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.

On revolving,

Triangle around side 5 cm it becomes a cone with height 12 cm.

From figure,

Radius of the cone (r) = 12 cm

Height of the cone (h) = 5 cm

By formula,

Volume of the cone (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times π \times 12^2 \times 5 \\[1em] = \dfrac{1}{3} \times π \times 144 \times 5 \\[1em] = \dfrac{1}{3} \times π \times 720 \\[1em] = 240π \text{ cm}^3.$

Volume of the cone = 240π cm³

Volume of the cone in question 7 = 100π cm³

Ratio = Volume of the cone in question 7 : Volume of the cone in question 8

= 100π : 240π

= 5 : 12

Hence, the volume of the cone is 240π cm³ and the required ratio is 5 : 12.