Class 9 Surface and Volumes ex 11.3

class 9 maths chapter 11 exercise 11.3

ncert class 9 maths chapter 11 solution, Surface Areas and Volumes Exercise 11.3 involve complete answers for each question in the exercise 11.3. The solutions provide students a strategic methods to prepare for their exam. Class 9 Maths Chapter 11 Surface Areas and Volumes exercise 11.3 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes prepared by www.skmath.in team in very delicate, easy and creative way.

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 (Assume \(\pi = {22\over7}\))

Solution:

Volume of cone = \({1\over 3}\times πr^2h\) cube units

Where r be radius and h be the height of the cone

(i) Radius of the cone, r = 6 cm

Height of the cone, h = 7 cm

Ley V be the volume of the cone, so we have

\[V = {1\over 3}×{22\over 7}×36×7= 12×22  = 264\]

The volume of the cone is 264 \(\text{cm}^3\).

(ii) Radius of the cone, r = 3.5cm

Height of the cone, h = 12cm

Volume of the cone = \({1\over 3}×{22\over 7}×3.52×7 = 154\)

The volume of the cone is 154 \(\text{cm}^3\)

2. Find the capacity in litres of a conical vessel with

(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

(Assume \(π = {22\over 7}\)

Solution:

(i) Radius of the cone, r =7 cm

Slant height of the cone, l = 25 cm

height of the cone,h=\(\sqrt{l^2-r^2}\)

h=\(\sqrt{25^2-7^2}\)

h=\(\sqrt{625-49}=\sqrt{576}=24\)

h = 24

Height of the cone is 24 cm

Volume of the cone, \(V = {1\over 3} πr^2h \)

\[V = {1\over 3}×{22\over 7}×72×24=154×8=1232\]

So, the volume of the vessel is 1232 \(\text{cm}^3\)

\(\therefore\), the capacity of the conical vessel = \(1232\over 1000\) liters (because 1L = 1000 \(\text{cm}^3\)) = 1.232 Liters.

(ii) Height of the cone, h = 12 cm

Slant height of the cone, l = 13 cm

height of the cone,r=\(\sqrt{l^2-h^2}\)

h=\(\sqrt{13^2-12^2}\)

h=\(\sqrt{169-144}=25\)

h = 5

Hence, the radius of the cone is 5 cm.

Now, Volume of the cone,  \(V = {1\over 3}\times πr^2h\)

\(V = {1\over 3}\times 52×12 \text{cm}^3= {2200\over 7}\)

Volume of the cone is \({2200\over 7} \text{cm}^3\)

Capacity of the conical vessel= \({2200\over 7000}\) litres (1L = 1000 \(\text{cm}^3\))

\(= {11\over 35}\) litres

3. The height of a cone is 15 cm. If its volume is \(1570 \text{cm}^3\), find the diameter of its base. (Use π = 3.14)

Solution:

Height of the cone, h = 15 cm

Volume of cone  \(=1570 \text{cm}^3\)

Let r be the radius of the cone

Volume of the cone, \(V = {1\over 3}\times πr^2h\)

\( {1\over 3}\times πr^2h = 1570\)

\({1\over 3}×3.14×r^2 ×15 = 1570\)

\(r^2 = 100\)

r = 10

Radius of the base of the cone 10 cm.

4. If the volume of a right circular cone of height 9 cm is 48π \(\text{cm}^3\), find the diameter of its base.

Solution:

Height of cone, h = 9 cm

Volume of cone =48π \(\text{cm}^3\)

Let r be the radius of the cone.

As we know, volume of the cone, \(V = {1\over 3}\times πr^2h\)

So, \(V = {1\over 3}\times πr^2\times 9 = 48 π \)

\(r^2 = 16\)

r = 4

Radius of the cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of the base is 8 cm.

5. A conical pit of a top diameter 3.5m is 12m deep. What is its capacity in kiloliters?
(Assume \(π = {22\over 7}\)

Solution:

Diameter of conical pit = 3.5 m

Radius of conical pit, \(r = {\text{diameter}\over 2} = {3.5\over 2} = 1.75 \) m

Height of pit, h = Depth of pit = 12m

Volume of cone, \(V = {1\over 3}\times πr^2h\)

\(V = {1\over 3}\times {22 \over 7} ×(1.75)^2×12 = 38.5\)

Volume of the cone is 38.5 \(\text{m}^3\)

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

6. The volume of a right circular cone is 9856 \(\text{m}^3\) . 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

(Assume \(\pi = {22\over7}\))

Solution:

Volume of a right circular cone = 9856 \(\text{m}^3\)

Diameter of the base = 28 cm

(i) Radius of cone, \(r = {28\over 2}\) cm = 14 cm

Let the height of the cone be \(h\)

Volume of cone, \(V ={1\over3} πr^2h\)

\({1\over3} πr^2h = 9856\)

\({1\over3}×{22\over 7}×14×14×h = 9856\)

\(h = 48\)

The height of the cone is 48 cm.

ii) Slant Height of the cone, \(l=\sqrt{r^2+h^2}\)

\(l=\sqrt{14^2+48^2}=\sqrt{196+2304}=50\)

Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl

= \(22\over 7\times 14\times 50\)

= 2200

Curved surface area of the cone is 2200 cm\(^2\).

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.

Solution:

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm



Volume of cone,  \(V ={1\over3} πr^2h\)

 \(V ={1\over3} π{5^2}\times12= 100π\)

Volume of the cone so formed is 100π \(\text{cm}^3\).

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

Solution:



A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone \(V ={1\over3} πr^2h\), where r is the radius and h is the height of the cone.

\(V ={1\over3}×π×12×12×5\)

= 240 π

The volume of the cones formed is 240π \(text{cm}^3\).

So, the required ratio = \({\text{(the result of question 7)} \over \text{(the result of question 8)}} = {100π\over 240π} = {5\over12} = 5:12\).

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.

(Assume \(\pi = {22\over7}\))

Solution:

Radius (r) of heap = \(10.5\over 2\) m = 5.25

Height (h) of heap = 3m

Volume of heap = \({1\over 3}×πr^2h\)

= \({1\over 3}×{22\over 7}×5.25×5.25×3\)

= 86.625

The volume of the heap of wheat is 86.625 \(text{m}^3\)

Area of Canvas required=CSAS=\(\pi rl\) ,where l=\(\sqrt{r^2+h^2}\)

CSA of cone=

\(={22\over 7}×5.25×6.05\)

= 99.825

Therefore, the area of the canvas is 99.825 \(text{m}^2\).