Question

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these three circle.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

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

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