The number of solid cones with integer radius | vedclass

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(B) Let the height and radius of the cone be $h$ and $r$ respectively,where $h, r \in \mathbb{Z}^+$.Given that the volume of the cone is numerically e

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$1$

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(B) Let the height and radius of the cone be $h$ and $r$ respectively,where $h, r \in \mathbb{Z}^+$.Given that the volume of the cone is numerically equal to its total surface area:$\frac{1}{3} \pi r^2 h = \pi r l + \pi r^2$,where $l = \sqrt{h^2 + r^2}$.Dividing by $\pi r$ (since $r 
eq 0$):$\frac{1}{3} r h = \sqrt{h^2 + r^2} + r$$\frac{1}{3} r h - r = \sqrt{h^2 + r^2}$$r(\frac{h}{3} - 1) = \sqrt{h^2 + r^2}$Squaring both sides:$r^2(\frac{h-3}{3})^2 = h^2 + r^2$$r^2(\frac{h^2 - 6h + 9}{9}) = h^2 + r^2$$r^2(h^2 - 6h + 9) = 9h^2 + 9r^2$$r^2 h^2 - 6h r^2 + 9r^2 = 9h^2 + 9r^2$$r^2 h^2 - 6h r^2 = 9h^2$Since $h 
eq 0$,divide by $h$:$r^2 h - 6r^2 = 9h$$h(r^2 - 9) = 6r^2$$h = \frac{6r^2}{r^2 - 9} = \frac{6(r^2 - 9) + 54}{r^2 - 9} = 6 + \frac{54}{r^2 - 9}$.For $h$ to be an integer,$r^2 - 9$ must be a divisor of $54$. Also,$r^2 - 9 > 0$ implies $r > 3$.The divisors of $54$ are $1, 2, 3, 6, 9, 18, 27, 54$.Testing $r^2 - 9 = k$:If $r^2 - 9 = 27$,$r^2 = 36 \Rightarrow r = 6$. Then $h = 6 + \frac{54}{27} = 6 + 2 = 8$.If $r^2 - 9 = 54$,$r^2 = 63$ (not a perfect square).Other values of $k$ do not yield integer $r$.Thus,only one such cone exists with $(r, h) = (6, 8)$.

The number of solid cones with integer radius and integer height each having its volume numerically equal to its total surface area is

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