Circle Theorems: WJEC GCSE Higher November 2022 Unit 2 Q4

Higher November 2022 Unit 2 Q4

WJECHigherCurrent spec9 marksCircle Theorems Circles & Sectors Pythagoras

4. In this question, you will be assessed on the quality of your organisation, communication and accuracy in writing.

\(A\), \(B\) and \(C\) are points on the circumference of a circle with centre \(O\).
The length of \(BC\) is 10 cm.
The diameter of the circle is 18 cm.

Circle centre O with diameter AC = 18 cm, B on the circumference with a right angle at B and BC = 10 cm; the circle is shaded except for triangle ABC

Diagram not drawn to scale

Calculate the shaded area.
You must show all your working. [7 + 2 OCW]

Mark scheme

Scheme	Marks	Comments
(Length of AB) \(18^2 = AB^2 + 10^2\) OR \((AB^2 =)\ 18^2 - 10^2\) or equivalent	M1	note: \((AB^2 =)\ 324 - 100\)
\((AB^2 =)\ 224\)	A1
(\(AB\) =) 14·9(66…) or 15 or \(4\sqrt{14}\) or \(\sqrt{224}\) (cm)	A1	FT √’their 224’ provided M1 gained for M1A0A1. Alternative method to find AB A correct and complete method that would lead to a correct answer (using trigonometric relationships). M2 (AB =) 14·9(66…) or 15 or \(4\sqrt{14}\) or \(\sqrt{224}\) (cm) A1
(Area of the circle =) \(\pi \times 9^2\)	M1
(Area of the triangle = ) \(\tfrac{1}{2} \times 10 \times 14{\cdot}9(66\ldots)\) or equivalent	M1	Award M1 for sight of 75 or \(20\sqrt{14}\). FT ½ × 10 × ‘their \(AB\)’, provided not 18 or 10. Accept any valid method that leads to a correct answer.
254.4(69…) (cm²) or 81π AND 74·8(33..) or \(20\sqrt{14}\) (cm²)	A1	An answer between 254·3 and 254·51 (cm²). Allow 254. Allow 75.
(Area of the shaded region = 254.4(69) – 74·8(33..) =) 179 to 180 (cm²)	B1	Allow rounded or truncated answers. FT ‘their area of a circle’ – ‘their triangle’ (not ‘their \(AB\)’) provided at least one area M1 awarded previously.
Organisation and Communication.	OC1	For OC1, candidates will be expected to: present their response in a structured way explain to the reader what they are doing at each step of their response lay out their explanation and working in a way that is clear and logical write a conclusion that draws together their results and explains what their answer means
Accuracy of writing.	W1	For W1, candidates will be expected to: show all their working make few, if any, errors in spelling, punctuation and grammar use correct mathematical form in their working use appropriate terminology, units, etc

Alternative method to first calculate angle BCA and then use ½absinC to calculate the area of the triangle

Scheme	Marks	Comments
(BCA =) \(\cos^{-1}\dfrac{10}{18}\)	M2	M1 for \(\cos BCA = \dfrac{10}{18}\) (= 0·555..) Note: An alternative correct use of a ‘two-step’ method that would lead to the correct answer for BCA (e.g. finding angle BAC first) is M2. A partial method is M0.
Correct evaluation in the range 56·2 to 56·3	A1	Allow 56 from correct working. Note: cos BCA = 0·55, BCA = 56·632… OR cos BCA = 0·56, BCA = 55·944… is awarded M2A0.
(Area of the triangle = ) \(\tfrac{1}{2} \times 10 \times 18 \times \sin(56{\cdot}251\ldots)\)	M1	FT ‘their BCA’ if previous M1 awarded.
(Area of the circle =) \(\pi \times 9^2\)	M1	An answer between 254·3 and 254·51 (cm²). Allow 254.
254.4(69…) (cm²) or 81π AND 74·8(33..) or \(20\sqrt{14}\) (cm²)	A1	Allow 75.
(Area of the shaded region = 254.4(69) – 74·8(33..) =) 179 to 180 (cm²)	B1	Allow rounded or truncated answers. FT ‘their area of a circle’ – ‘their triangle’ provided at least one area M1 awarded previously.