# Surface Areas and Volumes : Exercise 13.7 (Mathematics NCERT Class 9th)

Assume $\pi = {{22} \over 7}$, unless stated otherwise.
Q.1      Find the volume of the right circular cone with

(i) Radius 6 cm , height 7 cm
(ii) Radius 3.5 cm , height 12 cm
Sol.

(i) Here , r = 6 cm and h = 7 cm
Volume of the cone $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times {{22} \over 7} \times 6 \times 6 \times 7} \right)c{m^3}$
$= 264\,c{m^3}$
(ii) Here ,  r = 3.5 cm and h = 12 cm

Volume of the cone $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times {{22} \over 7} \times 3.5 \times 3.5 \times 12} \right)c{m^3}$

$= 154\,\,c{m^3}$.

Q.2      Find the capacity in litres of a conical vessel with
(i) Radius 7 cm , slant height 25 cm
(ii) Height 12 cm, slant height 13 cm
Sol.

(i) Here , r = 7 cm and l = 25 cm
Let the height of the cone be h cm. Then ,
${h^2} = {\ell ^2} - {r^2} = {25^2} - {7^2}$
$= 625 - 49 = 576$
$\Rightarrow$ $h = \sqrt {576} = 24\,cm$
Volume of the conical vessel $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times {{22} \over 7} \times 7 \times 7 \times 24} \right)c{m^3} = 1232\,c{m^3}$
Therefore , capacity of the vessel in litres $= \left( {{{1232} \over {1000}}} \right)\ell = 1.232\,\ell$

(ii) Here , h = 12 cm and l = 13 cm
Let the radius of the base of the cone be r cm . Then,
${r^2} = {\ell ^2} - {h^2} = {13^2} - {12^2}$
$= 169 - 144 = 25$
$\Rightarrow$ $r = \sqrt {25} = 5\,cm$
Volume of the conical vessel $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times {{22} \over 7} \times 5 \times 5 \times 12} \right)c{m^3} = {{2200} \over 7}c{m^3}$
Therefore Capacity of the vessel in litres  $= \left( {{{2200} \over 7} \times {1 \over {1000}}} \right)\ell = {{11} \over {35}}\ell$

Q.3    The height of a cone is 15 cm. If its volume is 1570 $c{m^3}$, find the radius of the base. (Use $\pi = 3.14$)
Sol.

Here , h = 15 cm and volume = 1570 $c{m^3}$
Let the radius of the base of cone be r cm.
Therefore Volume = $1570\,\,c{m^3}$
$\Rightarrow$ ${1 \over 3}\pi {r^2}h = 1570$
$\Rightarrow$ ${1 \over 3} \times 3.14 \times {r^2} \times 15 = 1570$
$\Rightarrow$ ${r^2} = {{1570} \over {3.14 \times 5}} = 100$
$\Rightarrow$ $r = \sqrt {100} = 10$

Thus , the radius of the base of cone is 10 cm.

Q.4      If the volume of a right circular cone of height 9 cm is $48\,\pi \,c{m^3},$ find the diameter of its base.
Sol.

Here , h = 9 cm and volume = $48\,\pi \,c{m^3},$
Let the radius of the base of the cone be r cm
Therefore Volume =$48\,\pi \,c{m^3},$
$\Rightarrow$ ${1 \over 3}\pi {r^2h} = 48\pi$
$\Rightarrow$ ${1 \over 3} \times {r^2} \times 9 = 48$
$\Rightarrow$ $3{r^2} = 48$
$\Rightarrow$ ${r^2} = {{48} \over 3} = 16$
$\Rightarrow$ $r = \sqrt {16} = 4$
Thus, the diameter of the base of the cone = 2X4=8 cm

Q.5      A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol.

Diameter of the top of the conical pit = 3.5 m
Therefore , radius $= \left( {{{3.5} \over 2}} \right)m = 1.75\,m$
Depth of the pit i.e., height = 12 m
Volume $= {1 \over 3}\pi {r^2}h$

$= \left( {{1 \over 3} \times {{22} \over 7} \times 1.75 \times 1.75 \times 12} \right){m^3}$
$= 38.5\,{m^3}$ [1$m^3$ = 1 kilolitres]
Therefore Capacity of pit = 38.5 kilolitres.

Q.6      The volume of a right circular cone is 9856 $c{m^3}$. If the diameter of the base is 28 cm, find

(i) Height of the cone.
(ii) Slant height of the cone,
(iii) Curved surface area of the cone.
Sol.

(i) Diameter of the base of the cone = 28 cm
Therefore  , radius = r = $\left( {{{28} \over 2}} \right)cm = 14\,cm$
Volume of the cone = $9856\,c{m^3}$
Let the height of the cone be h cm.

Now,    volume =$9856\,c{m^3}$
$\Rightarrow$ ${1 \over 3}\pi {r^2}h = 9856$
$\Rightarrow$ ${1 \over 3} \times {{22} \over 7} \times 14 \times 14 \times h = 9856$
$\Rightarrow$ $h = {{9856 \times 3 \times 7} \over {22 \times 14 \times 14}} = 48$ cm
Thus , the height of the cone = 48 cm

(ii) Here , r = 14 m and h = 48 cm

Let l be the slant height of the cone. Then,
${\ell ^2} = {h^2} + {r^2} = {48^2} + {14^2}$
$= 2304 + 196 = 2500$
$\Rightarrow$ $\ell = \sqrt {2500} = 50$
Thus, the slant height of the cone = 50 cm.

(iii) Here , r = 14 m and l = 50 cm.

Curved surface area = $\pi r\ell$
$= \left( {{{22} \over 7} \times 14 \times 50} \right)c{m^2}$
$= 2200\,\,c{m^2}$

Q.7     A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Sol.         On revolving the right $\Delta$ ABC about the side AB ( = 12 cm) , we get a cone as shown in the figure. Volume of solid so obtained $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times \pi \times 25 \times 12} \right)c{m^3}$
$= 100\,\pi \,c{m^3}$

Q.8      If the triangle ABC in the question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in question 7 and 8.
Sol.

On revolving the right $\Delta$ ABC about the side BC( = 5 cm), we get a cone as shown in the figure,
Volume of solid so obtained $= {1 \over 3}\pi {r^2}h$
$= {1 \over 3} \times \pi \times 12 \times 12 \times 5\,c{m^3}$   [h = 5 cm , r = 12 cm]
= $240\,\pi \,c{m^3}$

Therefore Ratio of their volumes = $100\,\pi \,\,:\,240\,\pi$ (i.e., of Q. 7 and Q. 8)  = 5 : 12

Q. 9    A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Sol.

Diameter of the base of the cone = 10.5 m
Therefore , radius = r = $\left( {{{10.5} \over 2}} \right)m = 5.25\,m$
Height of the cone = 3m
Therefore Volume of the cone (heap) $= {1 \over 3}\pi {r^2}h$
$= \left( {{1 \over 3} \times {{22} \over 7} \times 5.25 \times 5.25 \times 3} \right){m^3}$
$= 86.625\,{m^3}$
To find the slant height l :
We have, ${\ell ^2} = {h^2} + {r^2} = {3^2} + {\left( {5.25} \right)^2}$
$= 9 + 27.5625 = 36.5625$
$\Rightarrow$ $\ell = \sqrt {36.5625} = 6.0467\,\,\left( {approx} \right)$
Canvas required to protect wheat from rain = Curve surface area
$= \pi r\ell = \left( {{{22} \over 7} \times 5.25 \times 6.0467} \right){m^2}$

$= 99.77{m^2}\left( {approx} \right)$