# Class 9 NCERT Solutions Maths Chapter 13 - Surface Areas and Volumes

Revise the most important concepts in the chapter with NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes. In this chapter, TopperLearning gives you access to some of the best textbook solutions to understand the surface area of objects and spaces using the given data. Practise calculating the lateral surface area of a cubical box or the CSA of a given cylindrical pipe with our solutions.

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## Surface Areas and Volumes Exercise Ex. 13.1

### Solution 5

Length of the cuboidal box = 12.5 cm

Breadth of the cuboidal box = 10 cm

Height of the cuboidal box = 8 cm

Lateral surface area of cubical box = = =

Lateral surface area of cuboidal box = 2[lh + bh]

= [2(12.5 8 + 10 8)]

= 360

The lateral surface area of cubical box is greater than lateral surface area of cuboidal box.

Lateral surface area of cubical box - lateral surface area of cuboidal box = 400 - 360 = 40

Thus, the lateral surface area of cubical box is greater than that of cuboidal box by 40 cm

^{2}.

(ii) Total surface area of cubical box = = = 600 cm

^{2}

Total surface area of cuboidal box = 2[lh + bh + lb]

= [2(12.5 8 + 10 8 + 12.5 10] =

The total surface area of cubical box is smaller than that of cuboidal box

Total surface area of cuboidal box - total surface area of cubical box = - =

Thus, the total surface area of cubical box is smaller than that of cuboidal box by

### Solution 6

Breadth of green house = 25 cm

Height of green house = 25 cm

Total surface area of green house = 2[lb + lh + bh]

= [2(30 25 + 30 25 + 25 25)]

= [2(750 + 750 + 625)]

= (2 2125)

= 4250

Thus, the area of the glass is 4250 .

(ii) Total length of tape = 4(l + b + h)

= [4(30 + 25 + 25)] cm

= 320 cm

Therefore, 320 cm tape is needed for all the 12 edges.

### Solution 7

Breadth of bigger box = 20 cm

Height of bigger box = 5 cm

Total surface area of bigger box = 2(lb + lh + bh)

= [2(25 20 + 25 5 + 20 5)]

= [2(500 + 125 + 100)] cm2 = 1450

Extra area required for overlapping = = 72.5

Considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5) =1522.5

Area of cardboard sheet required for 250 such bigger box

= (1522.5 250) = 380625

Total surface area of smaller box = [2(15 12 + 15 5 + 12 5] = [2(180 + 75 + 60)] = (2 315) = 630

Extra area required for overlapping = = 31.5

Considering all overlaps, total surface area of 1 smaller box

= (630 + 31.5) = 661.5

Area of cardboard sheet required for 250 smaller box

= (250 661.5) = 165375

Total cardboard sheet required = (380625 + 165375) = 546000

Cost of 1000 cardboard sheet = Rs 4

Cost of 546000 cardboard sheet = Rs = Rs 2184

So, cost of cardboard sheet required for 250 boxes of each kind will be Rs 2184.

### Solution 8

Breadth of shelter = 3 m

Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + lb

= [2(4 2.5 + 3 2.5) + 4 3]

= [2(10 + 7.5) + 12]

= 47

### Solution 4

= [2(22.5 10 + 10 7.5 + 22.5 7.5)]

= 2(225 + 75 + 168.75)

= (2 468.75)

= 937.5

Let n number of bricks be painted by the container.

Area of n bricks = 937.5n

Area that can be painted by the container = 9.375 m

^{2}= 93750

93750 = 937.5n

n = 100

Thus, 100 bricks can be painted out by the container.

### Solution 1

Breadth of box = 1.25 m

Depth of box = 65 cm = 0.65 m

(i) The box is open at the top.

Area of sheet required = 2bh + 2lh + lb

= [2 1.25 0.65 + 2 1.5 0.65 + 1.5 1.25] m

^{2}

= (1.625 + 1.95 + 1.875) = 5.45

(ii) Cost of sheet of area 1 = Rs 20

### Solution 2

Breadth of room = 4 m

Height of room = 3 m

Area to be white washed = Area of walls + Area of ceiling of room

= 2lh + 2bh + lb

= [2 5 3 + 2 4 3 + 5 4]

= (30 + 24 + 20)

= 74

Cost of white washing 1 area = Rs 7.50

Cost of white washing 74 area = Rs (74 7.50) = Rs 555

### Solution 3

Area of four walls = 2lh + 2bh = 2(l + b) h

Perimeter of floor of hall = 2(l + b) = 250 m

Area of four walls = 2(l + b) h = 250h

Cost of painting 1 area = Rs 10

Cost of painting 250h area = Rs (250h 10) = Rs 2500h

It is given that the cost of paining the walls is Rs 15000.

15000 = 2500h

h = 6

Thus, the height of hall is 6 m.

## Surface Areas and Volumes Exercise Ex. 13.2

### Solution 8

Radius (r) of circular end of pipe = cm = 2.5 cm = 0.025 m

CSA of cylindrical pipe = = 4.4

Thus, the area of radiating surface of the system is 4.4 .

### Solution 9

Radius (r) of circular end of cylindrical tank =m = 2.1m

(i) Lateral or curved surface area of tank =

=

= 59.4 m

^{2}

(ii) Total surface area of tank = 2 (r + h)

=

= 87.12 m

^{2}

Let A m

^{2}steel sheet be actually used in making the tank.

Thus, 95.04 steel was used in actual while making the tank.

### Solution 10

Radius of the circular end of frame of lampshade = cm = 10cm

Cloth required for covering the lampshade =

= 2200

Thus, for covering the lampshade 2200 cloth will be required.

### Solution 11

Height of penholder = 10.5 cm

Surface area of 1 penholder = CSA of penholder + Area of base of

Area of cardboard sheet used by 1 competitor =

Area of cardboard sheet used by 35 competitors

= 7920 cm

^{2}

Thus, 7920 cardboard sheet will be bought for the competition.

### Solution 1

Let diameter of cylinder be d and the radius of its base be r.

Curved surface area of cylinder = 88

Thus, the diameter of the base of the cylinder is 2 cm.

### Solution 2

Base radius (r) of cylindrical tank = = 70 cm = 0.7 m

Area of sheet required = total surface area of tank =

So, it will require 7.48 area of sheet.

### Solution 3

Outer radius of cylindrical pipe = 2.2 cm

Height (h) of cylindrical pipe = length of cylindrical pipe = 77 cm

(i) CSA of inner surface of pipe =

(ii) CSA of outer surface of pipe =

=

Thus, the total surface area of cylindrical pipe is 2038.08 .

### Solution 4

Height of the roller = length of roller = 120 cm

Radius of the circular end of the roller =

CSA of roller =

Area of field = 500 CSA of roller = (500 31680) = 15840000

= 1584

### Solution 5

Radius of the circular end of the pillar = cm = 25 cm = 0.25 m

CSA of pillar = =

Cost of painting 1 area = Rs 12.50

Cost of painting 5.5 area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

### Solution 6

Radius of the base of the cylinder = 0.7 m

CSA of cylinder = 4.4

= 4.4

h = 1 m

Thus, the height of the cylinder is 1 m.

### Solution 7

Depth (h) of circular well = 10 m

= (44 0.25 10)

= 110

(ii) Cost of plastering 1 area = Rs 40

Cost of plastering 110 area = Rs (110 40) = Rs 4400

## Surface Areas and Volumes Exercise Ex. 13.3

### Solution 1

Slant height of cone = 10 cm

CSA of cone = =

Thus, the curved surface area of cone is 165 .

### Solution 2

Slant height of cone = 21 m

Total surface area of cone = (r + l)

### Solution 3

Let radius of circular end of cone be r.

CSA of cone =

Thus, the radius of circular end of the cone is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base

=

### Solution 4

Radius (r) of conical tent = 24 m

Let slant height of conical tent be l.

(ii) CSA of tent = =

Cost of 1 canvas = Rs 70

Cost of canvas = = Rs 137280

Thus, the cost of canvas required to make the tent is Rs 137280.

### Solution 5

Radius (r) of base of tent = 6 m

Slant height (l) of tent =

CSA of conical tent = = (3.14 6 10) = 188.4

Let length of tarpaulin sheet required be L.

As 20 cm will be wasted so, effective length will be (L - 0.2 m)

Breadth of tarpaulin = 3 m

Area of sheet = CSA of tent

[(L - 0.2 m) 3] m = 188.4

L - 0.2 m = 62.8 m

L = 63 m

Thus, the length of the tarpaulin sheet will be 63 m.

### Solution 6

Base radius (r) of tomb = = 7 m

CSA of conical tomb = =

Cost of white-washing 100 area = Rs 210

Cost of white-washing 550 area =Rs = Rs 1155

Thus, the cost of white washing the conical tomb is Rs 1155.

### Solution 7

Height (h) of conical cap = 24 cm

Slant height (l) of conical cap =

CSA of 1 conical cap = =

CSA of 10 such conical caps = (10 550) = 5500

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

### Solution 8

Height (h) of cone = 1 m

Slant height (l) of cone =

CSA of each cone = = (3.14 0.2 1.02) = 0.64056

CSA of 50 such cones = (50 0.64056) = 32.028

Cost of painting 1 area = Rs 12

Cost of painting 32.028 area = Rs (32.028 12) = Rs 384.336

Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.

## Surface Areas and Volumes Exercise Ex. 13.4

### Solution 2

(i) Radius of sphere

Surface area of sphere

(ii) Radius of sphere

Surface area of sphere

(iii) Radius of sphere

Surface area of sphere =

### Solution 3

Total surface area of hemisphere

### Solution 4

Radius of spherical balloon, when air is pumped into it = 14 cm

### Solution 5

Surface area of hemispherical bowl =

Cost of tin-plating 100 area = Rs 16

Cost of tin-plating 173.25 area = Rs 27.72

### Solution 6

Surface area of the sphere = 154

= 154 cm2

Thus, the radius of the sphere is 3.5 cm.

### Solution 7

Radius of earth =

Radius of moon =

Surface area of moon =

Surface area of earth =

Required ratio =

Thus, the required ratio of the surface areas is 1:16.

### Solution 8

Thickness of the bowl = 0.25 cm

Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm

Outer CSA of hemispherical bowl =

Thus, the outer curved surface area of the bowl is 173.25 .

### Solution 1

Surface area of sphere =

(ii) Radius of sphere = 5.6 cm

Surface area of sphere = =

(iii) Radius of sphere = 14 cm

Surface area of sphere = =

### Solution 9

(ii) Height of cylinder = r + r = 2r

Radius of cylinder = r

CSA of cylinder =

(iii) Required ratio =

## Surface Areas and Volumes Exercise Ex. 13.5

### Solution 1

Volume of 1 match box = l b h = (4 2.5 1.5) = 15

Volume of the packet containing 12 such matchboxes = 12 15 =180

### Solution 2

1 = 1000 litres

Thus, the tank can hold 135000 litres of water.

### Solution 3

Length (l) of vessel = 10 m

Width (b) of vessel = 8 m

Volume of vessel = 380

l b h = 380

10 8 h = 380

h = 4.75

Thus, the height of the vessel should be 4.75 m.

### Solution 4

Width (b) of the cuboidal pit = 6 m

Depth (h) of the cuboidal pit = 3 m

Volume of the cuboidal pit = l b h = (8 6 3) = 144

Cost of digging 1 = Rs 30

Cost of digging 144 = Rs (144 30) = Rs 4320

### Solution 5

Length (l) of the tank = 2.5 m

Depth (h) of the tank = 10 m

Capacity of tank = 25b = 25000 b litres

25000 b = 50000 (Given)

### Solution 6

Breadth (b) of the cuboidal tank = 15 m

Height (h) of the cuboidal tank = 6 m

Capacity of tank = l bh = (20 15 6) = 1800 = 1800000 litres

Water consumed by people of village in 1 day = 4000 150 litres = 600000 litres

Let water of this tank lasts for n days.

Water consumed by all people of village in n days = capacity of tank

n 600000 = 1800000

n = 3

Thus, the water of tank will last for 3 days.

### Solution 7

_{1}) of the godown = 40 m

Breadth (b_{1}) of the godown = 25 m

Height (h_{1}) of the godown = 15 m

Volume of the godown = l_{1} × b_{1} × h_{1} = 40 × 25 × 15 = 15000m^{3}

Length (l_{2}) of a wooden crate = 40 m

Breadth (b_{2}) of a wooden crate = 25 m

Height (h_{2}) of a wooden crate = 15 m

Volume of a wooden crate = l_{2} × b_{2} × h_{2} = 1.5 × 1.25 × 0.5 = 0.9375m^{3}

Let n wooden crates be stored in the godown

Volume of n wooden crates = volume of the godown

⇒0.9375 × n = 15000

⇒ n = 16000

Thus, the number of wooden crates that can be stored in the godown is 16000.

### Solution 8

Volume of the cube = a

^{3}= (12 cm)

^{3}= 1728

Let the side of each smaller cube be l.

l = 6 cm

Thus, the side of each smaller cube is 6 cm.

Ratio between surface areas of the cubes =

So, the required ratio between surface areas of the cubes is 4 : 1.

### Solution 9

Depth (h) of river = 3 m

Width (b) of river = 40 m

Volume of water flowed in 1 min

Thus, in 1 minute 4000 water will fall into the sea.

## Surface Areas and Volumes Exercise Ex. 13.6

### Solution 1

Height (h) of the vessel = 25 cm

Circumference of the vessel = 132 cm

2r = 132 cm

^{2}h

### Solution 2

Volume of pipe =

^{3}wood = 0.6 g

Mass of 5720 cm

^{3}wood = 5720 0.6 g = 3432 g = 3.432 kg

### Solution 3

Breadth (b) of tin can = 4 cm

Height (h) of tin can = 15 cm

Capacity of tin can = l b h = (5 4 15) cm

^{3}= 300 cm

^{3}

^{2}H ==385 cm

^{3}

Difference in capacity = (385 - 300) cm

^{3}= 85 cm

^{3}

### Solution 4

Let radius of cylinder be r.

CSA of cylinder = 94.2 cm

^{2}

2rh = 94.2 cm

^{2}

(2 3.14 r 5) cm = 94.2 cm

^{2}

r = 3 cm

^{2}h = (3.14 (3)

^{2}5) cm

^{3}= 141.3 cm

^{3}

### Solution 5

^{2}area = Rs 20

So, Rs 2200 is cost of painting area , i.e, 110 m

^{2}area.

Thus, the inner surface area of the vessel is 110 m

^{2}.

(ii) Let radius of base of vessel be r.

Height (h) of vessel = 10 m

Surface area = 2rh = 110 m

^{2}

^{2}h = = 96.25 m

^{3}

### Solution 6

Height (h) of the cylindrical vessel = 1 m

Volume of cylindrical vessel = 15.4 litres = 0.0154 m

^{3}

^{2}of metal sheet would be needed to make the cylindrical vessel.

### Solution 7

_{1}) of pencil = = 0.35 cm

Radius (r

_{2}) of graphite =

Height (h) of pencil = 14 cm

Volume of wood in pencil =

^{3}

### Solution 8

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

^{2}h=

Volume of soup in 250 bowls = (250 154) cm

^{3}= 38500 cm

^{3}= 38.5 litres

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

## Surface Areas and Volumes Exercise Ex. 13.7

### Solution 1

Height (h) of cone = 7 cm

Volume of cone

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

Height (h) of cone = 12 cm

Volume of cone

### Solution 2

Slant height (l) of cone = 25 cm

Height (h) of cone

Volume of cone

Capacity of the conical vessel = litres= 1.232 litres

Slant height (l) of cone = 13 cm

Radius (r) of cone

Volume of cone

Capacity of the conical vessel = litres = litres.

### Solution 3

Let radius of cone be r.

Volume of cone = 1570 cm

^{3}

Thus, the radius of the base of the cone is 10 cm.

### Solution 4

Let radius of cone be r.

Volume of cone = 48 cm3

Thus, the diameter of the base of the cone is 2r = 8 cm.

### Solution 5

Depth (h) of pit = 12 m

Volume of pit = = 38.5 m

^{3}

### Solution 6

Let height of cone be h.

Volume of cone = 9856 cm

^{3}

Thus, the slant height of the cone is 50 cm.

^{2}

### Solution 7

Volume of cone = 100 cm

^{3}

Thus, the volume of cone so formed by the triangle is 100 cm

^{3}.

### Solution 8

Volume of cone =

### Solution 9

Height (h) of heap = 3 m

Volume of heap=

Slant height (l) =

Area of canvas required = CSA of cone

## Surface Areas and Volumes Exercise Ex. 13.8

### Solution 1

Volume of sphere =

Volume of sphere =

^{3}(approximately)

### Solution 2

Volume of ball =

Thus, the amount of water displaced is .

Volume of ball =

Thus, the amount of water displaced is 0.004851 m

^{3}.

### Solution 3

Volume of metallic ball =

Mass = Density Volume = (8.9 * 38.808) g = 345.3912 g

Thus, the mass of the ball is approximately 345.39 g.

### Solution 4

Then, diameter of moon will be . So, radius of moon will be .

Volume of moon =

Volume of earth =

Thus, the volume of moon is of volume of earth.

### Solution 5

= 303.1875 cm

^{3}

Capacity of the bowl

= 0.3031875 litre = 0.303 litre (approximately)

Thus, the hemispherical bowl can hold 0.303 litre of milk.

### Solution 6

_{1}) of hemispherical tank = 1 m

Thickness of hemispherical tank = 1 cm = 0.01 m

Outer radius (r

_{2}) of hemispherical tank = (1 + 0.01) m = 1.01 m

### Solution 7

Surface area of sphere = 154 cm

^{2}

r

^{2}= 154 cm

^{2}

Volume of sphere

### Solution 8

Cost of white washing 1 m

^{2}area = Rs 2

CSA of inner side of dome = = 249.48 m

^{2}

CSA of inner side of dome = 249.48 m

^{2}

2r

^{2}= 249.48 m

^{2}

Volume of air inside the dome = Volume of the hemispherical dome

= 523.908 m

^{3}

Thus, the volume of air inside the dome is approximately 523.9 m

^{3}.

### Solution 9

Volume of 1 solid iron sphere

Volume of 27 solid iron spheres

It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of

Radius of the new sphere = r'.

Volume of new sphere

^{2}

Surface area of iron sphere of radius r' = 4 (r')

^{2}= 4 (3r)

^{2}= 36 r

^{2}

### Solution 10

Volume of spherical capsule

Thus, approximately 22.46 mm

^{3}of medicine is required to fill the capsule.

## Surface Areas and Volumes Exercise Ex. 13.9

### Solution 1

External breadth (b) of bookshelf = 25 cm

External height (h) of bookshelf = 110 cm

External surface area of shelf while leaving front face of shelf

= lh + 2 (lb + bh)

= [85 110 + 2 (85 25 + 25 110)] cm

^{2}

= 19100 cm

^{2}

Area of front face = [85 110 - 75 100 + 2 (75 5)] cm

^{2}

= 1850 + 750 cm

^{2}

= 2600 cm

^{2}

Area to be polished = (19100 + 2600) cm2 = 21700 cm

^{2}

Cost of polishing 1 cm

^{2}area = Rs 0.20

Cost of polishing 21700 cm

^{2}area = Rs (21700 0.20) = Rs 4340

Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and

Area to be painted in 1 row = 2 (l + h) b + lh

= [2 (75 + 30) 20 + 75 30] cm

^{2}

= (4200 + 2250) cm

^{2}

= 6450 cm

^{2}

Area to be painted in 3 rows = (3 6450) cm

^{2}= 19350 cm

^{2}

Cost of painting 1 cm

^{2}area = Rs 0.10

Cost of painting 19350 cm

^{2}area = Rs (19350 0.10) = Rs 1935

Total expense required for polishing and painting the surface of the bookshelf

### Solution 2

Radius (r') of cylindrical support = 1.5 cm

Height (h) of cylindrical support = 7 cm

Area of circular end of cylindrical support = r

^{2}

^{2}

Area to be painted silver = [8 (1386 - 7.07)] cm

^{2}

= (8 1378.93) cm

^{2}= 11031.44 cm

^{2}

Cost occurred in painting silver colour = Rs (11031.44 0.25) = Rs 2757.86

Area to painted black = (8 66) cm

^{2}= 528 cm

^{2}

Cost occurred in painting black colour = Rs (528 0.05) = Rs 26.40

Total cost occurred in painting = Rs (2757.86 + 26.40) = Rs 2784.26

### Solution 3

Radius (r

_{1}) of the sphere =

It is given that the diameter of the sphere is decreased by 25%.

New radus (r

_{2}) of the sphere =

_{1}) of the sphere =

CSA (S

_{2}) of the new sphere =

Decrease in CSA of sphere = S

_{1}- S

_{2}

Percentage decrease in CSA of sphere=