Slant height and diameter of a cone

Question

The height, slant height & diameter of a cone area consecutive even numbers. Its volume is $k \pi$.

Find $k$

Attemp:

Let the height be $h$, diameter be $d$, slant height be $s$. Either $s>h>d$ or $s>d>h$. For the first case, let $d = 2r$, because it is even. We then have a right triangle with side lengths $a,2a+2,2a+4$. This yields the quadratic equation $a^2 -8a-12 = 0$, which has no integer solutions.

For the second case, again let $d =2a$. We then have a right triangle with side lengths $a,2a-2,2a+2$. This yields the quadratic equation $a^2 = 16a$, or $a = 16$. This means the cone has height $30$. So the volume is $\frac{30\cdot16\cdot16}{3} = 2560$.

I think it's incorrect

Answer 1

Let they be $h=2a, s = 2a+2, d = 2a+4 \text{ or } r = a+2$

In a right-angled cone, $s^2 = h^2+r^2 \implies (2a+2)^2= 4a^2+(a+2)^2 \implies 4a^2+8a+4 = 4a^2+a^2+4a+4$
$\implies a^2 = 4a \implies \color{blue} {a = 4} $

So, $h = 8, r = 6$

Volume, $ V= \frac{1}{3}\pi rh^2 = 128\pi = k\pi \implies k = 128$