Higher June 2021 Paper 2 Q18

Diagram NOT accurately drawn

\(ABCD\) is a quadrilateral where \(A\), \(B\), \(C\) and \(D\) are points on a circle.

\(AB = 8\) cm
\(BC = 7.5\) cm
Angle \(ABC = 98°\)
Angle \(ACD = 35°\)

Work out the perimeter of quadrilateral \(ABCD\).
Give your answer correct to one decimal place.

(6)

Mark scheme

Scheme	Marks
\([ADC =]\ 180 - 98\ (= 82)\)	M1
\([AC^2 =]\ 8^2 + 7.5^2 - 2 \times 8 \times 7.5 \times \cos(98)\ (= 136.95\ldots)\)	M1
\([AC =]\ \sqrt{\text{``}{136.95}\text{''}}\) or \(\sqrt{64 + 56.25 + 16.7\ldots}\ (= 11.7\ldots)\) oe	M1
eg \([AD =]\ \dfrac{\text{``}{11.7}\text{''} \sin 35}{\sin\text{``}{82}\text{''}}\ (= 6.77\ldots)\) or \([DC =]\ \dfrac{\text{``}{11.7}\text{''} \times \sin\text{``}{63}\text{''}}{\sin\text{``}{82}\text{''}}\ (= 10.5\ldots)\) oe (where “82” = 180 − 98, “63” = 180 − “82” − 35)	M1
eg \([AD =]\ \dfrac{\text{``}{11.7}\text{''} \sin 35}{\sin\text{``}{82}\text{''}}\) and \([DC =]\ \dfrac{\text{``}{11.7}\text{''} \sin\text{``}{63}\text{''}}{\sin\text{``}{82}\text{''}}\) oe or \([AD =]\ \dfrac{\text{``}{11.7}\text{''} \sin 35}{\sin\text{``}{82}\text{''}}\) and \([DC =]\ \sqrt{\text{``}{11.7}\text{''}^2 + \text{``}{6.77}\text{''}^2 - 2 \times \text{``}{11.7}\text{''} \times \text{``}{6.77}\text{''} \times \cos\text{``}{63}\text{''}}\) \([DC =]\ \dfrac{\text{``}{11.7}\text{''} \sin\text{``}{63}\text{''}}{\sin\text{``}{82}\text{''}}\) and \([AD =]\ \sqrt{\text{``}{11.7}\text{''}^2 + \text{``}{10.5}\text{''}^2 - 2 \times \text{``}{11.7}\text{''} \times \text{``}{10.5}\text{''} \times \cos 35}\) Where “63” = 180 − “82” − 35	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 32.8	A1
	(6)
	(6 marks)

M1: may be seen on diagram

M1: correct equation for \(AC\) or \(AC^2\)

M1: complete method to find \(AC\) showing correct order of operations

M1: correct calculation for \(AD\) or \(DC\)
dep on 1^st M1 and 2^nd M1

M1: correct calculations for \(AD\) and \(DC\)
(\(AD = 6.77\ldots\)
\(DC = 10.5\ldots\))
dep on 1^st M1 and 2^nd M1

A1: accept 32.7 – 32.9