Question

A trapezium is such that three of its sides have lengths as $6 \, cm,$ then the length (in $cm$ ) of the fourth side such that the area of trapezium is maximum, is

Answer 1

From the figure above,
Area of the trapezium $=6\times 6sin \theta +2\times \frac{1}{2}\times 6cos ⁡ \theta \times 6sin ⁡ \theta $
$A=36sin \theta +36sin ⁡ \theta cos ⁡ \theta $
$A=36sin \theta +18sin ⁡ 2 \theta $
$\frac{d A}{d \theta }=0\Rightarrow 36cos \theta +36cos⁡2\theta =0$
$2cos^{2}\theta +cos \theta -1=0$
$cos \theta =\frac{1}{2}$ or $cos \theta =-1$
for $\theta =\frac{\pi }{3}, \, \frac{d^{2} A}{d \theta ^{2}} < 0$
So, area is maximum when $\theta =\frac{\pi }{3}$
Length of fourth side $=\left(6 c o s \frac{\pi }{3}\right)\times 2+6=12$ $cm$