A circle is inscribed in an equilateral trian | vedclass

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(B) The side length of the equilateral triangle is $a = 8 	ext{ cm}$.The inradius $r$ of an equilateral triangle is given by the formula $r = \frac{a

Vedclass · Accepted Answer

$10.95$

Vedclass · Answer

(B) The side length of the equilateral triangle is $a = 8 	ext{ cm}$.The inradius $r$ of an equilateral triangle is given by the formula $r = \frac{a}{2\sqrt{3}}$.Substituting the value of $a$,we get $r = \frac{8}{2\sqrt{3}} = \frac{4}{\sqrt{3}} 	ext{ cm}$.The area of the incircle is $A_c = \pi r^2 = \pi \left( \frac{4}{\sqrt{3}} ight)^2 = \pi 	imes \frac{16}{3} 	ext{ cm}^2$.The area of the equilateral triangle is $A_t = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (8)^2 = \frac{\sqrt{3}}{4} 	imes 64 = 16\sqrt{3} 	ext{ cm}^2$.The area of the portion between the triangle and the circle is $A = A_t - A_c = 16\sqrt{3} - \frac{16\pi}{3}$.Using $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14159$:$A = 16(1.732) - \frac{16(3.14159)}{3} \approx 27.712 - 16.755 = 10.957 	ext{ cm}^2$.Rounding to two decimal places,the area is approximately $10.95 	ext{ cm}^2$.

$A$ circle is inscribed in an equilateral triangle of side $8 \text{ cm}$. The area of the portion between the triangle and the circle is ...... $\text{cm}^2$.

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