The base of a right pyramid is an equilateral | vedclass

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Question

(C) Area of the base $= \frac{\sqrt{3}}{4} 	imes 4^2 = 4 \sqrt{3} 	ext{ cm}^2$.The distance from the centroid of the equilateral triangle to the mid

Vedclass · Accepted Answer

$\frac{8}{9} \sqrt{3}$

Vedclass · Answer

(C) Area of the base $= \frac{\sqrt{3}}{4} 	imes 4^2 = 4 \sqrt{3} 	ext{ cm}^2$.The distance from the centroid of the equilateral triangle to the midpoint of any side is $r = \frac{1}{3} 	imes 	ext{median} = \frac{1}{3} 	imes \frac{\sqrt{3}}{2} 	imes 4 = \frac{2 \sqrt{3}}{3} 	ext{ cm}$.Let the height of the pyramid be $h$ and the slant height be $l$. Given $h = \frac{l}{2}$,so $l = 2h$.In the right-angled triangle formed by the height,the distance $r$,and the slant height $l$,we have $l^2 = h^2 + r^2$.Substituting $l = 2h$,we get $(2h)^2 = h^2 + (\frac{2 \sqrt{3}}{3})^2$.$4h^2 - h^2 = \frac{4 	imes 3}{9} \Rightarrow 3h^2 = \frac{4}{3} \Rightarrow h^2 = \frac{4}{9} \Rightarrow h = \frac{2}{3} 	ext{ cm}$.Volume $= \frac{1}{3} 	imes 	ext{Area of base} 	imes h = \frac{1}{3} 	imes 4 \sqrt{3} 	imes \frac{2}{3} = \frac{8 \sqrt{3}}{9} 	ext{ cm}^3$.

The base of a right pyramid is an equilateral triangle of side $4 \text{ cm}$. The height of the pyramid is half of its slant height. Its volume is ........ $\text{cm}^3$.

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