Q.A right circular cylinder and a cone have equal bases, and the ratio of their volumes is 3:2. Prove that the height of the cone is half the height of the cylinder.

Let the radius of the base of both the right circular cylinder and the cone be \(r\) units. Let the height of the cylinder be \(h_1\) units and the height of the cone be \(h_2\) units.
According to the question, \[ \pi r^2 h_1 : \frac{1}{3} \pi r^2 h_2 = 3 : 2 \] Or, \[ h_1 : \frac{1}{3} h_2 = 3 : 2 \] Or, \[ \frac{h_1}{\frac{1}{3} h_2} = \frac{3}{2} \] Or, \[ \frac{h_1}{h_2} = \frac{3}{2} \times \frac{1}{3} \] Or, \[ \frac{h_1}{h_2} = \frac{1}{2} \] Therefore, \[ h_2 = 2h_1 \] ∴ The height of the cone is double the height of the cylinder. (Proved)
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