In the given diagram find the area of segment PRQS Side

SOLUTION

$O{Q}^2$ = $O{S}^2$ + $S{Q}^2$       (Using Pythagoras Theorem)

$O{Q}^2$ = $a^2$ + $b^2$

OQ is the radius of the circle.

=> OQ=$\sqrt{(a^2+b^2}$

Area of $\Delta$OPQ =  $\left(\frac{1}{2}\right)\times Base\times Height$

                                       = $\left(\frac{1}{2}\right)\times PQ\times OS$

                                        = $\frac{1}{2}\times 2b\times a$= (ab)  $\therefore$ PQ = 2SQ = 2b cm (OS is perpendicular bisector of PQ))

Area of sector POQR =$\frac{\angle POQ}{{360}^{\circ }}\times \pi r^2$  ($\angle POQ=2\left(\angle SOQ\right)=2\theta )$

                                        = $\left(\frac{2\theta }{{360}^{\circ }}\right)\times \pi (a^2+b^2)$  (Since $r^2=a^2+b^2$ )

Area of segment PRQS= Area of sector POQR - Area of triangle OPQ =  $\left(\frac{\theta }{{180}^{\circ }}\right)\times \pi (a^2+b^2)$ -(ab)

Also we know area of  circle = $\pi r^2$ 

                                                   = $\pi (a^2+b^2)$ 

                                                     = A(given)

Therefore;Area of segment PRQS =  $\left(\frac{\theta }{{180}^{\circ }}\right)\times A$ - (ab)