Question

In the given diagram, find the area of segment PRQS. Sides OS and SQ have lengths a and b respectively. Let the area of circle be A.

• A. $\left(\frac{\theta }{{180}^{\circ }}\right)\times A$ - ab
• B. $\left(\frac{2\theta }{{180}^{\circ }}\right)\times A$ - 2ab
• C. $\left(\frac{\theta }{{360}^{\circ }}\right)\times A$ -(ab)
• D. $\left(\frac{2\theta }{{360}^{\circ }}\right)\times A$ - $\frac{1}{2}$ (ab)

Answer 1

The correct option is A

$\left(\frac{\theta }{{180}^{\circ }}\right)\times A$ - ab

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