Question
The diagram below shows a plan for a window in the shape of a trapezium.
Three sides of the window are \(2{\text{ m}}\) long. The angle between the sloping sides of the window and the base is \(\theta \) , where \(0 < \theta < \frac{\pi }{2}\) .
Show that the area of the window is given by \(y = 4\sin \theta + 2\sin 2\theta \) .
Zoe wants a window to have an area of \(5{\text{ }}{{\text{m}}^2}\). Find the two possible values of \(\theta \) .
John wants two windows which have the same area A but different values of \(\theta \) .
Find all possible values for A .
Answer/Explanation
Markscheme
evidence of finding height, h (A1)
e.g. \(\sin \theta = \frac{h}{2}\) , \(2\sin \theta \)
evidence of finding base of triangle, b (A1)
e.g. \(\cos \theta = \frac{b}{2}\) , \(2\cos \theta \)
attempt to substitute valid values into a formula for the area of the window (M1)
e.g. two triangles plus rectangle, trapezium area formula
correct expression (must be in terms of \(\theta \) ) A1
e.g. \(2\left( {\frac{1}{2} \times 2\cos \theta \times 2\sin \theta } \right) + 2 \times 2\sin \theta \) , \(\frac{1}{2}(2\sin \theta )(2 + 2 + 4\cos \theta )\)
attempt to replace \(2\sin \theta \cos \theta \) by \(\sin 2\theta \) M1
e.g. \(4\sin \theta + 2(2\sin \theta \cos \theta )\)
\(y = 4\sin \theta + 2\sin 2\theta \) AG N0
[5 marks]
correct equation A1
e.g. \(y = 5\) , \(4\sin \theta + 2\sin 2\theta = 5\)
evidence of attempt to solve (M1)
e.g. a sketch, \(4\sin \theta + 2\sin \theta – 5 = 0\)
\(\theta = 0.856\) \(({49.0^ \circ })\) , \(\theta = 1.25\) \(({71.4^ \circ })\) A1A1 N3
[4 marks]
recognition that lower area value occurs at \(\theta = \frac{\pi }{2}\) (M1)
finding value of area at \(\theta = \frac{\pi }{2}\) (M1)
e.g. \(4\sin \left( {\frac{\pi }{2}} \right) + 2\sin \left( {2 \times \frac{\pi }{2}} \right)\) , draw square
\(A = 4\) (A1)
recognition that maximum value of y is needed (M1)
\(A = 5.19615 \ldots \) (A1)
\(4 < A < 5.20\) (accept \(4 < A < 5.19\) ) A2 N5
[7 marks]