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In figure 6, three circles each of radius 3.5 cm are drawn in such a way each of them touches the other two. Find the area enclosed between these three circles $(shaded\ \ region)$. $\left[ use\ \pi =\frac{22}{7}\right] \ $"

Given: Three circles with 3.5 cm radius each, in such a way that each of them touches the another two.

To do: To find the area of the shaded region.

Solution: 
Here as given in the question radius of each circle, r=3.5cm

And let us say that the circles touches each other on P, Q and R.

A, B and C are the centres of the circles.

And it is clear that ∠ABC=∠BCA=∠CAB=60°

$\therefore$ area of the sector PCQ$=\frac{\theta }{360^{o}} \times \pi r^{2}$

$=\frac{60^{o} }{360^{o} } \times \frac{22}{7} \times 3.5\times 3.5$

$=\frac{1}{6} \times \frac{22}{7} \times 3.5\times 3.5$

$=\frac{77}{12} \ cm^{2}$

Similiarly area of sector PBR and RAQ is $\frac{77}{12} \ cm^{2}$

Area of $\vartriangle ABC$

$=\frac{1}{2} \times AB\times BC\times sin60^{o}$

$=\frac{1}{2} \times 7\times 7\times \frac{\sqrt{3}}{2}$

$=\frac{\left( 49\sqrt{3}\right)}{4}$

Area of the shaded region$=$area of the triangle$-3( area\ of\ the\ sector\ PBR)$

$=\frac{\left( 49\sqrt{3}\right)}{4} -3\times \frac{77}{12} =1.96\ cm^{2}$
 

Updated on: 10-Oct-2022

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