Math, asked by AintRude, 12 days ago

\textsf{\large{\underline {Question}:}}

Three circles of radius 3 cm are drawn such that each of them touches the other two. The area enclosed between these circles is approximately?


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Answered by anshdiyasingh12
1

Three circles of radius 3 cm are drawn such that each of them touches the other two. The area enclosed between these circles is approximately?

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Answered by OoAryanKingoO78
2

Answer:

The area enclosed between these three circle is \bold{1.967\ \mathrm{cm}^{2}}.

The centre of all the three circles is joined by means of a line segment.  

We know that the radius of each circle is 3.5 cm.  

Therefore,

A B=2 \times \text { Radius of circle }=2 \times 3.5=7 \mathrm{cm}

\Rightarrow A C=B C=A B=7 \mathrm{cm}

This shows that, ΔABC is found to be an equilateral triangle whose side is 7 cm.  

It is clearly known that the angle between two adjacent sides of an equilateral triangle is 60°  

∴ Area of sector with angle \angle A=60^{\circ}.

Now,

Required area \begin{array}{l}{=(\text { area of an equilateral } \Delta \text { of side } 7 \mathrm{cm})-(3 \times} \\ {\text { area of a sector with } \theta=60^{\circ} \text { and } r=3.5 \mathrm{cm} ) ]}\end{array}

=\left(\frac{\sqrt{3}}{4} \times 7 \times 7\right)-\left(3 \times \frac{22}{7} \times 3.5 \times 3.5 \times \frac{60}{360}\right) \mathrm{cm}^{2}

\begin{array}{l}{=\left(\frac{49 \sqrt{3}}{4}-11 \times 0.5 \times 3.5\right)\ c m^{2}} \\ {=(22.217-19.25)\ \mathrm{cm}^{2}} \\ \bold{=1.967\ \mathrm{cm}^{2}}\end{array}

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