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# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids

Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 15 - Volume and Surface Area of Solids.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids Page/Excercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Correct option: (b)

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64

Solution 65

Solution 66

Solution 67

Solution 68

Solution 69

Solution 70

Solution 71

Solution 72

Solution 73

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids Page/Excercise 15A

Solution 1(i)

length =12cm, breadth = 8 cm and height = 4.5 cm

Volume of cuboid = l x b x h

= (12 x 8 x 4.5) cm3= 432 cm3

Lateral surface area of a cuboid = 2(l + b) x h

= [2(12 + 8) x 4.5] cm2

= (2 x 20 x 4.5) cm2 = 180 cm2

Total surface area cuboid = 2(lb +b h+ l h)

= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm2

= 2(96 +36 +54) cm2

= (2 x186) cm2

= 372 cm2

Solution 1(iv)

Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.

Volume of cuboid= l x b x h

= (24 x 0.25 x 6) m3.

= 36 m3.

Lateral surface area= 2(l + b) x h

= [2(24 +0.25) x 6] m2

= (2 x 24.25 x 6) m2

= 291 m2.

Total surface area =2(lb+ bh + lh)

=2(24 x 0.25+0.25x 6 +24 x 6) m2

= 2(6+1.5+144) m2

= (2 x151.5) m2=303 m2.

Solution 1(iii)

Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m

Volume of a cuboid = l x b x h

= (15 x 6 x 0.5) m3=45 m3.

Lateral surface area = 2(l + b) x h

= [2(15 + 6) x 0.5] m2

= (2 x 21x0.5) m2=21 m2

Total surface area =2(lb+ bh + lh)

= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m2

= 2(90+3+7.5) m2

= (2 x 100.5) m2

=201 m2

Solution 1(ii)

Length 26 m, breadth =14 m and height =6.5 m

Volume of a cuboid= l x b x h

= (26 x 14 x 6.5) m3

= 2366 m3

Lateral surface area of a cuboid =2 (l + b) x h

= [2(26+14) x 6.5] m2

= (2 x 40 x 6.5) m2

= 520 m2

Total surface area= 2(lb+ bh + lh)

= 2(26 x 14+14 x6.5 +26 x6.5)

= 2 (364+91+169) m2

= (2 x 624) m2= 1248 m2.

Solution 2

For a matchbox,

Length = 4 cm

Height = 1.5 cm

Volume of one matchbox = Volume of cuboid

= Length × Breadth × Height

= (4 × 2.5 × 1.5) cm3

= 15 cm3

Hence, volume of 12 such matchboxes = 12 × 15 = 180 cm3

Solution 3

For a cuboidal water tank,

Length = 6 m

Height = 4.5 m

Now,

Volume of a cuboidal water tank = Length × Breadth × Height

= (6 × 5 × 4.5) m3

= 135 m3

= 135 × 1000 litres

= 135000 litres

Thus, a tank can hold 135000 litres of water.

Solution 4

For a cuboidal water tank,

Length = 10 m

Volume = 50000 litres = 50 m3

Now,

Volume of a cuboidal tank = Length × Breadth × Height

50 = 10 × 2.5 × Height

Height = 2 m = Depth

Thus, the depth of a tank is 2 m.

Solution 5

For a godown,

Length = 40 m

Height = 15 m

Volume of a godown = Length × Breadth × Height

= (40 × 25 × 15) m3

For each wooden crate,

Length = 1.5 m

Height = 0.5 m

Volume of each wooden crate = Length × Breadth × Height

= (1.5 × 1.25 × 0.5) m3

Solution 6

Solution 7

Solution 8

Length of Cistern = 8 m

Breadth of Cistern = 6 m

And Height (depth) of Cistern =2.5 m

Capacity of the Cistern = Volume of cistern

Volume of Cistern = (l x b x h)

= (8 x 6 x2.5) m3

=120 m3

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

Total surface area = 2(lb +bh +lh)

= 2(8 x 6 + 6x2.5+ 2.5x8) m2

= 2(48 + 15 + 20) m2

= (2 x 83) m2=166 m2

Solution 9

Area of four walls of the room = 2(length + breadth) × Height

= [2(9 + 8) × 6.5] m2

= (34 × 6.5) m2

= 221 m2

Area of one door = Length × Breadth = (2 × 1.5) m2 = 3 m2

Area of two windows = 2 × (Length × Breadth)

= [2 × (1.5 × 1)] m2

= (2 × 1.5) m2

= 3 m2

Area to be whitewashed

= Area of four walls of the room - Area of one door - Area of two windows

= (221 - 3 - 3) m2

= 215 m2

Cost of whitewashing = Rs. 25 per square metre

Cost of whitewashing 215 m2 = Rs. (25 × 215) = Rs. 5375

Solution 10

L

Solution 11

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

External volume of the box = (36 × 25 × 16.5) cm3 = 14850 cm3

Internal length of the box = [36 - (1.5 × 2)] cm = 33 cm

Internal breadth of the box = [25 - (1.5 × 2)] cm = 22 cm

Internal height of the box = (16.5 - 1.5) cm = 15 cm

Internal volume of the box = (33 × 22 × 15) cm3 = 10890 cm3

Thus, volume of iron used in the box

= External volume of the box - Internal volume of the box

= (14850 - 10890) cm3

= 3960 cm3

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Let the edge of the cube = 'a' cm

Then, surface area of cube = 6a2 cm2

Solution 26

Solution 27

Solution 28

Volume of a cuboid = (9 × 8 × 2) m3 = 144 m3

Volume of each cube of edge 2 m = (2 m)3 = 8 m3

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids Page/Excercise 15B

Solution 1

Solution 2

Radius (r) of cylindrical bowl =

Height (h) up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = pr2h  = 154 cm3

Hence, volume of soup in 250 bowls = (250 × 154) cm3 = 38500 cm3 = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

Solution 3

Radius (r) of pillar = 20 cm =  m

Height (h) of pillar = 10 m

Solution 4

For a tin can of rectangular base,

Length = 5 cm

Height = 15 cm

Volume of a tin can = Length × Breadth × Height

= (5 × 4 × 15) cm3

= 300 cm3

For a cylinder with circular base,

Diameter = 7 Radius = r = cm

Height = h = 10 cm

Volume of plastic cylinder is greater than volume of a tin can.

Difference in volume = (385 - 300) = 85 cm3

Thus, a plastic cylinder has more capacity that a tin can by 85 cm3.

Solution 5

Radius (r) of 1 pillar =

Height (h) of 1 pillar = 4 m

Solution 6

Curved surface area of a cylinder = 4.4 m2

Radius (r) of a cylinder = 0.7 m

Solution 7

Lateral surface area of a cylinder = 94.2 cm2

Height (h) of a cylinder = 5 cm

Solution 8

Volume of a cylinder = 15.4 litres = 15400 cm3

Height (h) of a cylinder = 1 m = 100 cm

Solution 9

Internal diameter of a cylinder = 24 cm

Internal radius of a cylinder, r = 12 cm

External diameter of a cylinder = 28 cm

External radius of a cylinder, R = 14 cm

Length of the pipe, i.e height, h = 35 cm

Solution 10

Diameter of a cylindrical pipe = 5 cm

Radius (r) of a cylindrical pipe = 2.5 cm

Height (h) of a cylindrical pipe = 28 m = 2800 cm

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Diameter of a cylinder = 140 cm

Height (h) of a cylinder = 1 m = 100 cm

Solution 24

Radius (r) of cylindrical vessel = 15 cm

Height (h) of cylindrical vessel = 32 m

Radius of small cylindrical glass = 3 cm

Height of a small cylindrical glass = 8 cm

Solution 25

Radius of the well = 5 m

Depth of the well = 8.4 m

Width of the embankment = 7.5 m

External radius of the embankment, R = (5 + 7.5) m = 12.5 m

Internal radius of the embankment, r = 5 m

Area of the embankment = π (R2 - r2)

Volume of the embankment = Volume of the earth dug out = 660 m2

Solution 26

Speed of water = 30 cm/sec

Volume of water that flows out of the pipe in one second

= Area of cross-section × Length of water flown in one second

= (5 × 30) cm3

= 150 cm3

Hence, volume of water that flows out of the pipe in 1 minute

= (150 × 60) cm3

= 9000 cm3

= 9 litres

Solution 27

Suppose the tank is filled in x minutes. Then,

Volume of the water that flows out through the pipe in x minutes

= Volume of the tank

Hence, the tank will be filled in 28 minutes.

Solution 28

Let the rise in the level of water = h cm

Then,

Volume of the cylinder of height h and base radius 28 cm

= Volume of rectangular iron solid

Thus, the rise in the level of water is 4 cm.

Solution 29

Height, h = 280 m

Solution 30

Let the length of the wire = 'h' metres

Then,

Volume of the wire × 8.4 g = (13.2 × 1000) g

Thus, the length of the wire is 125 m.

Solution 31

Solution 32

Let R cm and r cm be the outer and inner radii of the cylindrical tube.

We have, length of tube = h = 14 cm

Now,

Outside surface area - Inner surface area = 88 cm2

2πRh - 2πrh = 88

2π(R - r)h = 88

It is given that the volume of the tube = 176 cm3

External volume - Internal volume = 176 cm3

πR2h - πr2h = 176

π (R2 - r2)h = 176

Adding (i) and (ii), we get

2R = 5

R = 2.5 cm

2.5 - r = 1

r = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm respectively.

Solution 33

When the sheet is folded along its length, it forms a cylinder of height, h1 = 18 cm and perimeter of base equal to 30 cm.

Let r1 be the radius of the base and V1 be is volume.

Then,

Again, when the sheet is folded along its breadth, it forms a cylinder of height, h2 = 30 cm and perimeter of base equal to 18 cm.

Let r2 be the radius of the base and V2 be is volume.

Then,

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids Page/Excercise 15C

Solution 1

Radius of a cone, r = 5.25 cm

Slant height of a cone, l = 10 cm

Solution 2

Radius of a cone, r = 12 m

Slant height of a cone, l = 21 cm

Solution 3

Radius of a conical cap, r = 7 cm

Height of a conical cap, h = 24 cm

Thus, 5500 cm2 sheet will be required to make 10 caps.

Solution 4

Let r be the radius of a cone.

Slant height of a cone, l = 14 cm

Curved surface area of a cone = 308 cm2

Solution 5

Radius of a cone, r = 7 m

Slant height of a cone, l = 25 m

Cost of whitewashing = Rs. 12 per m2

Cost of whitewashing 550 m2 area = Rs. (12 × 550) = Rs. 6600

Solution 6

Radius of a conical tent, r = 24 m

Height of a conical tent, h = 10 m

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Radius of a conical heap, r = 4.5 m

Height of a conical tent, h = 3.5 m

Solution 12

Radius of a conical tent, r = 7 m

Area of canvas used in making conical tent = (551 - 1) m2 = 550 m2

Curved surface area of a conical tent = 550 m2

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Curved surface area of the tent = Area of the cloth = 165 m2

## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 15 - Volume and Surface Area of Solids Page/Excercise 15D

Solution 1(iii)

Solution 1(ii)

Solution 1(i)

Solution 2

Solution 3

Solution 4

Surface area of sphere = 154 cm2

4πr2 = 154

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12