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# 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.