If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

When right-angled \(\triangle{ABC}\) is revolved about its side 5 cm, a cone will be formed having radius (r) as 12 cm, height (h) as 5 cm, and slant height (l) as 13 cm.

Now, thus, we know that,
Volume of cone
= \(\frac{1}{3} {\pi} {r}^2 h\)
= \(\frac{1}{3} × {\pi} × {12}^2 × 5\) \({cm}^3\)
= \(240 × {\pi}\) \({cm}^3\)
= \(240 {\pi}\) \({cm}^3\)

Therefore, the volume of the cone so formed is \(240 {\pi}\) \({cm}^3\)

Thus, the ratio of the volumes of the two solids obtained in Questions 7 and 8
= \(\frac{100}{240}\)
= \(\frac{5}{12}\)
\(\therefore \) 5 : 12.