Question

Three sides of a trapezium are each equal to $6cm$. Let $\alpha \in (0,\frac{\pi }{2})$ be the angle between a pair of adjacent sides. If the area of the trapezium is the maximum possible, then what is $\alpha$ equal to?

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{5}$

Answer 1

The correct option is C $\frac{\pi }{3}$

Assume the values as shown in the figure.

Area of the trapezium $=A=\frac{1}{2}(6+6+2x)\sqrt{36-x^2}=(6+x)\sqrt{36-x^2}$

Differentiating wrt $x$, we have

$\frac{dA}{dx}=(6+x)\frac{-2x}{2\sqrt{36-x^2}}+\sqrt{36-x^2}=0⟹x=3$

Also, at $x=3$, $\frac{d^{2}A}{d{x}^2}<0⟹$ maxima.

$\cos \alpha =\frac{x}{6}=\frac{1}{2}⟹\alpha =\frac{\pi }{3}$