 6^2}\) oe or M1 for 6² and 8² oe M1dep for \(2 \times \pi \times\) their r	Accept e.g. \(C = 20\pi\) M2 implied by 10 M1 dep on at least M1

Mark scheme (b)

Answer	Marks	Part marks and guidance
\(8\pi\) final answer	4	B2 for \(x = 45\) or \(\frac{45}{360}\) oe or \(\frac{2[\pi]}{2[\pi]8}\) oe or M1 for \(\frac{x}{360} \times 2 \times \pi \times 8 = 2\pi\) oe M1 for \(\frac{\textit{their } 45}{360} \times \pi \times 8^2\) oe or \(\frac{x}{360} \times \pi \times 8^2\)	0 < their 45 < 90 M1 for e.g. \(\frac{2[\pi]}{2[\pi]8} \times \pi \times 8^2\)

Foundation November 2024 Paper 1 Q6

OCRFoundationCurrent spec3 marksCircles & Sectors

6 The circumference of a circle is 17 cm.

Calculate the diameter of the circle.
Give your answer correct to 1 decimal place. [3]

Higher June 2024 Paper 5 Q20

OCRHigherCurrent spec7 marksCircles & Sectors Exact Trig & Graphs

20 The shape below is part of a circle, centre O and radius 12 cm.
Angle OAB = 30°.

Major sector of a circle centre O, radius 12 cm, with the chord AB forming the straight edge; dashed radii OA and OB; angle OAB = 30 degrees

Not to scale

Work out the perimeter of the shape.
Give your answer in its simplest terms in the form \(a\sqrt{b} + k\pi\).
You must show your working. [7]

Mark scheme

Answer	Marks	Part marks and guidance
\(12\sqrt{3} + 16\pi\) final answer with correct working	7		Correct working requires evidence of at least M1M1M2
		M1 for angle of major sector = 240° or \(\frac{240}{360}\) oe soi	For M1 accept e.g. \(\frac{2}{3}\) soi 240 may be on diagram
		M1 for \(\frac{\textit{their } 240}{360} \times 2 \times \pi \times 12\) A1 for \(16\pi\) AND	For M1 \(120 \leqslant \textit{their } 240 \leqslant 300\) but not 180 \(24\pi \div 3 \times 2\) oe eg done in stages implies M1M1
		M2 for [ 2 ×] 12 cos30° oe	M2 oe accept e.g. [\(\text{AB}^2 =\) ] \(12^2 + 12^2 - 2.12.12.\cos 120\) or \(\dfrac{12\sin 120}{\sin 30}\), [2×] \(\sqrt{12^2 - (12\sin 30)^2}\) oe
		or M1 for \(\frac{x}{12} = \cos 30\) oe	where \(x\) is ½ AB M1 for any correct implicit method for finding AB or ½ AB
		B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) soi A1 for \(12\sqrt{3}\)	In this method and other methods used, B1 is awarded for the correct trig value[s] associated with their method for find [½]AB, even if their method is incorrect, not just seen in a table of trig values unless selected Award maximum of 6 marks if answer incorrect
		If 0, 1 or 2, scored instead award SC3 for answer \(12\sqrt{3} + 16\pi\) If 0 or 1 scored, instead award SC2 for \(12\sqrt{3}\) or \(16\pi\) in answer	SEE AG

Additional guidance: Question 20

The length AB or \(\frac{1}{2}\) AB (\(12\sqrt{3}\) oe or \(6\sqrt{3}\) oe may be seen unsimplified e.g. \(2\sqrt{108}\) or \(\sqrt{108}\)) can be obtained by a number of different methods.
M2 is for the explicit method leading to AB or \(\frac{1}{2}\)AB and examples of this are in the guidance column of the scheme

M1 is for a correct implicit method

The B mark is awarded for a correct trig value[s] for their chosen method

Implicit methods may include

\(\text{AB}^2 + (12\sin 30)^2 = 12^2\) oe or better	\(\cos 120 = \dfrac{12^2 + 12^2 - \text{AB}^2}{2 \times 12 \times 12}\) oe or better	\(\dfrac{\sin 120}{AB} = \dfrac{\sin 30}{12}\) oe or better
Award B1 for \(\sin 30 = \frac{1}{2}\) soi	Award B1 for \(\cos 120 = -\frac{1}{2}\) soi	Award B1 for \(\sin 120 = \frac{\sqrt{3}}{2}\) and \(\sin 30 = \frac{1}{2}\) soi

Higher June 2024 Paper 4 Q14

OCRHigherCurrent spec6 marksAdvanced Trigonometry Circles & Sectors

14 The diagram shows a circle, centre O and radius 52 cm.

Circle centre O, radius 52 cm, with chord AB at the top; the segment above AB is shaded; a tangent touches the circle at T at the bottom; the perpendicular distance from the chord to the tangent is 72 cm

Not to scale

AB is a chord of the circle.
The line through T is a tangent to the circle.

The chord is parallel to the tangent.
The perpendicular distance between the chord and the tangent is 72 cm.

Calculate the area of the shaded segment.
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
2219.9… or 2220 with correct working	6		“Correct working” requires evidence of at least M2 or M1 M1 Accept answers in the range 2217 to 2226. Accept any correct method.
		M2 for \(\cos [\ldots] = \frac{72 - 52}{52}\) oe or M1 for 72 – 52 or 20 and	M2 implied by 67.38… or 134.76… rot to at least an integer e.g. 67, 67.3, 67.4, 67.38, 134, equivalent method includes \(\cos [\ldots] = \frac{52^2 + 52^2 - 96^2}{2 \times 52 \times 52}\) condone with 72 instead of 96
		M2 for \(\frac{1}{2} \times 52 \times 52 \times \sin\) their 134.76… oe or M1 for \(\sqrt{52^2 - (\textit{their } 20)^2}\) oe or 48 and M1 for \(\frac{1}{2} \times\) their96 × their20 oe and	M2 for finding the area of the triangle OAB implied by 960 their 134.76… must come from a correct attempt to find angle AOB and condone using AB = 72 for M2 and M1 including \(\sqrt{52^2 - 36^2}\) for M1
		M1 for \(\frac{\textit{their angle AOB}}{360} \times \pi \times 52^2\) or B1 for \(\pi \times 52^2\)	M1 implied by 3176 to 3186 or B1 implied by 2704\(\pi\), 8490 to 8496
		If 0, 1 or 2 scored award SC3 for the correct answer with insufficient working If 0 or 1 scored award SC2 for 3179.918… or 5314.718.. rot to at least an integer with insufficient working	alternative method using major sector: M2 for angle as before implied by 225, 225.2, 225.23 or 225.24 M2 for area of triangle as before or M1 for the third side e.g. 48 M1 for \(\left[\frac{\textit{their reflex angle AOB}}{360}\right] \times \pi \times 52^2\) or 5314.718..

Higher June 2024 Paper 6 Q9

OCRHigherCurrent spec7 marks2D Area & Perimeter Circles & Sectors Show That & Proof

9 The diagram shows a shaded sector inside a parallelogram.
The sector has an angle of 30°.
The parallelogram, ABCD, has length BC = 20 cm and AB = 12 cm.
The perpendicular distance between BC and AD is 6 cm.

Parallelogram ABCD with BC = 20 cm at the top, AB = 12 cm, perpendicular height 6 cm; a shaded sector centre A, radius AB, with angle 30 degrees at A

Not to scale

(a) Show that the area of the sector is \(37.7\,\text{cm}^2\), correct to 3 significant figures. [3]

(b) Work out the percentage of the parallelogram that is not shaded. [4]

Mark scheme (a)

Answer	Marks	Part marks and guidance
\(\frac{30}{360} \times \pi \times 12^2\) oe	M2	M1 for \(\pi \times 12^2\) implied by \(144\pi\) or 452.3 to 452.45	M2 oe e.g. \(\frac{360}{30} = 12\) and \(\frac{\pi \times 12^2}{12}\) Condone 3.14 or \(\frac{22}{7}\) for \(\pi\) in M marks M0 for \(12\pi\) without working
37.69 to 37.704	A1	A0 for just 37.7

Mark scheme (b)

Answer	Marks	Part marks and guidance
68.5 to 68.6 nfww	4	With area of parallelogram as 120 or from an attempt at 20 × 6:	If 120 not used, their (20 × 6) must come from attempt at 20 × 6 Note there are other methods for finding area of a parallelogram 37.7 may be their more accurate 37.69 to 37.704 from (a)
		M3 for \(\frac{\textit{their}(20 \times 6) - 37.7}{\textit{their}(20 \times 6)}\) [× 100] oe or M2 for their (20 × 6) – 37.7 implied by 82.3 or for \(\frac{37.7}{\textit{their}(20 \times 6)}\) [× 100] implied by 31.4 to 31.5 or M1 for 20 × 6 implied by 120	M3 oe e.g. \(100 - \frac{37.7 \times 100}{\textit{their}(20 \times 6)}\)
		If correct method for area of parallelogram is not shown or an incorrect value is used: If \(A \gt 37.7\) SC2 for \(\frac{A - 37.7}{A}\) [× 100] oe or SC1 for \(\frac{37.7}{A}\) [× 100]	e.g SC2 for \(\frac{240 - 37.7}{240}\) [× 100] oe For SC marks method must be shown; do not imply from an answer only

Foundation June 2024 Paper 3 Q7

OCRFoundationCurrent spec2 marksCircles & Sectors

7 A circle has radius 8 cm.

Work out the area of the circle. [2]

Mark scheme

Answer	Marks	Part marks and guidance
201 or 201.1 or 201.06 to 201.09	2	M1 for \(\pi \times 8^2\) oe Allow \(64\pi\) for 2 marks	\(\pi\) gives 201.0619... 3.142 gives 201.088 For M1 \(\pi\) may be evaluated allowing \(\frac{22}{7}\) or 3.14 or 3.142

Foundation June 2024 Paper 2 Q4

OCRFoundationCurrent spec1 markCircles & Sectors

4 The radius of a circle is 3 cm.

Find the length of the diameter of the circle. [1]

Foundation November 2023 Paper 1 Q26

OCRFoundationCurrent spec3 marksCircles & Sectors Show That & Proof

26 XOY is a sector of a circle, centre O.

Sector XOY with radius OX = 5 cm and angle XOY = 70 degrees

Not to scale

Show that the area of the sector is 15.3 cm\(^2\), correct to 3 significant figures. [3]

Mark scheme

Answer	Marks
\(\pi \times 5^2\)	M1
their\((\pi \times 5^2) \times \frac{70}{360}\) oe	M1	Accept \((78.5\) to \(78.6) \times \frac{70}{360}\)
15.26 to 15.284 [rounds to 15.3]	A1

Higher November 2023 Paper 5 Q23

OCRHigherCurrent spec9 marksCircles & Sectors Exact Trig & Graphs Show That & Proof

23 In the diagram,

ABC is a right-angled triangle
ACD is the sector of a circle with centre A.

Right-angled triangle ABC with the right angle at B, AB = 4 m and angle BAC = 30 degrees; sector ACD with centre A and angle CAD = 45 degrees

Not to scale

(a) Show that the area of the sector ACD is \(\frac{8}{3}\pi\) m². [6]

(b) Work out the total area of the shape ABCD.

Give your answer in the form \(\left(\dfrac{a\sqrt{k}}{b} + \dfrac{8}{3}\pi\right)\) m². [3]

Mark scheme (a)

Answer	Marks	Part marks and guidance
[\(AC =\)] \(\dfrac{4}{\cos 30}\) oe	M2	M1 for \(\frac{4}{AC} = \cos 30\) oe or for \(\frac{\sin 60}{4} = \frac{\sin 90}{AC}\) oe	Accept any variable for AC provided not incorrect length Accept longer methods using 4 tan 30 to find BC then Pythagoras’ or sine rule M2 for AC explicit
(\(AC =\)) \(\frac{8}{\sqrt{3}}\) or better	A2	B1 for \(\cos 30 = \frac{\sqrt{3}}{2}\) oe	oe for B1 e.g. \(\sin 60 = \frac{\sqrt{3}}{2}\)
\(\dfrac{45}{360} \times \pi \times \left(\textit{their}\ \dfrac{8}{\sqrt{3}}\right)^2\)	M1		dep on at least M1
\(\frac{45}{360} \times \pi \times \frac{64}{3}\) or better or \(\frac{1}{8} \times \pi \times \frac{8}{\sqrt{3}} \times \frac{8}{\sqrt{3}}\) \(= \frac{8}{3}\pi\)	A1	with no errors seen	Accept \(\frac{1}{8} \times \pi \times \frac{8^2}{3}\)

Mark scheme (b)

Answer	Marks	Part marks and guidance
\(\dfrac{8\sqrt{3}}{3} + \dfrac{8}{3}\pi\)	3	M1 for \(\frac{1}{2} \times 4 \times \frac{8}{\sqrt{3}} \times \sin 30\) FT their AC from (a) or for \(\frac{1}{2} \times 4 \times 4 \times \tan 30\) B1 for \(\sin 30 = \frac{1}{2}\) oe or for \(\tan 30 = \frac{\sqrt{3}}{3}\) oe	Accept e.g. \(\frac{16\sqrt{3}}{6} + \frac{8}{3}\pi\) for 3 marks

Foundation November 2023 Paper 3 Q20

OCRFoundationCurrent spec4 marksCircles & Sectors

20 The diagram shows a Penny Farthing bicycle.

Drawing of a Penny Farthing bicycle with a very large front wheel and a small rear wheel

Not to scale

The diameter of the large wheel is 130 cm.
The diameter of the small wheel is 46 cm.

On a short journey, the large wheel makes exactly 69 rotations.
The small wheel also makes an exact number of rotations.

Work out the number of rotations made by the small wheel. [4]

Mark scheme

Answer	Marks	Part marks and guidance
195	4	B3 for answer that rounds to 195 OR M3 for \(\dfrac{[\pi \times]130}{[\pi \times]46} \times 69\) oe or M2 for \(\dfrac{[\pi \times]130}{[\pi \times]46}\) or \([\pi \times]130 \times 69\) or M1 for \(\pi \times 130\) soi 408.4… or \(\pi \times 46\) soi 144.5…	May be in stages or in metres Implied by 2.82[6…] or 8970 or 28180. … If multiple calculations refer to general guidance

Higher November 2023 Paper 6 Q20

OCRHigherCurrent spec7 marksAngles (Including Parallel Lines)Circles & Sectors Show That & Proof

20 This shape is formed from a rectangle and two sectors of circles.

Shape made of rectangle BDEF with DE = 5t and EF = 2t, a sector BCD with centre B and angle CBD = 35 degrees, and a sector BAF with centre B; A, B and C lie on a straight line

Not to scale

Points A, B and C lie on a straight line.
Angle CBD = 35°.
DE = \(5t\) and EF = \(2t\).

(a) Explain why BC = \(2t\).
Give a reason for each step of your explanation. [2]

(b) Show that the perimeter of the shape is \(\frac{23}{12}\pi t + 14t\). [5]

Mark scheme (a)

Answer	Marks	Part marks and guidance
BD = EF or BD = \(2t\) and [opposite sides of a] rectangle [are equal]	1	For two marks, \(2t\) must be seen in at least one statement as BD or on the diagram as BD
BC = BD [= \(2t\)] and radii [of a sector/circle]	1

Mark scheme (b)

Answer	Marks
ABF = 55 and AB = \(5t\)	B1	Stated or seen on diagram
\(\dfrac{\textit{their}\ 55}{360} \times 2\pi \times \textit{their}\ 5t\)	M1	All M marks may be seen within a summarising expression
\(\dfrac{35}{360} \times 2\pi \times 2t\)	M1
\(5t + 2t + 5t + 2t\)	M1	Condone \(10t + 4t\), \(7t + 7t\) etc but not \(14t\)
\(\frac{35}{360} \times 2\pi \times 2t + \frac{55}{360} \times 2\pi \times 5t + 5t + 2t + 5t + 2t\) \(= \frac{23}{12}\pi t + 14t\)	A1

Foundation June 2023 Paper 1 Q27

OCRFoundationCurrent spec6 marks2D Area & Perimeter Circles & Sectors

27 The diagram shows a circle inside a square of side 12 cm.

Square of side 12 cm with a circle touching all four sides; the region between the circle and the bottom right corner of the square is shaded

Work out the percentage of the square that is shaded.
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
5.36 to 5.4 and correct working	6	B5 for the correct answer in the wrong format with correct working e.g. 0.0536 OR M5 for \(\dfrac{12^2 - \pi \times 6^2}{12^2 \times 4}\) [\(\times\) 100] oe OR Square M1 for \(12^2\) or 144 or \(6^2\) or 36 (must be consistent with \(\dfrac{1}{4} \times \pi \times 6^2\)) and Circle M2 for \(\pi \times 6^2\) or \(\dfrac{1}{4} \times \pi \times 6^2\) or M1 for radius of 6 may be implied e.g. \(2 \times \pi \times 6\) (with \(\pi\)) and M1 for (their \(12^2\) – their \((\pi \times 6^2)\)) [\(\div\)4] or their \((\pi \times 6^2) \div\) their \(12^2\) or their (\(6^2 -\) their \(\left(\dfrac{1}{4} \times \pi \times 6^2\right)\)) [\(\div\)their \(12^2\)] or their \(\left(\dfrac{1}{4} \times \pi \times 6^2\right) \div \dfrac{\textit{their}\ 12^2}{4}\) If 0 or M1 or M2 scored, instead award SC3 for answer 5.36 to 5.4 with no or insufficient working If 0 or M1 scored, instead award SC2 for 30.88 to 31 or 7.72 to 7.75 or 0.785 to 0.786 with no or insufficient working	“Correct working” requires evidence of at least M1 AND M2 AND M1 their 144 must come from attempt at area of a square M2 implied by 113.04 to 113.12 or 28.26 to 28.28 6 [cm] could be on diagram FT their incorrect 6 identified as radius implied by 30.88 to 31 or 7.72 to 7.75 implied by 0.785 to 0.786 implied by 7.72 to 7.75 implied by 0.785 or 0.7852 to 0.7856 their area of the circle has to be an attempt at \(\pi r^2\) not \(2\pi r\) so that, if they do not, the most they can be awarded is M1 for 144 and M1 for radius = 6 cm

Higher June 2023 Paper 4 Q8

OCRHigherCurrent spec6 marks2D Area & Perimeter Circles & Sectors

8 The diagram shows a circle inside a square of side 12 cm.

A circle touching all four sides of a square of side 12 cm; the region between the circle and the bottom right corner of the square is shaded

Work out the percentage of the square that is shaded.
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
5.36 to 5.4 and correct working	6	B5 for the correct answer in the wrong format with correct working e.g. 0.0536 OR M5 for \(\frac{12^2 - \pi \times 6^2}{12^2 \times 4}\) [× 100] oe OR Square M1 for \(12^2\) or 144 or \(6^2\) or 36 (must be consistent with \(\frac{1}{4} \times \pi \times 6^2\)) and Circle M2 for \(\pi \times 6^2\) or \(\frac{1}{4} \times \pi \times 6^2\) or M1 for radius of 6 may be implied e.g. 2×π×6 (with \(\pi\)) and M1 for (their \(12^2\) – their (\(\pi \times 6^2\))) [÷4] or their (\(\pi \times 6^2\)) ÷ their \(12^2\) or their (\(6^2\) – their (\(\frac{1}{4} \times \pi \times 6^2\))) [÷their \(12^2\)] or their (\(\frac{1}{4} \times \pi \times 6^2\)) ÷ \(\frac{their\ 12^2}{4}\) If 0 or M1 or M2 scored, instead award SC3 for answer 5.36 to 5.4 with no or insufficient working If 0 or M1 scored, instead award SC2 for 30.88 to 31 or 7.72 to 7.75 or 0.785 to 0.786 with no or insufficient working	“Correct working” requires evidence of at least M1 AND M2 AND M1 their 144 must come from attempt at area of a square M2 implied by 113.04 to 113.12 or 28.26 to 28.28 6 [cm] could be on diagram FT their incorrect 6 identified as radius implied by 30.88 to 31 or 7.72 to 7.75 implied by 0.785 to 0.786 implied by 7.72 to 7.75 implied by 0.785 or 0.7852 to 0.7856 their area of the circle has to be an attempt at \(\pi r^2\) not \(2\pi r\) so that, if they do not, the most they can be awarded is M1 for 144 and M1 for radius = 6 cm

Foundation June 2023 Paper 3 Q8

OCRFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors

(a) Work out the area of this trapezium.

Right trapezium with parallel sides 6 cm and 9 cm, perpendicular height 4 cm and slanted side 5 cm

Not to scale

[2]

(b) The circumference of a circle, in terms of \(\pi\), is \(100\pi\) cm.

Work out the radius of the circle. [2]

Mark scheme (a)

Answer	Marks	Part marks and guidance
30	2	M1 for \(\dfrac{4 \times (9 + 6)}{2}\) oe or \(4 \times 6 + \dfrac{4 \times 3}{2}\) oe or \(4 \times 9 - \dfrac{4 \times 3}{2}\)	Implied by 24 + 6 Implied or by 36 – 6

Foundation November 2022 Paper 3 Q17

OCRFoundationCurrent spec4 marksCircles & Sectors

17 Frankie draws a circle and works out its area, in cm², and circumference, in cm.
The answer for the area is two times the answer for the circumference.

Work out the diameter of the circle.
You must show your working.

.................... cm [4]

Mark scheme

Answer	Marks	Part marks and guidance
8 cao with correct working	4	M1 for \(\pi r^2\) [=] \(4\pi r\) or \(\pi r^2\) [=] \(2\pi d\) M1 for cancellation by \(\pi\) implied or factorising e.g. \(\pi r(r - 4) = 0\) or \(r(r - 4) = 0\) A1 for [radius =] 4 [and 0] Trials using \(\pi r^2\) and \(4\pi r\) or \(\pi r^2\) and \(2\pi d\) or \(\pi r^2\) and \(2\pi r\) or \(\pi r^2\) and \(\pi d\) M2 for two correct trials with \(r \ne 4\) or two correct trials with first \(r \ne 4\) and then \(r = 4\) or one correct trial with \(r = 4\) or M1 for one correct trial with \(r \ne 4\) A1 for [radius =] 4 If 0 or M1 scored, instead award SC2 for answer 8 cao If 0 scored, instead award SC1 for answer 4 [and 0]	“Correct working” requires evidence of at least M1M1 FT their initial statement if possible Accept \(\pi r^2 = 4\pi r\) followed by \(r^2 = 4r\) as showing cancelling Dependent on M1M1 Must use same value substituted for \(r\) in both formulas Answers required Dependent on M2 or M1 with no working or insufficient working with no working or insufficient working

Higher June 2022 Paper 5 Q18

OCRHigherCurrent spec5 marksCircles & Sectors

18 The diagram shows a shaded shape made by removing sector OAB from sector OCD.
Both sectors have an angle of 60°.
The radius, OA, of the smaller sector is \(r\) cm.
The ratio of radius OA to radius OC is 2 : 3.

Sector OCD with angle 60° at O; the smaller sector OAB, radius OA = r cm, is removed leaving a shaded band between arc AB and arc CD; A lies on OC and B lies on OD

Not to scale

Work out, in terms of \(\pi\) and \(r\), the total length of arc AB and arc CD.
Give your answer in its simplest form.
You must show your working. [5]

Mark scheme

Answer	Marks	Part marks and guidance
\(\frac{5\pi r}{6}\) with correct working	5		Condone ‘×’ sign oe in simplified answer if otherwise correct e.g. \(\frac{5}{6} \times \pi r\) “correct working” requires M1A1M1A1 Condone \(R\) for \(r\) throughout For method marks, allow use of 3.14, 3.142, 22/7 for \(\pi\)
		B4 for correct unsimplified answer with correct working
		OR
		M1 for \(\frac{60}{360} \times [2\times]\ \pi k\) oe	Where \(k\) is numeric or algebraic but does not come from squaring Allow e.g. \(k\) = 2, \(r\), \(d\), 0.4 , 0.4\(x\)
		A1 for \(\frac{60}{360} \times 2\pi r\) oe or better isw incorrect cancelling/simplification	For A1 accept e.g. \(0.333\pi r\) Correct expression implies M1A1
		AND
		M1 for \(\frac{60}{360} \times [2\times]\ \pi\frac{3k}{2}\)	For M1 must use their previous \(k\) e.g. uses \(k\) = 10 for first M1 then uses 15 here for \(\frac{3k}{2}\) gets 2^nd M1 unless the expression is correctly stated as \(\frac{60}{360} \times \pi 3r\) oe which gets M1A1
		A1 for \(\frac{60}{360} \times \pi 3r\) oe or better isw incorrect cancelling/simplification	Correct expression implies M1A1
		If 0 or 1 scored, instead award SC2 for final answer \(\frac{5\pi r}{6}\) oe simplified answer with no or insufficient working

Foundation June 2022 Paper 1 Q12

OCRFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors

12 The diagram shows Kai’s garden.
The garden is a rectangle, 16 m by 10 m.

It has a lawn and a flowerbed.
The flowerbed is a circle of diameter 6 m.

Rectangle 16 m by 10 m, shaded and labelled Lawn, containing an unshaded circle of diameter 6 m labelled Flowerbed

Not to scale

Work out the area of Kai’s lawn. [4]

Mark scheme

Answer	Marks	Part marks and guidance
131.7[…] or 132	4	M1 for [rectangle] \(16 \times 10\) soi 160 M1 for [circle] \(\pi \times 3^2\) oe soi 28.2 to 28.3 M1 for their rectangle area – their circle area If 0 or 1 scored, instead award SC2 answer 46.9[…]	Method for rectangle and circle areas spoilt by additional steps in initial area calculations e.g. M0 for \(16 \times 10 \times 2\) Do not lose M1 for [rectangle] or M1 for [circle] if further work does not include these, provided it is not contradicted by further method Their circle area should be from \(\pi \times 3^2\) or \(\pi \times 6^2\) eg soi by 113.[…] not \(2\pi r\)

Foundation June 2022 Paper 3 Q4

OCRFoundationCurrent spec2 marksCircles & Sectors

4 The diagram shows a circle, centre O, and a line that meets the circle twice.

Circle with centre O and a straight line joining two points on the circle, passing above O

(a) On the diagram, draw a diameter. [1]

(b) Write down the mathematical name of the line shown on the diagram. [1]

Mark scheme (a)

Answer	Marks	Part marks and guidance
Diameter drawn	1		Allow freehand intended straight No white paper seen between line and centre and line and circumference Ends no more than 2mm by eye beyond circumference If two lines mark the worst

Higher June 2022 Paper 6 Q2

OCRHigherCurrent spec4 marksCircles & Sectors Show That & Proof

2 The circumference of a circle is 23 cm.

Show that the area of the circle is 42.1 cm², correct to 3 significant figures. [4]

Mark scheme

Answer	Marks	Part marks and guidance
42.09[…] with correct working	4	M3 for \(\pi \times \left(\dfrac{23}{2\pi}\right)^2\)	Correct working requires evidence of at least M1 Do not accept starting with 42.1 and working backwards Candidates should use the π button or 3.142, accept 3.14 and \(\frac{22}{7}\) for max of M3
		OR M1 for \(23 \div 2\pi\) or \(23 = 2\pi r\) or \(23 = \pi d\) A1 for 3.66[…] M1 for \(\pi \times (\textit{their } 3.66)^2\)	M1 may be in two steps or seen as diameter here and diameter/2 in the area calculation their 3.66 must come from a calculation involving π
		If 0 or 1 scored SC2 for 42.09[…] with no or insufficient working If 0 scored SC1 for 3.66[…], 42.08[…], 42.10[…], 42.11[…] or 168[…] with no or insufficient working

Higher November 2021 Paper 5 Q21

OCRHigherCurrent spec6 marksCircles & Sectors

21 AOB is a sector of a circle, centre O and radius 10 cm.

Sector AOB of a circle centre O with radius OB = 10 cm

Not to scale

The area of the sector is \(40\pi\) cm².

Work out the perimeter of the sector.
Give your answer in the form \(a + b\pi\), where \(a\) and \(b\) are integers.
You must show your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
\(20 + 8\pi\) final answer with correct working	6		For full marks “correct working” requires evidence of at least M1 AND M1 ie use of formulas for sector area and arc length or alternate convincing approach
		B3 for [angle at centre ]= 144 with correct working or \(\frac{4}{10}\) [of circle] oe or M2 for \(\dfrac{[360 \times]\,40\pi}{\pi \times 10^2}\) oe or M1 for \(\dfrac{\theta}{360} \times \pi \times 10^2\) oe	For B3 “correct working” requires at least M1 for use of formula for sector area M2 method for finding fraction of circle
		AND M2 for answer \(20 + k\pi\) where \(k = \dfrac{\textit{their } \theta}{18}\) or M1 for \(\dfrac{\textit{their } \theta}{360} \times 2 \times \pi \times 10\) oe or answer \(20 + k\pi\)	M1 Implied by \(8\pi\) For M1 \(k \ne 0\)
		If 0 scored SC2 for answer \(20 + 8\pi\)	May be on diagram

Foundation November 2021 Paper 1 Q11

OCRFoundationCurrent spec4 marks2D Area & Perimeter Circles & Sectors

11 A rectangle, 6 cm by 3 cm, and a semi-circle are joined to make this shape.

Rectangle 3 cm wide and 6 cm tall with a semicircle on its right-hand 6 cm side; the join is shown dashed

Not to scale

Work out the area of the shape. [4]

Mark scheme

Answer	Marks	Part marks and guidance
32.1 to 32.14 nfww	4	M1 for \(3 \times 6\) may be implied by 18 AND M2 for \(\dfrac{3^2\pi}{2}\) soi by 14.1 to 14.15 or M1 for \(3^2\pi\) soi by 28.2 to 28.3	18 must not come from wrong working Not part of longer multiplication e.g. \(3 \times 6 \times 3\)

Foundation November 2020 Paper 1 Q25

OCRFoundationCurrent spec6 marks2D Area & Perimeter Circles & Sectors

25 The diagram shows Jane’s lawn.
It is in the shape of a square of side 36 m and three semi-circles.

Lawn made of a square with a semicircle on its top side and semicircles on its left and right sides

Not to scale

She is going to spread fertiliser on the lawn at a rate of 30 g per square metre.
The fertiliser is only sold in 10 kg bags costing £15.80 each.

Calculate the cost of buying the bags of fertiliser for her lawn.
You must show all your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
142.2[0] with correct working	6	M1 for \(36^2\) or 1296 M1 for \(k \times \pi \times 18^2\) oe where \(k = \tfrac{1}{2}\), 1, \(1\tfrac{1}{2}\) or 3 AND M1 for their area × 30 M1 for their mass ÷ 1000 and ÷ 10 or counting up in 10 000s to their mass AND M1 for their 9 × 15.8 If 0, 1 or 2 scored instead award SC3 for answer of 142.2[0] with insufficient working If 0 or 1 scored instead award SC2 for 2822 to 2823.02 If 0 scored award SC1 for 1526 to 1527.02, 1017 to 1018.008, 508 to 509.004, 3051 to 3054.024, \(162\pi\), \(324\pi\), \(486\pi\) rot to at least nearest integer	Correct working requires M1 AND M1 AND M1 M2 implied by 2822 to 2823.02 or M1 implied by 1526 to 1527.02, 1017 to 1018.008, 508 to 509.004, 3051 to 3054.024, \(162\pi\), \(324\pi\), \(486\pi\) rot to at least nearest integer their area cannot be 36 and M1 implied by 84660 to 84 690.6 or 84.66 to 84.7 their mass is attempt at (rectangle and circle(s)) × 30, M1 implied by 8.46 to 8.47 their 9 dep. on fourth M1 scored with a rounding up to next integer

Foundation November 2020 Paper 2 Q23

OCRFoundationCurrent spec4 marksCircles & Sectors

23 AOB is a sector of a circle, centre O.

Sector AOB with angle 120 degrees at O and radius OB labelled r cm

Not to scale

The area of the sector is 8 cm\(^2\).

Work out the exact value of the radius, \(r\) cm. [4]

Mark scheme

Answer	Marks	Part marks and guidance
\(\sqrt{\dfrac{24}{\pi}}\)	4	M2 for [\(r^2 =\)] \(\dfrac{360 \times 8}{120 \times \pi}\) or better or M1 for \(\dfrac{120}{360} \times \pi r^2\) [= 8] oe or better A1 for [\(r^2 =\)] \(\dfrac{24}{\pi}\)	For method condone use of 3.1, 3.14, etc used for \(\pi\) M1 implied by \(\pi r^2 = 24\) Implied by 7.6…

Higher November 2020 Paper 4 Q10

OCRHigherCurrent spec6 marks2D Area & Perimeter Circles & Sectors

10 The diagram shows Jane’s lawn.
It is in the shape of a square of side 36 m and three semi-circles.

A square, shown with dashed sides, with a semicircle on its top side, its left side and its right side

Not to scale

She is going to spread fertiliser on the lawn at a rate of 30 g per square metre.
The fertiliser is only sold in 10 kg bags costing £15.80 each.

Calculate the cost of buying the bags of fertiliser for her lawn.
You must show all your working. [6]

Mark scheme

Answer	Marks	Part marks and guidance
142.2[0] with correct working	6	M1 for 36² or 1296 M1 for k ×π × 18² oe where k = ½, 1, 1½ or 3 AND M1 for their area × 30 M1 for their mass ÷ 1000 and ÷ 10 or counting up in 10 000s to their mass AND M1 for their 9 × 15.8 If 0, 1 or 2 scored instead award SC3 for answer of 142.2[0] with insufficient working If 0 or 1 scored instead award SC2 for 2822 to 2823.02 If 0 scored award SC1 for 1526 to 1527.02, 1017 to 1018.008, 508 to 509.004, 3051 to 3054.024, 162π, 324π, 486π rot to at least nearest integer	Correct working requires M1 AND M1 AND M1 M2 implied by 2822 to 2823.02 or M1 implied by 1526 to 1527.02, 1017 to 1018.008, 508 to 509.004, 3051 to 3054.024, 162π, 324π, 486π rot to at least nearest integer their area cannot be 36 and M1 implied by 84660 to 84 690.6 or 84.66 to 84.7 their mass is attempt at (rectangle and circle(s)) × 30, M1 implied by 8.46 to 8.47 their 9 dep. on fourth M1 scored with a rounding up to next integer

Foundation November 2019 Paper 3 Q4

OCRFoundationCurrent spec4 marksCircles & Sectors

4 A circle has radius 5 cm.

(a) Work out the circumference of the circle. [2]

(b) Work out the area of the circle. [2]

Foundation June 2019 Paper 1 Q24

OCRFoundationCurrent spec3 marksCircles & Sectors Show That & Proof

24 AOB is a sector of a circle, centre O.

Sector AOB with centre O, radius OB = 6 cm and angle AOB = 50 degrees

Not to scale

Show that the length of arc AB is 5.24 cm, correct to 3 significant figures. [3]

Mark scheme

Answer	Marks	Part marks and guidance
\(2 \times \pi \times 6\) oe implied by \(12\pi\)	M1	Accept correct alternative methods
\(\dfrac{50}{360} \times\) their \((2 \times \pi \times 6)\)	M1
5.235[…] or 5.236[…] or 5.237 which rounds to 5.24	A1

Higher June 2019 Paper 5 Q15

OCRHigherCurrent spec6 marksCircles & Sectors Exact Trig & Graphs Show That & Proof

15 OAB is a sector of a circle, centre O.
OA = 6 cm and AX is perpendicular to OB.

Sector OAB with OA = 6 cm; AX is perpendicular to OB with X between O and B

Not to scale

The area of sector OAB is \(6\pi\) cm\(^2\).

Show that \(\text{AX} = 3\sqrt{3}\) cm. [6]

Mark scheme

Answer	Marks	Part marks and guidance
\(\dfrac{x}{360} \times \pi \times 6^2\) or \(\dfrac{[\pi \times]6^2}{[\pi \times]6}\) or shows \(\pi \times 6^2\) and \(\dfrac{1}{6}\) oe	M1	Accept 36 for \(6^2\)	\(x\) = angle AOX , condone any variable used For M1 may be seen in stages e.g. M1 for \(36\pi \div 6\)
\(\dfrac{x}{360}[\times \pi] \times 6^2 = 6[\pi]\) or 360 ÷ 6	M1dep	Dep on previous M1
[\(x\) = ] 60	A1		Must earn M1M1 before awarding A1
\(\dfrac{\text{AX}}{6} = \sin\) their 60 oe	M1	Dep on 0 < their 60 < 90 Accept use of cos 30 or cos 60 and Pythagoras’ or sine rule with 90	Do not accept assumption that OX = 3 without any evidence
\(\text{AX} = 6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3}\) or \(\dfrac{3\sqrt{3}}{6} = \dfrac{\text{AX}}{6}\), \(\text{AX} = 3\sqrt{3}\)	M2	or M1 for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) or \(\cos 30 = \dfrac{\sqrt{3}}{2}\) To award 6 marks, there must be no errors seen	Beware circular methods using \(3\sqrt{3}\) leading to 60, this can only score M1 maximum for \(\sin 60 = \dfrac{\sqrt{3}}{2}\) but ignore circular methods if alongside a correct method

Foundation June 2019 Paper 1 Q13

OCRFoundationCurrent spec2 marksCircles & Sectors

13 Calculate the area of a circle with radius 14 cm. [2]

Higher June 2019 Paper 4 Q7

OCRHigherCurrent spec6 marksCircles & Sectors Pythagoras

7 The diagram shows a shape ABCDE.
The shape is made from a rectangle, a right-angled triangle and a quarter of a circle.

Shape ABCDE: rectangle ABDE with AE = 18 m at the bottom; above BD a quarter circle from B up to C and a right-angled triangle CFD with F on BD below C; the perpendicular height from C to AE is 41 m

Not to scale

F is the mid-point of BD.
AE = 18 m and the perpendicular distance from C to AE is 41 m.

Work out the perimeter of the shape ABCDE. [6]

Mark scheme

Answer	Marks	Part marks and guidance
109 or 108.8 to 108.9	6	figures and working may be on diagram B1 for [radius of circle =] 9 B1 for [AB or ED =] 32 soi by 41 – their 9 M1 for \(\frac{1}{4} \times 2 \times \pi \times\) their 9 soi by \(\frac{9\pi}{2}\), 14.1 or 14.13 to 14.14 M2 for \(\sqrt{(\textit{their }9)^2 + (\textit{their }9)^2}\) soi by \(9\sqrt{2}\), 12.7[2…] or 12.73 or M1 for \((\textit{their }9)^2 + (\textit{their }9)^2\) soi by 162 M1 for their 12.72… + their 14.1 + 2 × their 32 + 18 seen to a maximum of 5 marks

Foundation November 2018 Paper 2 Q21

OCRFoundationCurrent spec2 marks2D Area & Perimeter Circles & Sectors

21 A circle, with centre O and radius 6 cm, contains another circle, with centre O and radius \(x\) cm.

Two circles with the same centre O. The inner circle has radius x cm and the outer circle has radius 6 cm. The ring between them is shaded. Not to scale

Write down an expression, in terms of \(\pi\) and \(x\), for the shaded area in cm\(^2\). [2]

Foundation November 2018 Paper 1 Q13

OCRFoundationCurrent spec2 marksCircles & Sectors

13 Calculate the circumference of a circle with diameter 10 cm. [2]

Higher November 2018 Paper 6 Q13

OCRHigherCurrent spec3 marks2D Area & Perimeter Circles & Sectors

13 The shape below is formed from two semicircles and a straight line.

A shaded shape between a large semicircle and a smaller semicircle with the same centre, standing on a common straight line; the diameter of the small semicircle is shown dashed

Not to scale

The radius of the large semicircle is 8 cm.
The radius of the small semicircle is \(t\) cm.

Find an expression, in terms of \(t\), for the exact perimeter of the shaded shape. [3]

Mark scheme

Answer	Marks	Part marks and guidance
\(8\pi + t\pi + 16 - 2t\) oe including unsimplified expressions	3	B2 for two of \(\dfrac{16\pi}{2}\) oe, \(\dfrac{2t\pi}{2}\) oe, or \(16 - 2t\) oe seen or B1 for one of \(\dfrac{16\pi}{2}\) oe, \(\dfrac{2t\pi}{2}\) oe, or \(16 - 2t\) oe seen	Mark final answer Equivalent simplified expressions include: \(\pi(8 + t) + 16 - 2t\) or \(t(\pi - 2) + 8(\pi + 2)\) Penalise use of 3.14 once e.g. 25.1… and 3.14\(t\) scores B1

Higher November 2018 Paper 5 Q12

OCRHigherCurrent spec5 marksCircles & Sectors

12 AOB is a sector of a circle, centre O and radius 6 cm.
The length of arc AB is \(5\pi\) cm.

Sector AOB with OB = 6 cm and an obtuse angle at O

Not to scale

Find the area of the sector.
Give your answer in terms of \(\pi\). [5]

Mark scheme

Answer	Marks	Part marks and guidance
\(15\pi\)	5	B3 for [angle at centre] = 150 or sector is \(\dfrac{5}{12}\) oe of circle or M2 for \(\dfrac{[360 \times]\,5\pi}{2 \times \pi \times 6}\) or better or M1 for \(\left[\dfrac{\theta}{360} \times\right] 2 \times \pi \times 6\) oe M1 for \(\dfrac{\textit{their } \theta}{360} \times \pi \times 6^2\) oe	Final M1 is independent

Foundation June 2018 Paper 2 Q21

OCRFoundationCurrent spec5 marksCircles & Sectors Show That & Proof

21 This shape consists of three semicircles.

Shape made of a large semicircle below the diameter PQ and two small semicircles above it, on PO and OQ, with O the midpoint of PQ

OP = OQ.
The length of PQ is \(4x\) cm.

Show that the area, in cm\(^2\), of the whole shape is \(3\pi x^2\). [5]

Mark scheme

Answer	Marks	Part marks and guidance
Radius C is \(2x\) Or radius A or B is \(x\)	B1		A and B are the small semicircles C is the large semicircle May be indicated on the diagram
Area C = \(\dfrac{\pi \times (2x)^2}{2}\) oe	M1
\(= 2\pi x^2\)	A1
Area A or B = \(\dfrac{\pi \times x^2}{2}\) oe	M1	or Area A + B = \(\pi x^2\) oe	\(\pi x^2\) must result from combining area A and area B
Area = \(2\pi x^2 + \dfrac{\pi x^2}{2} + \dfrac{\pi x^2}{2} = 3\pi x^2\)	A1	or Area = \(2\pi x^2 + \pi x^2 = 3\pi x^2\)	Addition must be seen with no errors or omissions but condone equivalent expressions for \(2\pi x^2\), \(\dfrac{\pi x^2}{2}\), \(\pi x^2\)

Foundation June 2018 Paper 1 Q12

OCRFoundationCurrent spec3 marksCircles & Sectors

12 A circle has radius 6 cm.

Calculate its circumference.
Give your answer in centimetres, correct to 1 decimal place. [3]

Foundation November 2017 Paper 2 Q15

OCRFoundationCurrent spec2 marksCircles & Sectors

15 The diagram shows a circle, centre O.

Circle with centre O. Line A touches the circle at one point. A chord cuts off the shaded region B at the bottom of the circle

Write down the mathematical name of

(a) line A, [1]

(b) shaded region B. [1]

Higher November 2017 Paper 6 Q8

OCRHigherCurrent spec5 marksCircles & Sectors

8 The design below is made from two sectors of circles, centre O.

Two sectors with centre O and angle 45°: the large sector has radius 6 cm; the small unshaded sector is cut out of it, leaving a shaded band 3.5 cm wide along the bottom edge

Calculate the perimeter of the shaded part.
Give your answer correct to 3 significant figures.

........................................ cm [5]

Mark scheme

Answer	Marks	Part marks and guidance
13.7	5	M4 for \(\dfrac{45}{360} \times 2 \times \pi \times 6 + \dfrac{45}{360} \times 2 \times \pi \times 2.5 + 2 \times 3.5\) oe soi	by 13.67… to 13.68
		OR M3 for \(\dfrac{45}{360} \times 2 \times \pi \times 6\) oe and \(\dfrac{45}{360} \times 2 \times \pi \times 2.5\) oe soi	by 4.71… and 1.96… or by 6.67… to 6.68
		OR M2 for \(\dfrac{45}{360} \times 2 \times \pi \times 6\) oe or \(\dfrac{45}{360} \times 2 \times \pi \times 2.5\) oe soi or \(2 \times \pi \times 6 + 2 \times \pi \times 2.5\) oe soi	by 4.71… or 1.96…or 53.4..
		OR M1 for \(2 \times \pi \times 6\) oe or \(2 \times \pi \times 2.5\) oe soi	by 37.699… to 37.7 or 15.7…
		If 0 scored SC2 for \(\dfrac{45}{360} \times \pi \times 6^2\) oe and \(\dfrac{45}{360} \times \pi \times 2.5^2\) oe soi by 14.1… and 2.45… or by 11.6 to 11.7 OR SC1 for \(\dfrac{45}{360} \times \pi \times 6^2\) oe or \(\dfrac{45}{360} \times \pi \times 2.5^2\) oe soi by 14.1… or 2.45…	Method marks may be awarded for multiples of \(\pi\) seen in correct working. Eg. \(\dfrac{45}{360} \times 2 \times \pi \times 6 = \dfrac{3}{2}\pi\) \(\dfrac{45}{360} \times 2 \times \pi \times 2.5 = \dfrac{5}{8}\pi\)

Higher November 2017 Paper 5 Q1

OCRHigherCurrent spec2 marksCircles & Sectors

1 The diagram shows a circle, centre O.

Circle with centre O; a straight line A touches the circle at one point; a chord near the bottom cuts off a shaded region B

Write down the mathematical name of

(a) line A, [1]

(b) shaded region B. [1]

Foundation June 2017 Paper 1 Q21

OCRFoundationCurrent spec5 marks2D Area & Perimeter Circles & Sectors

21 The diagram shows a semi-circle inside a rectangle of length 120 m.
The semi-circle touches the rectangle at A, B and C.

Rectangle with base AC of 120 m and a semi-circle on AC touching the top side at B; the region between the semi-circle and the top left corner of the rectangle is shaded

Not to scale

Calculate the perimeter of the shaded region.
Give your answer correct to 3 significant figures. [5]

Mark scheme

Answer	Marks	Part marks and guidance
214	5	B4 for 214.2 or 214.24 to 214.26 OR B1 for 60 marked or used as width of rectangle or distance from B to the corner AND M2 for \(\dfrac{1}{4} \times \pi \times 120\) soi by \(30\pi\), 94.2 or 94.24 to 94.26 or M1 for \(\pi \times 120\) soi by 376.8 to 377.1 or \(\dfrac{1}{2}\pi \times 120\) soi by 188.4 to 188.6 AND M1 for \(2 \times\) their 60 + their \(30\pi\) AND B1 for their final answer written to more than 3 figs correctly rounded to 3 s.f. to a max. of 4 marks	Accept \(120 + 30\pi\) for B4 Allow e.g. \(r = 60\) for B1

Higher June 2017 Paper 4 Q6

OCRHigherCurrent spec5 marks2D Area & Perimeter Circles & Sectors

6 The diagram shows a semi-circle inside a rectangle of length 120 m.
The semi-circle touches the rectangle at A, B and C.

Rectangle with base AC of 120 m containing a semicircle on AC that touches the top side at B; the region between the arc and the top left corner of the rectangle is shaded

Not to scale

Calculate the perimeter of the shaded region.
Give your answer correct to 3 significant figures. [5]

Mark scheme

Answer	Marks	Part marks and guidance
214	5	B4 for 214.2 or 214.24 to 214.26	Accept \(120 + 30\pi\) for B4
		OR
		B1 for 60 marked or used as width of the rectangle or distance from B to the corner	Allow eg \(r = 60\) for B1
		AND
		M2 for \(\frac{1}{4} \times \pi \times 120\) soi by \(30\pi\), 94.2 or 94.24 to 94.26 or M1 for \(\pi \times 120\) soi by 376.8 to 377.1 or \(\frac{1}{2}\pi \times 120\) soi by 188.4 to 188.56
		AND
		M1 for \(2 \times\) their 60 + their \(30\pi\)
		AND
		B1 for their final answer written to more than 3 figs correctly rounded to 3 s.f.
		to a max. of 4 marks