Surface Areas and Volumes – Exercise 13.7 – (MATHEMATICS) – 9th Class

0

 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.

Show Answer

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.

Show Answer

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).

Show Answer

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.

Show Answer

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?

Show Answer

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.

Show Answer

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.

Show Answer

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.

Show Answer

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.

Show Answer

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


Leave A Reply

Your email address will not be published.