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

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(B) Given that,three circles are drawn such that each of them touches the other two.Let the centers of the three circles be $A, B,$ and $C$. Since eac

Vedclass · Accepted Answer

$1.967$

Vedclass · Answer

(B) Given that,three circles are drawn such that each of them touches the other two.Let the centers of the three circles be $A, B,$ and $C$. Since each circle has a radius $r = 3.5\, cm$,the distance between the centers of any two touching circles is $2r = 2 	imes 3.5 = 7\, cm$.Thus,$AB = BC = CA = 7\, cm$. This forms an equilateral triangle $	riangle ABC$ with side length $a = 7\, cm$.The area of the equilateral triangle $	riangle ABC$ is given by:$	ext{Area} = \frac{\sqrt{3}}{4} 	imes a^{2} = \frac{\sqrt{3}}{4} 	imes (7)^{2} = \frac{49\sqrt{3}}{4} \approx \frac{49 	imes 1.732}{4} \approx 21.2176\, cm^{2}$.The area enclosed between the three circles is the area of the triangle minus the area of the three sectors inside the triangle. Each angle of the equilateral triangle is $60^{\circ}$.Total area of the three sectors $= 3 	imes \left( \frac{60^{\circ}}{360^{\circ}} 	imes \pi r^{2} ight) = 3 	imes \frac{1}{6} 	imes \pi 	imes (3.5)^{2} = \frac{1}{2} 	imes \frac{22}{7} 	imes 3.5 	imes 3.5 = 11 	imes 0.5 	imes 3.5 = 19.25\, cm^{2}$.Therefore,the area enclosed between the circles $= 	ext{Area of } 	riangle ABC - 	ext{Total area of the three sectors}$$= 21.2176 - 19.25 = 1.9676\, cm^{2}$.Rounding to three decimal places,the required area is $1.967\, cm^{2}$.

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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