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In fig. 4, the boundary of shaded region consists of four semicircular arcs, two being equals. If diameter of the largest is 14cm and that of the smallest is 3.5 cm, calculate the area of shaded region.$[Use\ \pi=\frac{22}{7}]$
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Given: A boundary region consisting of four semi-circular arcs.
Diameter of the largest arc $14\ cm$ and of the smallest arc $3.5\ cm$.

What to do: To find out the area of shaded region. 

Solution:
Here given diameter of the largest semi-circle $AB=14\ cm$

Diameter of the smallest semi-circle $AC=DB=3.5\ cm$

Diameter of the middle semi-circle $CD=AB-( AC+BD)$

$=14-( 3.5+3.5)$

$=7\ cm$
Area of largest semi-circle$A_{1} =\frac{\pi }{2}\left(\frac{AB}{2}\right)^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times \left(\frac{14}{2}\right)^{2}$

$=77\ cm^{2}$

Area of middle semi-circle $A_{2} =\frac{\pi }{2}\left(\frac{CD}{2}\right)^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times \left(\frac{3.5}{2}\right)^{2}$

$=19.25\ cm^{2}$

Area of both smallest semi circles $A_{3} =\frac{2\pi }{2} \times \left(\frac{3.5}{2}\right)^{2}$

$=\frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}$

$=9.625\ cm^{2}$

Area of the shaded region $A=( A_{1} +A_{2}) -A^{3}$

$=77+19.25-9.625$

$ =86.625\ cm^{2}$

Thus, the area of the shaded region is $86.625\ cm^{2}$. 

Updated on: 10-Oct-2022

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