Question

Figure shows a sector of a circle, centre O, containing an angle $\theta {}^{\circ }$. Prove that:

(i) Perimeter of the shaded region is $r(tan\theta +sec\theta +\frac{\pi \theta }{180}-1)$

(ii) Area of the shaded region is $\frac{r^2}{2}(tan\theta -\frac{\pi \theta }{180})$

Answer 1

(i) Length of arc AC $=2\pi r\times \frac{\theta }{{360}^o}=\frac{\pi r\theta }{180}$

Now, in $△$ OAB, $tan\theta =\frac{AB}{OA}$

$\Rightarrow tan\theta =\frac{AB}{r}$

$\Rightarrow AB=rtan\theta$

Also, $sec\theta =\frac{OB}{OA}=\frac{OB}{r}$

$\Rightarrow OB=rsec\theta$

$\Rightarrow BC=OB-OC=rsec\theta -r$

Now, perimeter of shaded region = arc AC + AB + BC

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

$=r(tan\theta +sec\theta +\frac{\pi \theta }{180}-1)$

Hence, proved.

(ii) Area of shaded region = Area of triangle OAB - Area of sector OAC

$=\frac{1}{2}\times OA\times AB-\pi r^2\times \frac{\theta }{{360}^o}$

$=\frac{1}{2}\times r\times rtan\theta -\pi r^2\times \frac{\theta }{{360}^o}$

$=\frac{r^2}{2}(tan\theta -\frac{\pi \theta }{180})$

Hence, proved.