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 maximize 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

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

Area of trapezium $=\frac{1}{2}\left(\text{sum of parallel sides}\right)\left(\text{distance between parallel sides}\right)$

For the given trapezium, area $=A=\frac{1}{2}(a+2a\text{cos}\theta +a)\left(a\text{sin}\theta \right)$ where $a=30$

$A=a^2(\text{cos}\theta +1)\left(\text{sin}\theta \right)=a^2(\frac{\text{sin}2\theta }{2}+\text{sin}\theta )$

To maximize the area $A,$

$\frac{dA}{d\theta }=0$

$\Rightarrow \text{cos}2\theta +\text{cos}\theta =0$

$\Rightarrow 2{\text{cos}}^2\theta +\text{cos}\theta -1=0$

$\Rightarrow \text{cos}\theta =\frac{1}{2},-1$

$\Rightarrow \theta =\frac{\pi }{3},\pi$

We can reject $\theta =\pi$ as the smallest angle of the trapezium must be less than $\pi$

So, the smallest angle is $\frac{\pi }{3}$