Higher June 2024 Paper 2R Q15
15
Diagram NOT accurately drawn
\(AB\), \(BC\) and \(CD\) are three sides of a regular pentagon and \(CDE\) is a triangle.
\(BCE\) is a straight line.
\(CD = 6.5\) cm \(\qquad CE = 3\) cm
Work out the area of triangle \(CDE\)
Give your answer correct to 3 significant figures.
(3)
| Scheme | Marks |
|---|---|
| 360 ÷ 5 (= 72) oe or (5 – 2)×180 ÷ 5 (= 108) oe or 540 ÷ 5 (= 108) oe | M1 |
\(\dfrac{1}{2} \times 6.5 \times 3 \times \sin[\textit{angle } DCE]\) oe or \((h =)\, 6.5 \times \sin[\textit{angle } DCE]\) (= 6.18….) and \(\dfrac{1}{2} \times 3 \times \text{``}{6.18...}\text{''}\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 9.27 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a method to find an exterior or interior angle for a regular pentagon
Do not award this mark if 108 is assigned as an exterior angle or 72 is assigned as an interior angle
Ignore angles on the diagram other than exterior/interior angles of the pentagon even if incorrectly labelled
M1: ft their angle \(DCE\) when substituting in
\(\dfrac{1}{2} \times 6.5 \times 3 \times \sin[\textit{angle } DCE]\)
[angle DCE] means their angle \(DCE\) provided it is less than 90°
A1: accept 9.26 – 9.28
SC B2 for
\(\dfrac{1}{2} \times 6.5 \times 3 \times \sin \text{``}{108}\text{''} = 9.27\ldots\)