In fig., AB and CD are two diameters of a circle O, which are perpendicular to each other. OB is the diameter of the smaller circle. If $OA=7\ cm$, find the area of the shaded region.[use $\pi =\frac{22}{7}$].

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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Updated on: 10-Oct-2022

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