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
$AD = B C = CD = 30$
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 $\theta = \frac{\pi}{3}$