# In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm.

Question:

In fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

Solution:

(i) Area of the quadrant OACB $\left(\operatorname{rar}^{-} \mathrm{is}=\frac{\mathbf{7}}{\mathbf{2}} \mathbf{c m}\right)$

$=\frac{1}{4} \times \pi \times r^{2}=\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \mathrm{~cm}^{2}$

$=\frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} \mathrm{~cm}^{2}=\frac{11 \times 7}{8} \mathrm{~cm}^{2}$

$=\frac{77}{8} \mathrm{~cm}^{2}$

(ii) In right angled $\Delta \mathrm{OBD}$,

$\mathrm{OB}=\frac{\boldsymbol{7}}{\boldsymbol{2}} \mathrm{cm}, \mathrm{OD}=2 \mathrm{~cm}$

The area of $\Delta \mathrm{OBD}=\frac{\mathbf{1}}{\mathbf{2}} \times \mathrm{OB} \times \mathrm{OD}$

$=\frac{1}{2} \times \frac{7}{2} \times 2 \mathrm{~cm}^{2}=\frac{7}{2} \mathrm{~cm}^{2}$

Then area of the shaded region $=$ The area of

quadrant $\mathrm{OACB}-$ The area of $\Delta \mathrm{OBD}$

$=\frac{77}{8} \mathrm{~cm}^{2}-\frac{7}{2} \mathrm{~cm}^{2}=\frac{77-28}{8} \mathrm{~cm}^{2}=\frac{49}{8} \mathrm{~cm}^{2}$