Question

In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to $30$. In order to maximise the area of the trapezium, the smallest angle should be

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

Given $ABCD$ is trapezium
$AD=BC=CD=30$
Let the smallest angle be $\theta$.

$\angle DAB=\theta$
In $\Delta AMD$,
$\cos \theta =\frac{AM}{30}$
$\Rightarrow AM=30\cos \theta$
$\sin \theta =\frac{DM}{30}=DM=30\sin \theta$
Area of trapezium $=\frac{1}{2}(AB+CD)DM$
$\Rightarrow A=\frac{1}{2}(60+60\cos \theta )30\sin \theta$
$\Rightarrow A=900(\sin \theta +\sin \theta \cos \theta )$
$\Rightarrow \frac{dA}{d\theta }=900(\cos \theta -{\sin }^2\theta +{\cos }^2\theta )$
For maximum or minimum, put $\frac{dA}{d\theta }=0$
$\therefore \cos \theta -{\sin }^2\theta +{\cos }^2\theta =0$
$\Rightarrow \cos \theta +\cos 2\theta =0$
$\Rightarrow 2\cos \frac{3\theta }{2}\cos \frac{\theta }{2}=0$
$\Rightarrow \frac{3\theta }{2}=\frac{\pi }{2}$ or $\frac{\theta }{2}=\frac{\pi }{2}$
$\Rightarrow \theta =\frac{\pi }{3}$ or $\theta =\pi$
For maximum $\theta =\frac{\pi }{3}$