A wire when bent in the form of equilateral triangle encloses a area of 121 under root 3 cm if the wire is bent in the form of a circle find the area enclosed by a circle
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area of the equilateral triangle =121√3cm²
area of a equilateral triangle = √3a²/4 Unit²
therefore,121√3=√3a²/4
=a²=484cm²
a=22cm
now parameter of the triangle
=22cm*3=66cm
again,
parameter of the circle formed =parameter of the equilateral triangle
66=2πr
(66*7/(2*22)=r
21/2=r
now area of the circle=πr²=(22/7)*(21/2)*(21/2)
=693/2cm²
hope it helps
if any doubts feel free to ask
if u feel like the answer was adequate then give a brainlist answer..thanku
area of a equilateral triangle = √3a²/4 Unit²
therefore,121√3=√3a²/4
=a²=484cm²
a=22cm
now parameter of the triangle
=22cm*3=66cm
again,
parameter of the circle formed =parameter of the equilateral triangle
66=2πr
(66*7/(2*22)=r
21/2=r
now area of the circle=πr²=(22/7)*(21/2)*(21/2)
=693/2cm²
hope it helps
if any doubts feel free to ask
if u feel like the answer was adequate then give a brainlist answer..thanku
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