Question

The height of a cone is 15 cm. If its volume is \[500\pi {\text{ c}}{{\text{m}}^3}\], then find the radius of its base.

Answer 1

Hint: The formula to calculate the volume of a cone is, \[{\text{Volume of Cone = }}\dfrac{1}{3}\pi {r^2}h\], where \[r\] is the radius of the base of a cone and \[h\] is the height of a cone. Apply this formula, and then use the given conditions to find the required value.

Complete step by step answer:
Given that the height \[h\] of a cone is 15 cm and the volume of a cone is \[500\pi {\text{ c}}{{\text{m}}^3}\].

Let us assume that the radius of the base of a cone is \[r\].

We will use the formula of the volume of a cone is, \[{\text{Volume of Cone = }}\dfrac{1}{3}\pi {r^2}h\], where \[r\] is the radius of the base of a cone and \[h\] is the height of a cone.

Substituting the values of \[r\] and \[h\] in the above formula of volume of a cone, we get

\[
  {\text{Volume of a cone = }}\dfrac{1}{3}\pi \times {r^2} \times 15 \\
   = 5\pi {r^2} \\
 \]

Since we know that the volume of the given cone is \[500\pi {\text{ c}}{{\text{m}}^3}\].

Substituting the value of volume of a cone in the above equation, we get

\[500\pi = 5\pi {r^2}\]

Dividing the above equation by \[5\pi \]on each of the sides, we get

\[
   \Rightarrow \dfrac{{500\pi }}{{5\pi }} = \dfrac{{5\pi {r^2}}}{{5\pi }} \\
   \Rightarrow 100 = {r^2} \\
   \Rightarrow {r^2} = 100 \\
 \]

Taking square root on each of the sides in the above equation, we get

\[
   \Rightarrow {r^2} = 100 \\
   \Rightarrow r = \pm 10{\text{ }} \\
 \]

Since the radius of a cone can never be negative, the negative value of \[r\] is discarded.

Therefore, the radius of a cone is 10 cm.

Note: In solving these types of questions, you should be familiar with the formula of the volume of a cone. To make the calculations easier, the value of \[\pi \] has not been substituted. Since \[\pi \] occurs in the expression for the volume and its formula, hence, cancels out when they are equated and will help in reducing the calculations to a great extent.