A circular garden of radius 10 m is divided into two pa

SOLUTION

In $△ORQ,$
${10}^2$ = ${OR}^2$ + ${RQ}^2$
${10}^2$ = $6^2$ +${RQ}^2$
$\Rightarrow$ ${RQ}^2$ = 64
$\Rightarrow$ RQ = 8 m
$\therefore$ PQ = 2RQ = 16 m  [OR is perpendicular bisector of PQ]

Area of $\Delta$OPQ = $\frac{1}{2}$ × Base × Height  
= $\frac{1}{2}$ × PQ × OR  
= $\frac{1}{2}$ × 16 × 6  
= 48 $m^2$ 

Area of sector OPSQ = $\frac{\angle POQ}{{360}^{\circ }}$ × $\pi \times r^2$
In  ORQ,
cos($\angle$ROQ) =  $\frac{\text{adjacent side}}{\text{hypotenuse}}$
                          = $\frac{OR}{OQ}$
                          = $\frac{6}{10}$
                          = $\frac{3}{5}$
$\Rightarrow$ cos( $\angle$ROQ) = $\frac{3}{5}$
It is given that cos(${53}^{\circ }$) = $\frac{3}{5}$.
Hence  $\angle$ROQ = ${53}^{\circ }$

$\angle$POQ = 2($\angle$ROQ) = ${106}^{\circ }$ 

Area of sector POQS = $\frac{POQ}{360}$ ×$\pi \times r^2$
= $\frac{{106}^{\circ }}{{360}^{\circ }}$ ×$\pi \times {10}^2$ = 92.5 $m^2$

Area of segment PRQS = Area of sector POQS – Area of $△$OPQ  = 92.5 – 48 = 44.5 $m^2$