Three circles each of radius 3.5, cm are draw | vedclass

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Question

Given that, three circles are in such a way that each of them touches the other two.

Now, we join centre of all three circles to each other by a li

Vedclass · Accepted Answer

$1.967$

Vedclass · Answer

Given that, three circles are in such a way that each of them touches the other two.

Now, we join centre of all three circles to each other by a line segment. since, radius of each circle is $3.5\, cm$

So; $AB =2 	imes$ Radius of circle

$=2 	imes 3.5=7 \,cm$

$\Rightarrow \quad A C=B C=A B=7 \,cm$

which shows that, $	riangle A B C$ is an equilateral triangle with side $7\, cm$.

We know that, each angle between two adjacent sides of an equilateral triangle is $60^{\circ}$

$	herefore$ Area of sector with angle $\angle A =60^{\circ}$

$=\frac{\angle A}{360^{\circ}} 	imes \pi r^{2}=\frac{60^{\circ}}{360^{\circ}} 	imes \pi 	imes(3.5)^{2}$

So, area of each sector $=3 	imes$ Area of sector with angle $A$

$=3 	imes \frac{60^{\circ}}{360^{\circ}} 	imes \pi 	imes(35)^{2}$

$=\frac{1}{2} 	imes \frac{22}{7} 	imes 35 	imes 35$

$=11 	imes \frac{5}{10} 	imes \frac{35}{10}=\frac{11}{2} 	imes \frac{7}{2}$

$=\frac{77}{4}=19.25 \,cm ^{2}$

and Area of $	riangle A B C=\frac{\sqrt{3}}{4} 	imes(7)^{2} \quad\left[\becauseight.$ area of an equilateral triangle $\left.=\frac{\sqrt{3}}{4}(	ext { side })^{2}ight]$

$=49 \frac{\sqrt{3}}{4} \,cm ^{2}$

$	herefore$ Area of shaded region enciosed between these circles $=$ Area of $	riangle ABC$ $-$ Area of each sector

$=49 \frac{\sqrt{3}}{4}-19.25=1225 	imes \sqrt{3}-19.25$

$=21.2176-19.25=1.9676 \,cm ^{2}$

Hence, the required area enclosed between these circles is $1.967 \,cm ^{2}$ (approx).

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 circles. (in $cm ^{2}$)

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