Higher January 2021 Paper 2 Q22

22 The diagram shows a sector \(OBC\) of a circle with centre \(O\) and radius \((6 + x)\) cm.

Diagram NOT accurately drawn

\(A\) is the point on \(OB\) and \(D\) is the point on \(OC\) such that \(OA = OD\) = 6 cm

Angle \(BOC = 50^\circ\)

Given that

the perimeter of sector \(OBC\) = 2 × the perimeter of triangle \(OAD\)

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

(6)

Mark scheme

Scheme	Marks
eg \((AD =)\sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(50)}\) (= 5.07…) or 2 × 6sin25 (=5.07...) or \(\dfrac{6\sin 50}{\sin 65}\) (= 5.07...) oe	M1
eg \(6 + 6 + \sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(50)}\) or 12 + “5.07…” (= 17.07... or 17.1)	M1
eg (arc \(BC\)=) \(\dfrac{50}{360} \times \pi \times 2 \times (6 + x)\) oe	M1
eg \(2 \times \text{“}17.1\text{”} = 12 + 2x + \dfrac{50}{360} \times \pi \times 2 \times (6 + x)\) oe	M1
eg \(2 \times 17.1 - 12 - \dfrac{30}{18}\pi = 2x + \dfrac{5x}{18}\pi\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 5.89	A1
	(6)
	(6 marks)

M1: Correct expression for \(AD\)
ie \(AD\) = … or \(x\) = oe

M1: A correct statement of perimeter of triangle \(OAD\)
(= 17.07... corrected from the printed mark scheme: “(=17.0)7...”)

M1: A correct statement for arc \(BC\) (condone missing brackets around \((6 + x)\) for this mark only)

M1: dep on M3 for a correct equation for \(x\)

M1: isolating terms in \(x\) in a correct equation

A1: 5.88 – 5.89