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Question

(A) The centers of the three circles form an equilateral triangle with side length $s = 21 + 21 = 42 \, cm$.The area of an equilateral triangle is giv

Vedclass · Accepted Answer

$441 \sqrt{3} - 693$

Vedclass · Answer

(A) The centers of the three circles form an equilateral triangle with side length $s = 21 + 21 = 42 \, cm$.The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} 	imes s^2 = \frac{\sqrt{3}}{4} 	imes 42^2 = \frac{\sqrt{3}}{4} 	imes 1764 = 441 \sqrt{3} \, sq \cdot cm$.Each interior angle of the equilateral triangle is $60^\circ$.The area of the three sectors inside the triangle is $3 	imes (\frac{60}{360} 	imes \pi 	imes r^2) = 3 	imes (\frac{1}{6} 	imes \frac{22}{7} 	imes 21^2) = \frac{1}{2} 	imes \frac{22}{7} 	imes 441 = 11 	imes 63 = 693 \, sq \cdot cm$.The area of the portion enclosed by the three circles is the area of the triangle minus the area of the three sectors: $441 \sqrt{3} - 693 \, sq \cdot cm$.

Three circles of radius $21 \, cm$ are placed in such a way that each circle touches the other two. What is the area of the portion enclosed by the three circles? (in $sq \cdot cm$)

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