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In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If $OA = 7\ cm$, find the area of the shaded region.
Given:
AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle.
$OA = 7\ cm$.
To do:
We have to find the area of the shaded region
Solution:
AB and CD are the diameters of a circle with centre O.
Radius of the circle $OA=OB=OC=OD=7\ cm$
Area of the circle with diameter $AB=2\pi r^{2}$
$=2\times \frac{22}{7} \times 7\times 7$
$=308\ cm^{2}$
Another shaded circle with diameter $OB=7\ cm$
Radius of shaded circle$=\frac{7}{2}\ cm$
Area of the shaded circle with diameter $OB=\pi r^{2}$
$=2\times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$=\frac{77}{2} \ cm^{2}$
Area of $\vartriangle ACD=\frac{1}{2} \times base\times height$
$=\frac{1}{2} \times 14\times 7$
$=49\ cm^{2}$
Area of semi-circle with diameter AB,
$=\frac{308}{2}$
$=154\ cm^{2}$
Therefore area of the shaded region$=$Area of the circle with diameter OB$+$Area of the semi-circle-area of $\vartriangle ACD$
$=\frac{77}{2} +154-49$
$=66.5\ cm^{2}$
Therefore, area of the shaded region is $66.5\ cm^{2}$.
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