Question

The above figure shows a sector $OAP$ of a circle with centre $O,$ containing $\angle \theta$. $AB$ is perpendicular to the radius $OA$ and meets $OP$ produced at $B.$ Prove that the perimeter of shaded region is $r[\tan \theta +\sec \theta +\frac{\pi \theta }{180}-1]$.

Answer 1

In $△OAB$,

$\tan \theta =\frac{AB}{OA}$

$OA=r$

$⟹AB=r\tan \theta$

$\cos \theta =\frac{r}{OB}$

$OB=\frac{r}{\cos \theta }=r\sec \theta$

$OP+PB=OB$

Therefore, $BP=r\sec \theta –r$

$arc\left(PA\right)=\pi r.\frac{\theta }{180}$ [Arc length formula]

Therefore required perimeter :
$=AB+BP+arc\left(PA\right)$

$=rtan\theta +\pi r.\frac{\theta }{180}+rsec\theta -r$

$=r[tan\theta +sec\theta +\frac{\pi \theta }{180}-1]$

Hence proved.