The lateral edge of a regular hexagonal pyram | vedclass

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(C) Let $x$ be the side length of the regular hexagonal base and $h$ be the height of the pyramid. The lateral edge $l$ is given as $1 	ext{ cm}$.Fro

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) Let $x$ be the side length of the regular hexagonal base and $h$ be the height of the pyramid. The lateral edge $l$ is given as $1 	ext{ cm}$.From the geometry of the pyramid,the distance from the center of the hexagon to a vertex is $x$. Thus,by the Pythagorean theorem,$x^2 + h^2 = l^2 = 1^2 = 1$,so $x^2 = 1 - h^2$.The area of the regular hexagonal base is $A = 6 	imes \frac{\sqrt{3}}{4} x^2 = \frac{3\sqrt{3}}{2} x^2$.The volume $V$ of the pyramid is $V = \frac{1}{3} A h = \frac{1}{3} \left( \frac{3\sqrt{3}}{2} x^2 ight) h = \frac{\sqrt{3}}{2} x^2 h$.Substituting $x^2 = 1 - h^2$,we get $V(h) = \frac{\sqrt{3}}{2} (1 - h^2) h = \frac{\sqrt{3}}{2} (h - h^3)$.To find the maximum volume,differentiate $V$ with respect to $h$ and set it to zero:$V'(h) = \frac{\sqrt{3}}{2} (1 - 3h^2) = 0$.$1 - 3h^2 = 0 \implies h^2 = \frac{1}{3} \implies h = \frac{1}{\sqrt{3}}$.Thus,the height for maximum volume is $\frac{1}{\sqrt{3}} 	ext{ cm}$.

The lateral edge of a regular hexagonal pyramid is $1 \text{ cm}$. If the volume is maximum,then its height must be equal to

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