# RS Aggarwal Class 9 Solutions Chapter 13 - Volume And Surface Area

## RS Aggarwal Class 9 Chapter 13 - Volume And Surface Area Solutions Free PDF

Volume and Surface Area are different metrics that are used for calculation of different shapes and figures. The area of a simple curve can be defined as the amount of space inside the object and the volume of a 3D shape is the amount of space displaced by an object. Knowing the formulas for the calculation of different shapes is useful to calculate the volume and areas with the given metrics.

The different geometrical shapes that we can calculate for Volume and Surface Area are:

1. Circle
2. Triangle
3. Right Triangle
4. Square
5. Rectangle
6. Trapezoid
7. Cube
8. Prism
9. Pyramid
10. Sphere
11. Parallelogram

## Download PDF of RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area

We have provided here step by step RS Aggarwal Class 9 Solutions Chapter 13 for all exercise questions given in pdf format. Students can refer to these solutions whenever they get stuck or have any doubt. All the exercise questions with solutions of Chapter 13: Volume And Surface Area is given below:

Q.1: Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:

(i) Â length = 12 cm, breadth = 8 cm and height = 4.5 cm.

âˆ´ Volume of cuboid = l Ã— b Ã— h

= (12 Ã—8 Ã—4.5) cm3

âˆ´ Lateral surface area of a cuboid = 2(l + b) Ã— h

= [2(12 + 8) Ã— 4.5] cm2

= ( 2 Ã— 20 Ã— 4.5) cm2 = 180 cm2

âˆ´ Total surface area of cuboid = 2(lb + bh + lh)

= 2(12 Ã— 8) = (8 Ã— 4.5) + (4.5 Ã— 12) cm2

= 2(96 + 36 + 54) cm2 = 2(186) cm2 = 372 cm2

(ii) Length = 26 m, breadth = 14 m and height = 6.5 m

âˆ´ Volume of cuboid = l Ã— b Ã— h = (26 Ã— 14 Ã— 6.5) m3 = 2366 m3

âˆ´ Lateral surface area of a cuboid = 2(l + b) Ã— h

= [2(26 + 14) Ã— 6.5] m2 =(2 Ã— 40 Ã— 6.5) m2 = 520 m2

âˆ´Total surface area = 2(lb + bh + hl)

= 2((26 Ã— 14) + (14 Ã— 6.5) + (6.5 Ã— 26)) m2 = 2(364 + 91 + 169) m2 = (2 Ã— 624) m2 = 1248 m2

(iii) Length = 15 m, breadth = 6 m and height = 5 dm = 0.5 m.

âˆ´ Volume of the cuboid = l Ã— b Ã— h = (15 Ã— 6 Ã— 0.5) m3

âˆ´ Lateral surface area of a cuboid = 2(l + b) Ã— h

= [2(15 + 6) Ã— 0.5] m2 = (2 Ã— 21 Ã— 0.5) m2 = 21 m2

âˆ´Total surface area = 2(lb + bh + hl)

= 2((15 Ã— 6) + (6 Ã— 0.5) + (0.5 Ã— 15)) m2 = 2(90 + 3 + 7.5) m2 = (2 Ã— 100.5) m2 = 201 m2

(iv) Â Length = 24 m, breadth = 0.25 m and height = 6 m.

âˆ´ Volume of the cuboid = l Ã— b Ã— h = (24 Ã— 0.25 Ã— 6) m3

âˆ´ Lateral surface area of a cuboid = 2(l + b) Ã— h

= [2(24 + 0.25) Ã— 6] m2 = (2 Ã— 24.25 Ã— 6) m2 =291 m2

âˆ´Total surface area = 2(lb + bh + hl)

= 2((24 Ã— 0.25) + (0.25 Ã— 6) + (24 Ã— 6)) m2 = 2(6 + 1.5 + 144) m2 = (2 Ã— 151.5) m2 = 303 m2

Q.2: Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet requires making the cistern.

Solution:

Length of Cistern = 8 m

Breadth of Cistern = 6 m

Height of Cistern = 2.5 m

Therefore, the capacity of the cistern = Volume of the cistern

Volume of cistern = l Ã— b Ã— h = (8 Ã— 6 Ã— 2.5) m3 =120 m3

Area of the iron sheet required = Total surface area of the cistern.

âˆ´Total surface area = 2(lb + bh + hl)

= 2((8 Ã— 6) + (6 Ã— 2.5) + (2.5 Ã— 8)) m2

= 2(48 + 15 + 20) m2 = (2 Ã— 83) m2 = 166 m2

Q.3: The dimensions of a room are (9 m Ã— 8 m Ã— 6.5 m). It has one door of dimensions (2 m Ã— 1.5 m) and two windows, each of dimensions (1.5 m Ã— 1 m). Find the cost of whitewashing the walls at Rs. 6.40 per square meter.

Solution:

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Therefore, area of 4 walls =Lateral surface area

Lateral surface area of the room = 2(l + b) Ã— h

= [2(9 + 8) Ã— 6.5] m2 = (2 Ã— 17 Ã— 6.5) m2 = 221 m2

âˆ´ Area not white washed = (area of one door) + (area of 2 windows)

= (2 Ã— 1.5) m2 + (2 Ã— 1.5 Ã— 1) m2 = 3 + 3 = 6 m2

âˆ´ Area whitewashed = (221 â€“ 6) m2 = 215 m2

âˆ´ Cost of whitewashing the walls at the rate of Rs.6.40 per square meter = Rs. (6.40 Ã— 215) = Rs. 1376

Q.4: How many planks of dimensions ( 5 m Ã— 25 cm Ã— 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?

Solution:

Length of plank = 5 m = 500 cm

Breadth of plank = 25 cm

Height of plank = 10 cm

Hence, volume = l Ã— b Ã— h = (500 Ã— 25 Ã— 10) cm3

Now,

Length of pit = 20 m = 2000 cm

Breadth of pit = 6 m = 600 cm

Height of pit = 80 cm

Hence, volume =Â l Ã— b Ã— hÂ = ( 2000 Ã— 600 Ã— 80) cm3

âˆ´ Number of planks that can be stored = VolumeofpitVolumeofplank$\frac{Volume\, of\, pit}{Volume\, of\, plank}$

= (2000Ã—600Ã—80)(500Ã—25Ã—10)$\frac{( 2000 Ã— 600 Ã— 80)}{(500 Ã— 25 Ã— 10)}$= 768

Q.5: How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm Ã— 11.25 cm Ã— 6 cm)?

Solution:

Length of wall = 8 m = 800 cm

Breadth of wall = 6 m = 600 cm

Height of wall = 22.5 cm

Hence, volume =Â l Ã— b Ã— h = (800 Ã— 600 Ã— 22.5) cm3

Length of brick = 25 cm

Breadth of brick = 11.25 cm

Height of brick = 6 cm

Hence, volume = l Ã— b Ã— h = (25 Ã— 11.25 Ã— 6) cm3

âˆ´ Number of bricks required =Â VolumeofwallVolumeofbrick$\frac{Volume\, of\, wall}{Volume\, of\, brick}$

= (800Ã—600Ã—22.5)(25Ã—11.25Ã—6)$\frac{(800 Ã— 600 Ã— 22.5)}{(25 Ã— 11.25 Ã— 6)}$= 6400

Q.6: A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm Ã— 12.5 cm Ã— 7.5 cm). If (1/12) of the total volume of the wall consists of the motor, how many bricks are there in the wall?

Solution:

Length of wall = 15 m

Breadth of wall = 0.3 m

Height of wall = 4 m

Hence, volume =Â l Ã— b Ã— h = (15 Ã— 0.3 Ã— 4) m3 = 18 m3

Volume of motor = 112Ã—18=1.5m3$\frac{1}{12} \times 18 = 1.5 m^{3}$

Volume of wall = (18 â€“ 1.5) m3 = 16.5 = 332m3$\frac{33}{2} m^{3}$

Length of brick = 22 cm

Breadth of brick = 12.5 cm

Height of brick = 7.5 cm

Hence, volume =Â l Ã— b Ã— h

= [22100Ã—12.5100Ã—7.5100]m3$[ \frac{22}{100} \times \frac{12.5}{100} \times \frac{7.5}{100} ] m^{3}$ = 3316000m3$\frac{33}{16000} m^{3}$

Therefore, the number of bricks = VolumeofbricksVolumeof1brick$\frac{Volume\, of\, bricks}{Volume\, of \, 1\, brick}$ = 332Ã—1600033$\frac{33}{2} \times \frac{16000}{33}$Â = 8000

Q.7: An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep. It is made up of iron, which is 2.5 cm thick. Find the capacity of the cistern and the volume of the iron used.

Solution:

External Length of Cistern = 1.35 m = 135 cm

External Breadth of Cistern = 1.08 m = 108 cm

External Height of Cistern = 90 cm

External Volume of cistern = l Ã— b Ã— h = (135 Ã— 108 Ã— 90) cm3 =1312200 cm3

Internal Length of Cistern = (135 â€“ 2 * 2.5) cm = 130 cm

Internal Breadth of Cistern = (108 â€“ 2 * 2.5) cm = 103 cm

Internal Height of Cistern = (90 â€“ 2.5) cm = 87.5 cm

Therefore, the Internal capacity of the cistern = Volume of the Internal cistern

Volume of cistern = l Ã— b Ã— h

= (130 Ã— 103 Ã— 87.5) cm3 =1171625 cm3

Volume of iron used = External Volume of the cistern. â€“ Internal Volume of the citern.

= (1312200 â€“ 1171625) cm3 = 140575 cm3

Q.8: A river 2 m deep and 45 m wide is flowing t the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.

Solution:

Depth of the river = 2 m

Breadth of the river = 45 m

Length of the river = 3 km/h = 3Ã—100060m/min$\frac{3 \times 1000}{60} m/min$ = 50 m/min.

âˆ´ Volume of water running into the sea per minute = (50 Ã— 45 Ã— 2) m3 = 4500 m3

Q.9: A box made of sheet metal costs Rs 1620 at Rs. 30 per square metre. If the box is 5 m long and 3 m wide, find its height.

Solution:

Total cost of sheet = Rs. 1620

Cost of metal sheet per square meter = Rs. 30

âˆ´ Area of the sheet required = [Totalcostrate/m2]sq.m.$[\frac{Total \, cost}{rate/m^{2}}]\, sq.m.$ = [162030]sq.m.$[\frac{1620}{30}]\, sq.m.$ = 54sq.m.$54 sq.m.$

Length of box = 5 m

Breadth of box = 3 m

Now, let the height of the box be x meters.

âˆ´ Area of the sheet = Total surface area of the box = Â 2(lb + bh + hl)

54 =Â 2((5 Ã— 3) + (3 Ã— x) + (x Ã— 5)) m2 = 2(15 + 3x + 5x)

54 = (2 Ã— (15 + 18x))

Solving for x, we get, x = 1.5 m

âˆ´ The height of the box = 1.5 m.

Q.10: Find the length of the longest pole that can be put in a room of dimensions (10 m Ã— 10 m Ã— 5 m).

Solution:

Length of room = 10 m

Breadth of room = 10 m

Height of room = 5 m

âˆ´ Length of the longest pole = Length of diagonal

= l2+b2+h2âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš$\sqrt{l^{2} + b^{2} + h^{2}}$ = 102+102+52âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš$\sqrt{10^{2} + 10^{2} + 5^{2}}$

= 100+100+25âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆ’âˆš$\sqrt{100 + 100 + 25}$ = 225âˆ’âˆ’âˆ’âˆš$\sqrt{225}$ = 15 m

âˆ´ The length of the longest pole that can be put in a room with given dimensions = 15 m.

Q.11: How many persons can be accommodated in a dining hall of dimensions (20 m Ã— 16 m Ã— 5 m)

Solution:

Length of hall = 20 m

Breadth of hall = 16 m

Height of hall = 4.5 m

âˆ´ Volume of hall =Â l Ã— b Ã— h =Â (20 Ã— 16 Ã— 4.5) m3

Volume of air needed per person = 5 m3

âˆ´ Number of persons = VolumeofthehallVolumeofairneededperperson$\frac{Volume\, of\, the\, hall}{Volume\, of\, air\, needed\, per\, person}$

= [(20Ã—16Ã—4.5)5]=288$[\frac{(20 Ã— 16 Ã— 4.5)}{5}] = 288$

Q.12: A classroom is 10 m long, 6.4 m wide and 5m high. If each student is given 1.6 m2 of the floor area, how many cubic meters of air would each student get?

Solution:

Length of classroom = 10 m

Breadth of classroom = 6.4 m

Height of classroom = 5 m

Each student is given 1.6 m2 of the floor area.

Number of students = Areaoftheroom1.6$\frac{Area\, of\, the\, room}{1.6}$ = (10Ã—6.4)1.6=641.6=40$\frac{(10 Ã— 6.4)}{1.6} = \frac{64}{1.6} = 40$

Therefore, the number of students = 40

Hence, air required by each student = Volumeoftheroomnumberofstudents$\frac{Volume\, of\, the\, room}{number\, of\, students}$m3

= 10Ã—6.4Ã—540=32040=8m3$\frac{10 Ã— 6.4 Ã— 5}{40} = \frac{320}{40} = 8 m^{3}$

Q.13: The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

Solution:

Volume of a cuboid = 1536 m3

Length of the cuboid = 16 m

Let the breadth and height of the cuboid be 3x and 2x.

Therefore, the volume of the cuboid = Â Â l Ã— b Ã— h

â‡’$\Rightarrow$ 1536 = (16 Ã— 3x Ã— 2x)

â‡’$\Rightarrow$ 1536 = 96x2

â‡’x2=153696=16$\Rightarrow x^{2} = \frac{1536}{96} = 16$

âˆ´ x=16âˆ’âˆ’âˆš=4m$x Â = \sqrt{16} = 4 m$

âˆ´ Breadth of the cuboid = 3x = 3 Ã— 4 = 12 m

And height of the cuboid = 2x = 2 Ã— 4 = 8 m

Q.14: The surface area of a cuboid is 758 cm2. Its length and breadth are 14 cm and 11 cm respectively. Find its height.

Solution:

Surface area of a cuboid = 758 cm2

Length of the cuboid = 14 cm

Breadth of the cuboid = 11 cm

Let the height of the cuboid be h cm

âˆ´ Surface area of cuboid = 2(lb + bh + hl)

758 = 2((14 Ã— 11) + (11 Ã— h) + (h Ã— 14)) cm2

758 =Â 2(154 + 11h + 14h) cm2

758 = (154 + 25h) cm2

758 = (308 + 50h) cm2

50h = 758 â€“ 308

h=45050=9cm$h = \frac{450}{50} = 9 cm$

âˆ´ The height of the cuboid = 9 cm.

Q.15: Find the volume, lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures (a) 9 m, (b) 6.5 cm [Take 3â€“âˆš=1.73$\sqrt{3} = 1.73$ ]

Solution:

(a) Each edge of a cube = 9 m

âˆ´ Volume of the cube = a3 = (9)3 m3 = 729 m3

âˆ´ Lateral surface area of the cube = 4a2 = 4 * (9)2 = (4 * 81) = 324 m2

âˆ´ Total surface area of the cube = 6a2 = 486 m2

âˆ´ Diagonal of the cube = 3â€“âˆša=3â€“âˆšâˆ—9=15.57m$\sqrt{3} a = \sqrt{3} * 9 = 15.57 m$

(b) Each edge of a cube = 6.5 cm

âˆ´ Volume of the cube = a3 = (6.5)3 m3 = 274.625 cm3

âˆ´ Lateral surface area of the cube = 4a2 = 4 * (6.5)2 = (4 * 42.25) = 169 cm2

âˆ´ Total surface area of the cube = 6a2 = 253.5 cm2

âˆ´ Diagonal of the cube = 3â€“âˆša=3â€“âˆšâˆ—6.5=11.245cm$\sqrt{3} a = \sqrt{3} * 6.5 = 11.245 cm$

Q.16: The total surface area of a cube is 1176 cm2. Find its volume.

Solution:

Let each side of the cube be â€˜aâ€™ cm.

Then, the total surface area of the cube = 6a2 cm2

Given,

6a2 = 1176

â‡’a2=11766=196$\Rightarrow a^{2} = \frac{1176}{6} = 196$
â‡’a=196âˆ’âˆ’âˆ’âˆš=14cm$\Rightarrow a = \sqrt{196} = 14 cm$

Volume of the cube = a3 = (14)3 = 2744 cm3

Q.17: The lateral surface area of a cube is 900 cm2. Find its volume.

Solution:

Let each side of the cube be â€˜aâ€™ cm

Then, the lateral surface area of the cube = 4a2

âˆ´ 4a2 = 900

â‡’a2=9004=225$\Rightarrow a^{2} = \frac{900}{4} = 225$
â‡’a=225âˆ’âˆ’âˆ’âˆš=15cm$\Rightarrow a = \sqrt{225} = 15 cm$

Volume of the cube = a3 = (15)3 = 3375 cm3

Q.18: The volume of a cube is 512 cm3. Find its surface area.

Solution:

Volume of the cube = 512 cm3 [Volume = a3]

âˆ´ Each edge of the cube = 512âˆ’âˆ’âˆ’âˆš3=8cm$\sqrt[3]{512} = 8 cm$

âˆ´ Surface area of the cube = 6a2 = 6(8)2 cm2 = 6(64) cm2 = 384 cm2

Q.19: Three cubes of metal with edges 3 cm, 4 cm, 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

Solution:

Volume of the new cube = [(3)3 + (4)3 + (5)3] cm3 = [27 + 64 + 125] cm3 = 216 cm3

Now, edge of this cube = a cm

And, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a3 = 216

Hence, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a = 6 cm

Lateral surface area of the new cube = 4a2 cm2 = 4 (6)2 cm2 = 144 cm2

âˆ´ The lateral surface area of the new cube formed = 144 cm2

Q.20: In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.

Solution:

1 hectare = 10000Â m2

Area = 2 hectares = 2 Ã— 10000 m2

Depth of the ground = 5 cm = 5100m$\frac{5}{100} m$

Volume of water = (area Ã— depth) = (2Ã—10000Ã—5100)m3$(2 \times 10000 \times \frac{5}{100}) m^{3}$ = 1000 m3

âˆ´ Volume of water that falls = 1000 m3

### RS Aggarwal Class 9 Solutions Chapter 13 – Volume And Surface Area

Students should refer to RS Aggarwal Maths Solutions of Class 9 prepared by subject experts which covers the complete syllabus of the exam and prepared in accordance with the latest curriculum prescribed by CBSE. By referring to these solutions students get to know different methods to solve difficult and tricky questions. The solutions are extremely helpful for performing and scoring good marks in the exam. So, students are advised to practice these solutions before the final exam.