The semivertical angle of a cone is $45^{\circ}$. If the height of the cone is $20.025 \mathrm{~cm}$, then the approximate value of its lateral surface area (in sq. $\mathrm{cm}$ ) is

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Aicharya · Accepted Answer

In $	riangle A O B$, $\begin{aligned} & 	an 45^{\circ}=\frac{r}{h} \ & 1=\frac{r}{h} \Rightarrow r=h \end{aligned}$ and $l^2=r^2+h^2$ $\begin{aligned} & l=\sqrt{h^2+h^2} \ & l=\sqrt{2 h^2}=\sqrt{2} h \end{aligned}$ $[\because r=h]$ Lateral surface area $\begin{array}{lll} (S) & =\pi r l & \ & =\pi h \cdot \sqrt{2} h & \ S & =\sqrt{2} \pi h^2 & {[\because h=20.025 \mathrm{~cm}]} \ & =\sqrt{2} \pi(20.025)^2 & \ & =\sqrt{2} \pi 	imes 401 & \ S & =401 \sqrt{2} \pi & \end{array}$ $	herefore$ Hence, option (a) is correct.

The semivertical angle of a cone is $45^{\circ}$. If the height of the cone is $20.025 \mathrm{~cm}$, then…

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