Question

If the radius of the base of a cone is doubled keeping the height same. What is the ratio of the volume of the larger cone to the smaller cone

Answer 1

Let, Radius of smaller cone be r
And, Radius of larger cone be 2r
Let, Height of both cones be h
To find,
Ratio between volumes of larger and smaller cone
ATQ,
Volume of smaller cone=⅓πr²h
Again,
Volume of larger cone=⅓π(2r)²h
=⅓π4r²h
Now,
Ratio between both=
(⅓πr²h)/(⅓π4r²h)=1/4

Hope this helps you!

Answer 2

Ratio of volume of the larger cone to the smaller cone is 4:1.

• Let the radius of the base of the cone be r and the height of the cone be h.
• Volume of cone is given by $\frac{1}{3} \pi r^2 h$.
• Now the radius of the  base of the cone is doubled, radius = 2r and the height of the cone remains the same, height = h.
• Now, Volume of the cone will be $\frac{1}{3} \pi (2r)^2 h$.
• Ratio of volume of larger cone to smaller cone =

$\frac{[tex]\frac{1}{3} \pi (2r)^2 h$}{$\frac{1}{3} \pi r^2 h$}[/tex]

Ratio = $\frac{2^2}{1}$

Ratio = $\frac{4}{1}$