In an equilateral triangle of side 24 cm,a c | vedclass

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(A) The side of the equilateral triangle is $a = 24 \ cm$.The in-radius $r$ of an equilateral triangle is given by $r = \frac{a}{2\sqrt{3}} = \frac{24

Vedclass · Accepted Answer

$98.55$

Vedclass · Answer

(A) The side of the equilateral triangle is $a = 24 \ cm$.The in-radius $r$ of an equilateral triangle is given by $r = \frac{a}{2\sqrt{3}} = \frac{24}{2\sqrt{3}} = 4\sqrt{3} \ cm$.The area of the equilateral triangle is $A_t = \frac{\sqrt{3}}{4} 	imes a^2 = \frac{\sqrt{3}}{4} 	imes 24 	imes 24 = 144\sqrt{3} \ cm^2$.Using $\sqrt{3} = 1.732$,$A_t = 144 	imes 1.732 = 249.408 \ cm^2$.The area of the inscribed circle is $A_c = \pi r^2 = \frac{22}{7} 	imes (4\sqrt{3})^2 = \frac{22}{7} 	imes 48 = \frac{1056}{7} \approx 150.857 \ cm^2$.The area of the remaining portion is $A_t - A_c = 249.408 - 150.857 = 98.551 \ cm^2$.Rounding to two decimal places,the area is $98.55 \ cm^2$.

In an equilateral triangle of side $24 \ cm$,a circle is inscribed touching its sides. The area of the remaining portion of the triangle is (in $cm^2$) $(\sqrt{3} = 1.732)$

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