Higher June 2025 Paper 2R Q11

11 The diagram shows two triangles, \(ADE\) and \(CDB\)

Diagram NOT accurately drawn

\(ABD\) is a straight line.

\(AE = 28\) cm \(ED = 45\) cm \(AB = 21\) cm \(CD = 35\) cm

angle \(AED\) = angle \(CBD = 90^\circ\)

Work out the area of triangle \(CDB\)
Give your answer correct to 3 significant figures.

(5)

Mark scheme

Scheme	Marks
\((AD^2 =)\;28^2 + 45^2\;(= 784 + 2025 = 2809)\) or \((EDA =)\;\tan^{-1}\left(\dfrac{28}{45}\right)\;(= 31.8(9\ldots))\) or \((EAD =)\;\tan^{-1}\left(\dfrac{45}{28}\right)\;(= 58.1(09\ldots))\)	M1
eg \((AD =)\sqrt{28^2 + 45^2}\;\left(= \sqrt{784 + 2025} = \sqrt{2809} = 53\right)\) oe or \((AD =)\;\dfrac{28}{\sin 31.8(9\ldots)}\;(= 53\ldots)\) or \((AD =)\;\dfrac{45}{\cos 31.8(9\ldots)}\;(= 53\ldots)\) oe or \((AD =)\;\dfrac{45}{\sin 58.1\ldots}\;(= 53\ldots)\) or \((AD =)\;\dfrac{28}{\cos 58.1\ldots}\;(= 53\ldots)\) oe	M1
eg \((BC =)\sqrt{35^2 - (\text{``}{53}\text{''} - 21)^2}\;\left(= \sqrt{1225 - 1024} = \sqrt{201} = 14.1(7\ldots)\right)\) or \((BDC =)\;\cos^{-1}\left(\dfrac{\text{``}{53}\text{''} - 21}{35}\right)\;(= 23.8(9\ldots))\) or \((BCD =)\;\sin^{-1}\left(\dfrac{\text{``}{53}\text{''} - 21}{35}\right)\;(= 66.1\ldots)\)	M1
eg \(\dfrac{1}{2} \times \text{``}{14.1(7\ldots)}\text{''} \times (\text{``}{53}\text{''} - 21)\;(= 16\sqrt{201})\) or \(\dfrac{1}{2} \times 35 \times (\text{``}{53}\text{''} - 21) \times \sin(\text{``}{23.8(9\ldots)}\text{''})\) or \(\dfrac{1}{2} \times 35 \times \text{``}{14.1(7\ldots)}\text{''} \times \sin(\text{``}{66.1\ldots}\text{''})\)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 227	A1
	(5)
	(5 marks)

M1: for a correct method to find \(AD^2\) or angle \(EDA\) or angle \(EAD\)

M1: for a correct method to find \(AD\)

M1: for correct method to find \(BC\) or angle \(BDC\) or angle \(BCD\)

M1: for a correct method to find the area of triangle \(CDB\)

A1: awrt 227

accept \(16\sqrt{201}\)