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In figure, is shown a sector $OAP$ of a circle with center O, containing $\angle \theta $. $AB$ is perpendicular to the radius $OQ$ and meets OP produced at B. Prove that the perimeter of shaded region is $r\left( tan\theta +sec\theta +\frac{\pi \theta }{180^{o} } -1\right)$.
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Given: In figure, is shown a sector $OAP$ of a circle with center O, containing $\angle \theta $. $AB$ is perpendicular to the radius $OQ$ and meets OP produced at B.

To do:

Solution:

Perimeter of shaded region $= AB +PB +arc\ length\ AP\ \  \ \ \ .........( 1)$

Arc length of $AP =\frac{\theta }{360^{o}} \times 2\pi r\ \ \ \ \  \ \ \ \ ........( 2)$

In right angled $\vartriangle OAB$,

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

$AB\ =\ r\ tan\ \theta \ \ \ \ \ \ \ \ \ \ ....( 3)$

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

$OB=r\ sec\ \theta$

$OB=OP\ +\ PB$

$\therefore \ r\ sec\ \theta=\ r\ +\ PB$

$\therefore  PB =r\ sec\ \theta \ –\ r\ \ \ \ \ \ .....( 4)$

Substitute $( 2) ,\ ( 3)$ and $( 4)$ in $( 1)$ , we get

Perimeter of shaded region $=\ AB\ +\ PB\ +\ arc\ length AP$

$=rtan\theta +r sec\theta –r+\frac{\theta }{360^{o}} \times 2\pi r$

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

Hence proved.

 

Updated on: 10-Oct-2022

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