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Question 31 : A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of

the remaining portion of the triangle?

- 144√3 - 48π cm
^{2} - 121√3 - 36π cm
^{2} - 144√3 - 36π cm
^{2} - 121√3 - 48π cm
^{2}

\(144 \sqrt{3}-48 \pi \mathrm{cm}^{2}\)

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When a circle is inscribed in an equilateral triangle, the radius of the circle will be \\frac{1}{3}\\) rd of the height of the triangle.

Height of an equilateral triangle(h) = \\frac{√3}{2}\\) * a

(where a is the length of a side of the equilateral triangle)

h = √\\frac{√3}{2}\\) * 24 = 12 √3.

Radius of the Circle(r) = \\frac{1}{3}\\) * h = 12 \\frac{√3}{3}\\) = 4 √3.

Area of an equilateral triangle (A) = \\frac{√3}{4}\\) * a^{2}

(where a is the length of a side of the equilateral triangle)

A = \\frac{√3}{4}\\) * (24)^{2} = √3 * 144.

Area of the Circle (a) = π r^{2}.

(where r is the radius of the circle)

a = π (4 √3)^{2} = 48 π.

Area of the remaining portion of the triangle= A – a = √3 * 144 - 48 π

Area of the remaining portion of the triangle= 144 √3 - 48 π.

The question is **"A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area ofthe remaining portion of the triangle?" **

Choice A is the correct answer

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