Circles & Sectors: WJEC GCSE Intermediate June 2023 Unit 1 Q15

Intermediate June 2023 Unit 1 Q15

WJECIntermediateCurrent spec5 marksCircles & Sectors Pythagoras

15. The shape below consists of a semicircle attached to one side of a right-angled triangle.
\(A\hat{B}C = 90^\circ\), AB = 8 cm, BC = 6 cm.
BC is the diameter of the semicircle.

Right-angled triangle ABC with the right angle at B, AB = 8 cm horizontal, BC = 6 cm vertical (dashed); a semicircle on diameter BC lies outside the triangle

Diagram not drawn to scale

Calculate the perimeter of the shape.
Use \(\pi = 3{\cdot}14\).
You must show all your working. [5]

Mark scheme

Scheme	Marks	Comments
\((AC^2 =)\ \ 8^2 + 6^2\)	M1	Check diagram. Note: \((AC^2 =)\ \ 64 + 36\)
\((AC =)\ \sqrt{8^2 + 6^2}\) or equivalent	m1	FT from \(\sqrt{\text{‘their }8^2\text{’} + \text{‘their }6^2\text{’}}\)
(AC =) 10 (cm)	A1	CAO. Final answer of AC = 100 is M1m0A0.
(Curved length =) \(\dfrac{6 \times 3{\cdot}14}{2}\) \(\dfrac{2 \times 3 \times 3{\cdot}14}{2}\) 3 × 3·14 \(3\pi\) 9·42 or equivalent	B1	Do not ignore subsequent working e.g. 3 × 3·14 = 9·42, then 9·42 × 2 or 9·42 ÷ 2 would gain B0.
(Perimeter of shape = 8 + 10 + 9·42 = ) 27·42(cm)	A1	Allow 27·4(cm). Award A0 for 18 + \(3\pi\). FT ‘their AC’ only if M1 gained. FT ‘their curved length’ only if B0 awarded and for one of the following: 3·14 used to find the circumference of a circle area of a semicircle used. For example, M1m1A1 awarded for 10 (cm), B1 for 3 × 3·14 = 9 (cm) A0 for 8 + 10 + 9 = 27 (cm) as B1 previously awarded. Note, if a final answer of: 33·4(2)(cm) is given (6cm also included) award M1m1A1B1A0 36·8(4)(cm) is given (full circumference used) award M1m1A1B0A1 32·1(3) (cm) is given (area semicircle used) award M1m1A1B0A1.