Question

a sphere of radius 8cm is melted and recast into a right circular cone of height 32cm find the radius of the base of the cone

Answer 1

since a sphereis melt and recast into a cone their volumes must be equal.

=4/3$\pi$r³=1/3$\pi$r²h

=4/3×22/7×8×8×8=1/3×22/7×r²×32

=64=r²

=so,r=8cm.

Answer 2

Given :

A sphere of radius 8cm is melted and recast into a right circular cone of height 32cm

To Find :

Find the radius of the base of the cone

Solution:

Radius of sphere = 8 cm

Volume of sphere =$\frac{4}{3} \pi r^3 = \frac{4}{3} \times \gfrac{22}{7} \times 8^3$

Height of cone = 32 cm

Let the radius of cone be r

Volume of cone =$\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 32$

We are given that sphere is melted into cone

So, Volume of sphere = Volume of cone

So,$\frac{4}{3} \times \frac{22}{7} \times 8^3 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 32$

$4 \times 8^3 = r^2 \times 32$

$\frac{ 4 \times 8^3}{32}=r^2$

$\sqrt{\frac{ 4 \times 8^3}{32}}=r$

8=r

Hence The radius of base of cone is 8 cm