Question

A trapezium $ABCD$ is inscribed into a semi-circle of radius $l$ so that the base $AD$ of the trapezium is diameter and the vertices $B$ and $C$ lie on the circumference. Then the value of base angle $\theta$ (in degree) of the trapezium $ABCD$ which has the greatest perimeter, is

Answer 1

Given, $AD=2l$
From $△ABD,$
$\cos \theta =\frac{AB}{AD}$
$\Rightarrow AB=AD\cos \theta$
$\Rightarrow AB=2l\cos \theta$
and $x=AB\cos \theta =2l{\cos }^2\theta$

Now perimeter, $P=2l+2l-2x+2AB$
$\Rightarrow P=4l-4l{\cos }^2\theta +4l\cos \theta$
$\Rightarrow P=4l(1-{\cos }^2\theta +\cos \theta )$
Differentiating w.r.t. to $\theta ,$ we get
$\frac{dP}{d\theta }=4l(2\cos \theta \sin \theta -\sin \theta )$
For max/min, $\frac{dP}{d\theta }=0$
$\Rightarrow 2\cos \theta \sin \theta -\sin \theta =0$
$\Rightarrow \cos \theta =\frac{1}{2}(\because \theta \ne 0)$
$\Rightarrow \theta ={60}^{\circ }$
$\frac{d^{2}P}{d{\theta }^2}=4l(2\cos 2\theta -\cos \theta )$
$\frac{d^{2}P}{d{\theta }^2}{|}_{\theta ={60}^{\circ }}<0$
$\therefore$ Perimeter is maximum when $\theta ={60}^{\circ }$