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Exercise 13.7

Assume π = ${{22} \over 7}$ unless stated otherwise.

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.

Sal. (i) Radius (r) = 6 cm
Height (h) = 7 cm

Volume = ${1 \over 3}\pi {r^2}h$

Volume =  $= {1 \over 3} \times {{22} \over 7} \times {(6)^2} \times 7c{m^3} = 264c{m^3}$

(ii) Radius (r) = 3.5 ern
Height (h) = 12 cm
Volume = $= {1 \over 3}\pi {r^2}h$
$= {1 \over 3} \times {{22} \over 7} \times {(3.5)^2} \times 12c{m^3} = 154c{m^3}$

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) r = 7 cm, l = 25 cm.

∴    $h = \sqrt {{l^2} - {r^2}} = \sqrt {{{(125)}^2} - {{(7)}^2}} cm$

$= \sqrt {625 - 49} cm = \sqrt {576} cm = 24cm$

∴  Capacity = 1 22 x x 7 x 7 x 24 cm3
= 1232 cm³ = 1.232 l.

(ii) h = 12 cm, l = 13 cm.

∴      $r = \sqrt {{l^2} - {h^2}} = \sqrt {{{(13)}^2} - {{(12)}^2}} cm$

$= \sqrt {169 - 144} cm = \sqrt {25} cm = 5cm$

Capacity = $= {1 \over 3} \times {{22} \over 7} \times 5 \times 5 \times 12c{m^3}$

$= {{2200} \over 7}c{m^3} = {{11} \over {35}}litre$

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14).

Sol. Let r be the required radius.

Height (h) = 15 cm

Volume = 1570 ⇒    ${1 \over 3} \times 3.14 \times {r^2} \times 15 = 1570$

$\Rightarrow {r^2} = {{1570 \times 3} \over {3.14 \times 15}} = 100 \Rightarrow r = 10cm$

4. If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.

Sol. Height (h) = 9 cm, radius (r) = ?

Volume = 48π cm³ ⇒ ${1 \over 3} \times \pi \times {r^2} \times 9 = 48\pi$

⇒    r2  =  16 ⇒    r  = 4  cm.

Diameter = 2r = 8 cm.

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

Sol. Diameter (d) = 3.5 m

$Radius(r) = {{diameter} \over 2} = {{3.5} \over 2}m = {{35} \over {20}}m$

Deep, i.e., height (h) = 12 m

Capacity of the pit = ${1 \over 3}$πr²h

$= {1 \over 3} \times {{22} \over 7} \times {{35} \over {20}} \times {{35} \over {20}} \times 12{m^3}$

= 38.5 m³ = 38.5 kilolitres.  [1 m³ = 1 kl]

6. The volume of a right circular cone is 9856 cm³. 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. Diameter = 28 cm, radius =  ${{28} \over 2} = 14cm$

(i) Volume of cone = 9856 cm³

${1 \over 3} \times {{22} \over 7} \times 14 \times 14 \times h = 9856$

$\Rightarrow h = {{4856 \times 3} \over {44 \times 14}} \Rightarrow h = 48$

∴   Height (h) of the cone = 48 cm.

(ii) Slant height of cone (l)

$= \sqrt {{{(14)}^2} + {{(48)}^2}} cm$

$= \sqrt {196 + 2304} cm = \sqrt {2500} cm = 50cm.$

(iii) Curved surface area of the cone = πrl

$= {{22} \over 7} \times 14 \times 50 = 2200c{m^2}$

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. As sides are 5 cm, 12 cm, 13 cm. Hence, triangle is right angled.

We have r = 5 cm, h = 12 cm.

∴   $Volume = {1 \over 3}\pi {r^2}h = {1 \over 3} \times \pi \times 5$

× 5  ×  12  cm3

= 100π cm3

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 Questions 7 and 8.

Sol. In this case, r = 12 cm, h = 5 cm.

∴     $Volume = {1 \over 3}\pi {r^2}h = {1 \over 3} \times \pi \times 12 \times 12 \times 5c{m^3}$

= 240 π cm3

Ratio of volumes obtained in Question 7 and Question 8:

${{{V_1}} \over {{V_2}}} = {{100\pi } \over {240\pi }} = {5 \over {12}},i.e,{V_1}:{V_2} = 5:12$

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. 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 (d) = 10.5 m  ⇒   radius (r) = ${{10.5} \over 2} = 5.25m$
Height (h) = 3 m
Volume of the heap of wheat = ${1 \over 3}\pi {r^2}h$

$= {1 \over 3} \times {{22} \over 7} \times 5.25 \times 5.25 \times 3{m^3}$

= 86.625 m3

Slant height of the cone = $\sqrt {{{(5.25)}^2} + {{(3)}^2}} m$

$= \sqrt {27.5625 + 9} m$

$= \sqrt {36.5625} m \approx 6.05m.$

Area of canvas required = ${{22} \over 7} \times 5.25 \times 6.05{m^2}$

= 99.825 m²