Question

In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB
(ii) shaded region.

Answer 1

Radius of the quadrant $=3.5cm=\frac{7}{2}cm$
Area of quadrant = $\frac{\left(\pi R^2\right)}{4}c{m}^2$
$=\frac{(\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2})}{4}c{m}^2$
$=\frac{77}{8}c{m}^2$

(ii) Radius of the quadrant
$=3.5cm=\frac{7}{2}cm$

Area of $\Delta BDO$
$=\frac{1}{2}\times \left(base\right)\times \left(height\right)$

Area of triangle BOD
$=\frac{1}{2}\times \frac{7}{2}\times 2c{m}^2$
$=\frac{7}{2}c{m}^2$

Area of shaded region = Area of quadrant - Area of triangle BOD
$=(\frac{77}{8}-\frac{7}{2})c{m}^2=\frac{49}{8}c{m}^2$
$=6.125c{m}^2$