# Ex.13.5 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 10

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## Question

Derive the formula for the volume of the frustum of a cone, given to you in Section $$13.5$$, using the symbols as explained.

Video Solution
Surface Areas And Volumes
Ex 13.5 | Question 7

## Text Solution

What is Known?

A frustum of a cone with $$h$$ as height, $$l$$ as the slant height, $$r1$$ and $$r2$$ radii of the ends where $$r1 > r2$$

To prove:

Volume of the frustum of a cone $$= \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)$$

Construction:

Extend side $$BC$$ and $$AD$$ of the frustum of cone to meet at $$O.$$

Proof:

The frustum of a cone can be viewed as a difference of two right circular cones $$OAB$$ and $$OCD.$$

Let $$h1$$ and$$l1$$ be the height and slant height of cone $$OAB$$ and $$h2$$ and $$l2$$ be the height and slant height of cone $$OCD$$ respectively.

In $$\Delta APO$$ and $$\Delta DQO$$

$$\Delta APO = \Delta DQO = 90^\circ$$ (Since both cones are right circular cones)

$$\angle AOP = \angle DOQ$$ (Common)

Therefore,$$\Delta APO \sim \Delta DQO$$ ( A.A criterion of similarity)

\begin{align}\frac{{AP}}{{DQ}} = \frac{{AO}}{{DO}} = \frac{{OP}}{{OQ}}\end{align} (Corresponding sides of similar triangles are proportional)

\begin{align} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{l_1}}}{{{l_2}}} = \frac{{{h_1}}}{{{h_2}}}\end{align}

\begin{align} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{h_1}}}{{{h_2}}}\end{align} or \begin{align} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{h_2}}}{{{h_1}}}\end{align}

Subtracting 1 from both sides

\begin{align}\frac{{{r_1}}}{{{r_2}}} - 1 &= \frac{{{h_1}}}{{{h_2}}} - 1\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{{{h_1} - {h_2}}}{{{h_2}}}\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{h}{{{h_2}}}\\{h_2} &= \frac{{h{r_2}}}{{{r_1} - {r_2}}}{\rm\qquad{ (i)}}\end{align}

or

\begin{align} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{h_2}}}{{{h_1}}}\end{align}

Subtracting 1 from both sides we get

\begin{align}\frac{{{r_2}}}{{{r_1}}} - 1 &= \frac{{{h_2}}}{{{h_1}}} - 1\\\frac{{{r_2} - {r_1}}}{{{r_1}}} &= \frac{{{h_2} - {h_1}}}{{{h_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{{{h_1} - {h_2}}}{{{h_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{h}{{{h_1}}}\\{h_1} &= \frac{{h{r_1}}}{{{r_1} - {r_2}}}{\rm\qquad{ (ii)}}\end{align}

Volume of frustum of cone $$=$$ Volume of cone $$OAB$$ $$-$$ Volume of cone $$OCD$$

\begin{align}&= \frac{1}{3}\pi {r_1}^2{h_1} - \frac{1}{3}\pi {r_2}^2{h_2}\\&= \frac{1}{3}\pi \left( {{r_1}^2{h_1} - {r_2}^2{h_2}} \right)\\&= \frac{1}{3}\pi \left[{{r_1}^2 \times \frac{{h{r_1}}}{{{r_1} - {r_2}}} - {r_2}^2 \times \frac{{h{r_2}}}{{{r_1} - {r_2}}}} \right]\\&\qquad\left[ {{\rm{using (i) and (ii)}}} \right]{\rm{ }}\\&= \frac{1}{3}\pi \left[ {\frac{{h{r_1}^3}}{{{r_1} - {r_2}}} - \frac{{h{r_2}^3}}{{{r_1} - {r_2}}}} \right]\\&= \frac{1}{3}\pi h\left[{\frac{{{r_1}^3 - {r_2}^3}}{{{r_1} - {r_2}}}} \right]\\& = \frac{1}{3}\pi h\left[ {\frac{{\left( {{r_1} - {r_2}} \right]\left[ {{r_1}^2 + {r_1}^2{r_2}^2 + {r_2}^2}\right)}}{{{r_1} - {r_2}}}} \right]\\&\qquad\left[ {\left( {{a^3} - {b^3}} \right] = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right]\\&= \frac{1}{3}\pi h\left( {{r_1}^2 + {r_2}^2 + {r_1}^2{r_2}^2} \right)\end{align}

Volume of the frustum of a cone\begin{align} = \frac{1}{3}\pi h\left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)\end{align}

Hence Proved

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