Question

In the figure, $PQRS$ is a square with side $10cm$. The sectors drawn with P and R as centres from the shaded figure. Find the area of the shaded figure $(\pi =3.14)$

Answer 1

R.E.F image

Solution: Draw hypotenuse $SQ$

$A(△PSQ)=\frac{1}{2}\times$ multiplication of perpendicular sides

$=\frac{1}{2}\times PS\times PQ$

$=\frac{1}{2}\times 10\times 10=50sqcm.$

$(P-SXQ)=\frac{\theta }{360}\times \pi r^2$

$=\frac{90}{360}\times 3.14\times 10\times 10$

$=\frac{314}{4}$

$=78.5sqcm$

Area of segment $SXQ=A(P-SXQ)-A$ $(△PSQ)$

$=78.5-50$

$=28.5Sq.cm$

Similarly,area of segment $SYQ=28.5Ssq.cm$

$\therefore$ Area of shaded figure

=Area of segment $SXQ$ $+$ Area of segment $SYQ$

$=28.5+28.5$

$=57sq.cm$

Ans. The area of the shaded figure is $57sq.cm.$