Question

In fig.3, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20cm, find the area of the shaded region (Use $\pi =3.14$)

Answer 1

Let us join OB.

In $\Delta$ OAB :
$O{B}^2=O{A}^2+A{B}^2=(20{)}^2+(20{)}^2=2\times (20{)}^2$
$\Rightarrow OB=20\sqrt{2}$
Radius of the circle, $r=20\sqrt{2}cm$
Area of quadrant OPBQ $=\frac{\theta }{{360}^{\circ }}\times \pi r^2$
$=\frac{{90}^{\circ }}{{360}^{\circ }}\times 3.14\times (20\sqrt{2}{)}^{2}c{m}^2$

= $\frac{1}{4}$ × 3.14 ×800 $c{m}^2$

= 628 $c{m}^2$

Area of square OABC = $(Side{)}^2$ = $(20{)}^{2}c{m}^2$ = 400 $c{m}^2$

∴ Area of the shaded region = Area of quadrant OPBQ – Area of square OABC

= (628 - 400) $c{m}^2$

= 228 $c{m}^2$