Question

In the figure AB and CD are two diameter of a circle with centre O are perpendicular to each other and OD is the diameter of the smaller circle. If OA=7 cm find the area of the shaded region.

Answer 1

Here, area of sector $OBQC=\frac{{90}^o}{{360}^o}\times \pi \times (7{)}^2$

$=\frac{1}{4}\times \frac{22}{7}\times 7\times 7=\frac{77}{2}c{m}^2$

And, Area of $△OBC=\frac{1}{2}\times OC\times OB=\frac{1}{2}\times 7\times 7=\frac{49}{2}c{m}^2$

$\therefore$ Area of the segment $BQC=$ Area of sector $OBQC-$ Area of $△OBC$

$=\frac{77}{2}-\frac{49}{2}=\frac{28}{2}=14c{m}^2$

Similarly, area of segment $APC=14c{m}^2$

Now, the area of the circle with $OD$ as diameter $=\pi r^2=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=\frac{77}{2}c{m}^2$

Hence, the total area of the shaded region

$=(14+14+\frac{77}{2})c{m}^2=(28+\frac{77}{2})c{m}^2$

$=\left(\frac{56+77}{2}\right)c{m}^2=\frac{133}{2}c{m}^2=66.5c{m}^2$