The area of the circle inscribed in an equila | vedclass

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Question

(C) For an equilateral triangle with side $a = 24 \, cm$,the height $h$ is given by $h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} 	imes 24 = 12 \sqr

Vedclass · Accepted Answer

$48 \pi \, cm^2$

Vedclass · Answer

(C) For an equilateral triangle with side $a = 24 \, cm$,the height $h$ is given by $h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} 	imes 24 = 12 \sqrt{3} \, cm$.The radius $r$ of the inscribed circle (inradius) of an equilateral triangle is given by $r = \frac{h}{3}$.Substituting the value of $h$,we get $r = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3} \, cm$.The area of the circle is given by $A = \pi r^2$.Substituting the value of $r$,$A = \pi (4 \sqrt{3})^2 = \pi (16 	imes 3) = 48 \pi \, cm^2$.

The area of the circle inscribed in an equilateral triangle of side $24 \, cm$ is

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