• # question_answer In Fig. 9, is shown a sector OAP of a circle with centre O, containing $\angle \theta .\text{ }AB$ is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is $r\left[ \tan \,\theta +\sec \,\theta +\frac{\pi \theta }{180}-1 \right]$

 Given, the radius of circle with centre O is r. $\angle POA=\theta$. then, length of the arc $\overset\frown{PA}=\frac{2\pi r\theta }{360{}^\circ }=\frac{\pi r\theta }{180{}^\circ }$ And                              $\tan \theta =\frac{AB}{r}$ $AB=r\,\tan \theta$ And                              $\sec \theta =\frac{OB}{r}$ $OB=r\,\sec \theta$ Now,                             $PB=OB-OP$ $=r\,\sec \,\theta -r$ $\therefore$ Perimeter of shaded region $=AB+PB+\overset\frown{PA}$. $=r\,\tan \,\theta +r\,\sec \,\theta -r+\frac{\pi r\theta }{180{}^\circ }$ $=r\left[ \tan \theta +\sec \theta +\frac{\pi \theta }{180}-1 \right]$                           Hence Proved.