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 maximum, what is the length of the fourth side?

• A. $8cm$
• B. $9cm$
• C. $10cm$
• D. $12cm$

Answer 1

The correct option is D $12cm$

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.

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

fourth side length=$6+2x=12$