Ex.13.5 Q6 Surface Areas and Volumes Solution - NCERT Maths Class 10

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Question

Derive the formula for the curved surface area and total surface area 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:

(i) CSA of the frustum of the cone \( = \pi l\left( {{r_1} + {r_2}} \right)\)

(ii)TSA of the frustum of the cone \(\begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}\)

where \(r1, r2, h\) and \(l\) are the radii height and slant height of the frustum of the cone respectively.

Construction:

Extended 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)

\(\angle APO = \angle DQO\) (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 in proportion)

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

\(\begin{align} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{l_1}}}{{{l_2}}} \\\text{or}\\\Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{l_2}}}{{{l_1}}}\end{align}\)

Subtracting \(1\) from both sides we get

\[\begin{align}\frac{{{r_1}}}{{{r_2}}} - 1 &= \frac{{{l_1}}}{{{l_2}}} - 1\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{{{l_1} - {l_2}}}{{{l_2}}}\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{l}{{{l_2}}}\\{l_2} &= \frac{{l{r_2}}}{{{r_1} - {r_2}}}{\rm\qquad{ (i)}}\end{align}\]

or

\[\begin{align} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{l_2}}}{{{l_1}}}\end{align}\]

Subtracting \(1\) from both sides we get

\[\begin{align}\frac{{{r_2}}}{{{r_1}}} - 1 &= \frac{{{l_2}}}{{{l_1}}} - 1\\\frac{{{r_2} - {r_1}}}{{{r_1}}} &= \frac{{{l_2} - {l_1}}}{{{l_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{{{l_1} - {l_2}}}{{{l_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{l}{{{l_1}}}\\{l_1} &= \frac{{l{r_1}}}{{{r_1} - {r_2}}}{\rm\qquad{ (ii)}}\end{align}\]

(i) CSA of frustum of cone \(=\) CSA of cone \(OAB\, –\) CSA of cone \(OCD\)

\[\begin{align}&= \pi {r_1}{l_1} - \pi {r_2}{l_2}\\&= \pi \left( {{r_1}{l_1} - {r_2}{l_2}} \right){\rm{ }}\\& = \pi \left( {{r_1} \times \frac{{l{r_1}}}{{{r_1} - {r_2}}} - {r_2} \times \frac{{l{r_2}}}{{{r_1} - {r_2}}}} \right){\rm{         }}\left[ {{\rm{using (i)and (ii)}}} \right]\\&= \pi \left( {\frac{{l{r_1}^2 - l{r_2}^2}}{{{r_1} - {r_2}}}} \right)\\&= \pi \left( {\frac{{l\left( {{r_1}^2 - {r_2}^2} \right)}}{{{r_1} - {r_2}}}} \right)\\&= \pi \left( {\frac{{l\left( {{r_1} - {r_2}} \right)\left( {{r_1} + {r_2}} \right)}}{{{r_1} - {r_2}}}} \right){\rm{   }}\left[ {{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \right]\\ &= \pi l\left( {{r_1} + {r_2}}\right)\end{align}\]

(ii)TSA of frustum of cone \(=\) CSA of frustum \(+\) Area of lower circular end \(+\) Area of top circular end

\[\begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}\]

Therefore, CSA of the frustum of the cone \( \begin{align}= \pi l\left( {{r_1} + {r_2}} \right)\end{align}\)

TSA of the frustum of the cone\(\begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}\)

Hence Proved.

  
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