Question

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

Answer 1

Answer: 12

From the figure above,
Area of the trapezium $=6\times 6sin\theta +2\times \frac{1}{2}\times 6cos\theta \times 6sin\theta$
$A=36sin\theta +36sin\theta cos\theta$
$A=36sin\theta +18sin2\theta$
$\frac{dA}{d\theta }=0\Rightarrow 36cos\theta +36cos2\theta =0$
$2co{s}^2\theta +cos\theta -1=0$
$cos\theta =\frac{1}{2}$ 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 $=\left(6cos\frac{\pi }{3}\right)\times 2+6=12$ $cm$