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(B) The volume of a right circular cone is given by $V = \frac{1}{3} \pi r^{2} h$.The lateral surface area $(LSA)$ of a right circular cone is given b

Vedclass · Accepted Answer

$\frac{1}{9}$

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(B) The volume of a right circular cone is given by $V = \frac{1}{3} \pi r^{2} h$.The lateral surface area $(LSA)$ of a right circular cone is given by $LSA = \pi r l$,where $l = \sqrt{r^{2} + h^{2}}$ is the slant height.According to the problem,the numerical values of the volume and the lateral surface area are equal:$\frac{1}{3} \pi r^{2} h = \pi r \sqrt{r^{2} + h^{2}}$Dividing both sides by $\pi r$ (assuming $r 
eq 0$):$\frac{1}{3} r h = \sqrt{r^{2} + h^{2}}$Squaring both sides:$\frac{r^{2} h^{2}}{9} = r^{2} + h^{2}$Dividing both sides by $r^{2} h^{2}$:$\frac{1}{9} = \frac{r^{2} + h^{2}}{r^{2} h^{2}}$$\frac{1}{9} = \frac{r^{2}}{r^{2} h^{2}} + \frac{h^{2}}{r^{2} h^{2}}$$\frac{1}{9} = \frac{1}{h^{2}} + \frac{1}{r^{2}}$Therefore,the value of $\frac{1}{h^{2}} + \frac{1}{r^{2}}$ is $\frac{1}{9}$.

The numerical values of the volume and the area of the lateral surface of a right circular cone are equal. If the height of the cone is $h$ and the radius is $r,$ the value of $\frac{1}{h^{2}}+\frac{1}{r^{2}}$ is

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