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.

 

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 \(OCD\)

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

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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