The curved surface area of a right circular cone is half of another right circular cone. If the ratio of their slant heights is 2 : 1 and that of their volumes is 3 : 1, find ratio of their:

(a) radii

(b) heights

(a) Let radius of smaller and larger cone be r and R, respectively.

Let height of smaller and larger cone be h and H, respectively.

Let slant height of smaller and larger cone be l and L respectively.

Given,

Ratio of their slant heights is 2 : 1.

∴ l : L = 2 : 1.

Given,

Curved surface area of a right circular cone is half of the other right circular cone.

∴ πrl = $\dfrac{1}{2}πRL$

⇒ rl = $\dfrac{1}{2}RL$

⇒ $\dfrac{r}{R} = \dfrac{L}{2l} = \dfrac{1}{2} \times \dfrac{1}{2}$

⇒ $\dfrac{r}{R} = \dfrac{1}{4}$

⇒ r : R = 1 : 4

Hence, ratio of the radii = 1 : 4.

(b) Let volume of cone with smaller curved surface area be v and that with larger curved surface area be V.

Given,

Volumes are in the ratio 3 : 1.

∴ v : V = 3 : 1

$\Rightarrow \dfrac{\text{Volume of larger CSA cone}}{\text{Volume of smaller CSA cone}} = \dfrac{V}{v} \\[1em] \Rightarrow \dfrac{\dfrac{1}{3}πR^2H}{\dfrac{1}{3}πr^2h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{4^2}{1^2} \times \dfrac{H}{h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{16}{1} \times \dfrac{H}{h} = \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{3 \times 16} \\[1em] \Rightarrow \dfrac{H}{h} = \dfrac{1}{48} \\[1em] \Rightarrow \dfrac{h}{H} = \dfrac{48}{1}.$

Hence, ratio of heights = 48 : 1.