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# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes| PDF Download

On this page you will find Chapter 13 Surface Areas and Volumes Class 9 Maths NCERT Solutions which are very important in knowing the essential topics present in the chapter. You can download PDF of NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes that will make learning exciting and cover the syllabus in less time. It will help you in rectifying all those mistakes that can happen in the exams.

These NCERT Solutions for Class 9 which are accurate and detailed are prepared by Studyrankers experts help you in building your own answers for homework and also make you aware of the difficulty of questions. Page No: 213

Exercise 13.1

1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs Rs 20.

Length of plastic box (l) = 1.5 m
Width of plastic box (b) = 1.25 m
Depth of plastic box (h) = 0.65 m
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(l+b)×h + (l×b)
= 2[(1.5 + 1.25)×0.65] + (1.5 × 1.25) m
= (3.575 + 1.875) m
= 5.45 m
The sheet required required to make the box is 5.45 m

(ii) Cost of 1 mof sheet = Rs 20
∴ Cost of 5.45 mof sheet = Rs (20 × 5.45) = Rs 109

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 per m2.

length of the room = 5m
breadth of the room = 4m
height of the room = 3m
Area of four walls including the ceiling = 2(l+b)×h + (l×b)
= 2(5+4)×3 + (5×4) m
= (54 + 20) m
= 74 mCost of white washing = ₹7.50 per m2
Total cost = ₹ (74×7.50) = ₹ 555

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m2 is ₹15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]

Perimeter of rectangular hall = 2(l + b) = 250 m
Total cost of painting = 15000
Rate per m= ₹10
Area of four walls = 2(l + b) h m= (250×h) m2
A/q,
(250×h)×10 = 15000
⇒ 2500×h = 15000
⇒ h = 15000/2500 m
⇒ h = 6 m
Thus the height of the hall is 6 m.

4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Volume of paint = 9.375 m= 93750 cm2
Dimension of brick = 22.5 cm×10 cm×7.5 cm
Total surface area of a brick = 2(lb + bh + lh) cm2
= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm2
= 2(225 + 75 + 168.75) cm2
= 2×468.75 cm2 = 937.5 cm2
Number of bricks can be painted = 93750/937.5 = 100

5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?

(i) Lateral surface area of cubical box of edge 10cm = 4×102 cm2 = 400 cm2
Lateral surface area of cuboid box = 2(l+b)×h
= 2×(12.5+10)×8 cm2
= 2×22.5×8 cm2 = 360 cm2
Thus, lateral surface area of the cubical box is greater by (400 – 360) cm2 = 40 cm2

(ii) Total surface area of cubical box of edge 10 cm =6×102cm2=600cm2
Total surface area of cuboidal box = 2(lb + bh + lh)
= 2(12.5×10 + 10×8 + 8×12.5) cm2
= 2(125+80+100) cm2
= (2×305) cm= 610 cm2
Thus, total surface area of cubical box is smaller by 10 cm2

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?

(i) Dimensions of greenhouse:
l = 30 cm, b = 25 cm, h = 25 cm
Total surface area of green house = 2(lb + bh + lh)
= 2(30×25 + 25×25 + 25×30) cm2
= 2(750 + 625 + 750) cm2
= 4250 cm2

(ii) Length of the tape needed = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4×80 cm = 320 cm

7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is 4 for 1000 cm2 , find the cost of cardboard required for supplying 250 boxes of each kind.

Dimension of bigger box = 25 cm × 20 cm × 5 cm
Total surface area of bigger box = 2(lb + bh + lh)
= 2(25×20 + 20×5 + 25×5) cm2
= 2(500 + 100 + 125) cm2
= 1450 cm2

Dimension of smaller box = 15 cm × 12 cm × 5 cm
Total surface area of smaller box = 2(lb + bh + lh)
= 2(15×12 + 12×5 + 15×5) cm2
= 2(180 + 60 + 75) cm2
= 630 cm2

Total surface area of 250 boxes of each type = 250(1450 + 630) cm2
= 250×2080 cm= 520000 cm2
Extra area required = 5/100(1450 + 630) × 250 cm= 26000 cm2

Total Cardboard required = 520000 + 26000 cm2 = 546000 cm2
Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 2184

8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3m?

Dimensions of the box- like structure = 4m × 3m × 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(l+b)×h + lb = [2(4+3)×2.5 + 4×3] m2
= (35+12) m2
= 47 m2
Page No: 216

Exercise 13.2

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.

Let r be the radius of the base and h = 14 cm be the height of the cylinder.
Curved surface area of cylinder = 2πrh = 88 cm2
⇒ 2 × 22/7 × r × 14 = 88
⇒ r = 88/ (2 × 22/7 × 14)
⇒ r = 1 cm
Thus, the diameter of the base = 2r = 2×1 = 2cm

2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm and Height (h) = 1m
Radius of base (r) = 140/2 = 70 cm = 0.7 m
Metal sheet required to make a closed cylindrical tank = 2πr(h + r)
= (2 × 22/7 × 0.7) (1 + 0.7) m2
= (2 × 22 × 0.1 × 1.7) m2
=7.48 m2

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area. Let R be external radius and r be the internal radius h be the length of the pipe.
R  = 4.4/2 cm = 2.2 cm
r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm2
= 2 × 22/7 × 2 × 77cm2
= 968 cm2

(ii) Outer curved surface = 2πRh cm2
= 2 × 22/7 × 2.2 × 77 cm2
= 1064.8 cm2

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh + 2πRh + 2π(R2 - r2)
= [968 + 1064.8 + (2 × 22/7) (4.84 - 4)] cm2
= (2032.8 + 44/7 × 0.84) cm2
= (2032.8 + 5.28) cm= 2038.08 cm2

Page No: 217

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.

Length of the roller (h) = 120 cm = 1.2 m
Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2πrh
= (2 × 22/7 × 0.42 × 1.2) m2 = 3.168 m2
Area of the playground = (500 × 3.168) m= 1584 m2

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of 12.50 per m2.

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m
Height of the pillar (h) = 3.5 m.
Rate of painting = 12.50 per m2
Curved surface = 2πrh
= (2 × 22/7 × 0.25 × 3.5) m2
=5.5 m2
Total cost of painting = (5.5 × 12.5) = 68.75

6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.

Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2πrh = 4.4 m2
⇒ 2 × 22/7 × 0.7 × h = 4.4
⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m
⇒ h = 1m

7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of 40 per m2.

Radius of circular well (r) = 3.5/2 m = 1.75 m
Depth of the well (h) = 10 m
Rate of plastering = 40 per m2
(i) Curved surface = 2πrh
= (2 × 22/7 × 1.75 × 10) m2
= 110 m2

(ii) Cost of plastering = (110 × 40) = 4400

8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m
Length of the pipe (h) = 28/2 m = 14 m
Total radiating surface = Curved surface area of the pipe = 2πrh
= (2 × 22/7 × 0.025 × 28) m2 = 4.4 m2

9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m
Height of the tank (h) = 4.5 m
Curved surface area = 2πrh m2
= (2 × 22/7 × 2.1 × 4.5) m2
= 59.4 m2

(ii) Total surface area of the tank = 2πr(r + h) m2
= [2 × 22/7 × 2.1 (2.1 + 4.5)] m2
= 87.12 m2

Let x be the actual steel used in making tank.
∴ (1 - 1/12) × x = 87.12
⇒ x = 87.12 × 12/11
⇒ x = 95.04 m2

10. In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. Radius of the frame (r) = 20/2 cm = 10 cm
Height of the frame (h) = 30 cm + 2×2.5 cm = 35 cm
2.5 cm of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2πrh
= (2 × 22/7 × 10 × 35)cm2
= 2200 cm2

11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Radius of the penholder (r) = 3cm
Height of the penholder (h) = 10.5cm
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2πrh + πr2
= [(2 × 22/7 × 3 × 10.5) + 22/7 × 32] cm2
= (198 + 198/7) cm2
= 1584/7 cm2
Cardboard required for 35 competitors = (35 × 1584/7) cm2
= 7920 cm2

Page No: 221

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Radius (r) = 10.5/2 cm = 5.25 cm
Slant height (l) = 10 cm
Curved surface area of the cone = (πrl) cm2
= (22/7 × 5.25 × 10) cm2
=165 cm2

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Radius (r) = 24/2 m = 12 m
Slant height (l) = 21 m
Total surface area of the cone = πr (l + r) m2
= 22/7 × 12 × (21 + 12) m2
= (22/7 × 12 × 33) m2
= 1244.57 m2

3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.

(i) Curved surface of a cone = 308 cm2
Slant height (l) = 14cm
Let r be the radius of the base
∴ πrℓ = 308
⇒ 22/7 × r × 14 = 308
⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm2
= 22/7 × 7 ×(14 + 7) cm2
= (22 × 21) cm2
= 462 cm2

4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is 70.

Radius of the base (r) = 24 m
Height of the conical tent (h) = 10 m
Let l be the slant height of the cone.
∴ l= h+ r2
⇒ l = √h+ r2
⇒ l = √10+ 242
⇒ l = √100 + 576
⇒ l = 26 m
(ii) Canvas required to make the conical tent = Curved surface of the cone
Cost of 1 m2 canvas = 70
∴ πrl = (22/7 × 24 × 26) m2 = 13728/7 m2
∴ Cost of canvas = ₹ 13728/7 × 70 = 137280

5. What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π = 3.14).

Radius of the base (r) = 6 m
Height of the conical tent (h) = 8 m
Let l be the slant height of the cone.
∴ l = √h+ r2
⇒ l = √8+ 62
⇒ l = √100
⇒ l = 10 m
CSA of conical tent = πrl
= (3.14 × 6 × 10) m2 = 188.4 m2
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be x.
20 cm will be wasted in cutting.
So, the length will be (x - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(x - 0.2 m) × 3] m = 188.4 m2
⇒ x - 0.2 m = 62.8 m
⇒ x = 63 m

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of 210 per 100 m2.

Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = πrl m2
=(227×25×7) m2
=550  m2
Rate of white- washing = 210 per 100 m2
Total cost of white-washing the tomb = (550 × 210/100) = 1155

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Radius of the cone (r) = 7 cm
Height of the cone (h) = 24 cm
Let l be the slant height
∴ l = √h+ r2
⇒ l = √24+ 72
⇒ l = √625
⇒ l = 25 m
Sheet required for one cap = Curved surface of the cone
= πrl cm2
= (22/7 × 7 × 25) cm2
= 550 cm2
Sheet required for 10 caps = 550 × 10 cm= 5500 cm2

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is 12 per m2 , what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m
Height of the cone (h) = 1 m
Let l  be the slant height of a cone.
∴ l = √h+ r2
⇒ l = √1+ 0.22
⇒ l = √1.04
⇒ l = 1.02 m
Rate of painting = 12 per m2

Curved surface of 1 cone = πrl m2
= (3.14 × 0.2 × 1.02) m2
= 0.64056 m2
Curved surface of such 50 cones = (50 × 0.64056) m2 = 32.028 m2
Cost of painting all these cones = (32.028 × 12) = 384.34

Page No: 225

Exercise 13.4

1. Find the surface area of a sphere of radius:
(i) 10.5 cm

(ii) 5.6 cm
(iii) 14 cm

(i) Radius of the sphere (r) = 10.5 cm
Surface area = 4πr2
= (4 × 22/7 × 10.5 × 10.5) cm2
= 1386 cm2

(ii) Radius of the sphere (r) = 5.6 cm
Surface area = 4πr2
= (4 × 22/7 × 5.6 × 5.6) cm2
= 394.24 cm2

(iii) Radius of the sphere (r) = 14 cm
Surface area = 4πr2
= (4 × 22/7 × 14 × 14) cm2
= 2464 cm2

2. Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m

(i) r = 14/2 cm = 7cm
Surface area = 4πr2
= (4 × 22/7 × 7 × 7)cm2
=616cm2

(ii) r = 21/2 cm = 10.5 cm
Surface area = 4πr2
= (4 × 22/7 × 10.5 × 10.5) cm2
= 1386 cm2

(iii) r = 3.5/2 m = 1.75 m
Surface area = 4πr2
= (4 × 22/7 × 1.75 × 1.75) m2
= 38.5 m2

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

r = 10 cm
Total surface area of hemisphere = 3πr2
= (3 × 3.14 × 10 ×10) cm2
= 942 cm2

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Let r be the initial radius and R be the increased radius of balloons.
r = 7cm and R = 14cm
Ratio of the surface area =4πr2/4πR2
= r2/R2
= (7×7)/(14×14) = 1/4
Thus, the ratio of surface areas = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm2.

Radius of the bowl (r) = 10.5/2 cm = 5.25 cm
Curved surface area of the hemispherical bowl = 2πr2
= (2 × 22/7 × 5.25 × 5.25) cm2
= 173.25 cm2
Rate of tin - plating is = ₹16 per 100 cm2
Therefor, cost of 1 cm= ₹16/100
Total cost of tin-plating the hemisphere bowl = 173.25 × 16/100
= ₹27.72

6. Find the radius of a sphere whose surface area is 154 cm2.

Let r be the radius of the sphere.
Surface area = 154 cm2
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= 154/(4 × 22/7)
⇒ r= 49/4
⇒ r = 7/2 = 3.5 cm

7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = r/2
Radius of the moon = r/8
Ratio of their surface area = 4π(r/8)2/4π(r/2)2
= (1/64)/(1/4)
= 4/64 = 1/16
Thus, the ratio of their surface areas is 1:16

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Inner radius of the bowl (r) = 5 cm
Thickness of the steel = 0.25 cm
∴ outer radius (R) = (r + 0.25) cm
= (5 + 0.25) cm  = 5.25 cm
Outer curved surface = 2πR2
= (2 × 22/7 × 5.25 × 5.25) cm2
= 173.25 cm2

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii). (i) The surface area of the sphere with raius r = 4πr2

(ii) The right circular cylinder just encloses a sphere of radius r.
∴ the radius of the cylinder = r and its height = 2r
∴ Curved surface of cylinder =2πrh
= 2π × r × 2r
= 4πr2
(iii) Ratio of the areas = 4πr2:4πr2 = 1:1

Page No: 228

Exercise 13.5

1. A matchbox measures 4cm × 2.5cm × 1.5cm. What will be the volume of a packet containing 12 such boxes?

Dimension of matchbox = 4cm × 2.5cm × 1.5cm
l = 4 cm, b = 2.5 cm and h = 1.5 cm
Volume of one matchbox = (l × b × h)
= (4 × 2.5 × 1.5)  cm= 15 cm3
Volume of a packet containing 12 such boxes = (12 × 15)  cm= 180 cm3

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1m3 = 1000 l)

Dimensions of water tank = 6m × 5m × 4.5m
l = 6m , b = 5m and h = 4.5m
Therefore Volume of the tank =ℓbh m3
=(6×5×4.5)m3=135 m3
Therefore , the tank can hold = 135 × 1000 litres  [Since 1m3=1000litres]
= 135000 litres of water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Length = 10 m , Breadth = 8 m and Volume = 380 m3
Volume of cuboid = Length × Breadth × Height
⇒ Height = Volume of cuboid/(Length × Breadth)
= 380/(10×8) m
= 4.75m

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹30 per m3.

l = 8 m, b = 6 m and h = 3 m
Volume of the pit = lbh m3
= (8×6×3) m3
= 144 m3
Rate of digging = ₹30 per m3
Total cost of digging the pit = ₹(144 × 30) = ₹4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

length = 2.5 m, depth = 10 m and volume = 50000 litres
1m3 = 1000 litres

∴ 50000 litres = 50000/1000 m3 = 50 m3
= 50/(2.5×10) m
= 2 m

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last?

Dimension of tank = 20m × 15m × 6m
l = 20 m , b = 15 m and h = 6 m
Capacity of the tank = lbh m3
= (20×15×6) m3
= 1800 m3
Water requirement per person per day =150 litres
Water required for 4000 person per day = (4000×150) l
= (4000×150)/1000
= 600 m3
Number of days the water will last = Capacity of tank Total water required per day
=(1800/600) = 3
The water will last for 3 days.

7. A godown measures 40m × 25m × 15m. Find the maximum number of wooden crates each measuring 1.5m × 1.25m × 0.5m that can be stored in the godown.

Dimension of godown = 40 m × 25 m × 15 m
Volume of the godown = (40 × 25 × 15) m3 = 10000 m3
Dimension of crates = 1.5m × 1.25m × 0.5m
Volume of 1 crates = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375 = 10666.66 = 10666

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Edge of the cube = 12 cm.
Volume of the cube = (edge)3 cm3
= (12 × 12 × 12) cm3
= 1728 cm3
Number of smaller cube = 8
Volume of the 1 smaller cube =1728/8 cm3 = 216 cm3
Side of the smaller cube = a
a= 216
⇒ a = 6 cm
Surface area of the cube = 6 (side)2
Ratio of their surface area = (6 × 12 × 12)/(6 × 6 × 6)
= 4/1 = 4:1

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Depth of river (h) = 3 m
Width of river (b) = 40 m
Rate of flow of water (l) = 2 km per hour = (2000/60) m per minute
= 100/3 m per minute
Volume of water flowing into the sea in a minute = lbh m3
= (100/3 × 40 × 3) m3
= 4000 m3

Page No: 230

Exercise 13.6

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm 3 = 1l)

Let the base radius of the cylindrical vessel be ‘r’ cm.
∴ Circumference = 2πr
⇒ 2πr = 132     [ Circumference = 132 cm]
⇒ 2× 22/7 × r = 132 cm
R=(132×7)/(2×22) cm = 21 cm
∵ Height of the vessel = 25 cm
∴ Volume = πr2 x h    [ Volume of a cylinder = πr2h]
= (22/7) (21)2 × 25 cm3
= (22/7) × 21 × 21 × 25 cm3
= 22 × 3 × 21 × 25 cm3
= 34650 cm
∵ Capacity of the vessel = Volume of the vessel
∴ Capacity of cylindrical vessel = 34650 cm
Since 1000 cm= 1 litre
⇒ 34650 cm3 = (34650/1000) litres = 34.65 l

2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.

Here, Inner diameter of the cylindrical pipe = 24 cm
⇒ Inner radius of the pipe (r) = (24/2) cm = 12 cm
Outer diameter of the pipe = 28 cm
⇒ Outer radius of the pipe (R) = (28/2)cm = 14 cm
Length of the pipe (h) = 35 cm
∵ Inner volume of the pipe = πr2h
Outer volume of the pipe = πr2h
∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume
= πR2h - πr2h
= πh(R2-r2)
= πh(R+r)(R-r)  a2 - b2 = (a+b)(a-b)]
= 22/7 x 35 x (14+12) x (14-12) cm3
= 22 x 5 x 26 x 2 cm3

Mass of the wood in the pipe = [Mass of wood in 1 m3 of wood] x [Volume of wood in the pipe]
= [0.6g] x [22 x 5 x 26 x 2] cm3
= (6/10)x 22 x 10 x 26 g = 6 x 22 x 26 g
= 3432 g = (3432/1000)= 3.432 kg   [ 1000 g = 1 kg]
Thus, the required mass of the pipe is 3.432 kg.

3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

For rectangular pack: Length (l) = 5 cm
Breadth (b) = 4 cm Height (h) = 15 cm
∴ Volume = l x b x h = 5 x 4 x 15 cm3
= 300 cm3
⇒ Capacity of the rectangular pack = 300 cm3     ...(1)
For cylindrical pack: Base diameter = 7 cm
⇒ Radius of the base (r) = (7/2)cm
Height (h) = 10 cm
∴ Volume = πr2h = (22/7) x (7/2)2 x 10 cm3
= (22/7) x (7/2) x (7/2) x 10 cm3
= 11 x 7 x 5 cm3 = 385 cm3
⇒ Volume of the cylindrical pack = 385 cm3   ...(2)
From (1) and (2),
we have 385 cm3 – 300 cm3 = 85 cm3
⇒ The cylindrical pack has the greater capacity by 85 cm3.

4. If the lateral surface of a cylinder is 94.2 cmand its height is 5 cm, then find: (i) radius of its base (ii) its volume. (Use π = 3.14)

(i) Since lateral surface of the cylinder = 2 πrh
But lateral surface of the cylinder = 94.2 cm2
2πrh = 94.2
2×3.14×r×5 =942/10
⇒{(10×314)/100} × r = 942/10
⇒ r=(942/10)×{100/(10×314)}cm
⇒ r =471/157 cm
Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = πr2h
⇒ Volume of the given cylinder = 3.14 x (3)2 x 5 cm3
=314×100×3×3×5cm3
=(157×3×3)/10
= 1413/10 = 141.3 cm3

Thus, the required volume = 141.3 cm3

5. It costs ₹ 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2; find: (i) inner curved surface of the vessel (ii) radius of the base (iii) capacity of the vessel.

(i) To find inner curved surface
Total cost of painting = ₹ 2200
Rate of painting = ₹ 20 per m2
∴  Area =cost/rate = 2200/20 = 110 m2
⇒ Inner curved surface of the vessel = 110 m2

(ii) To find radius of the base Let the base radius of the cylindrical vessel.
∵ Curved surface of a cylinder = 2 πrh
∴  2πrh = 110
⇒ 2× 22/7 ×r×10 = 110 [∵Height=10 m ]
⇒ r= (110×7)/(2×22×10)m = 7/4 m
= 1.75 m
⇒ The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel
Since, volume of a cylinder = πr2h
∴ Volume (capacity) of the vessel =22/7 × (7/2)2 × 10 m3
= 22/7 × 7/4 × 7/4 × 10 m
= (11×7×5)/4 m3 = 385/4 m3 = 96.25 m3

Since, 1 m3 = 1000000 cm3 = 1000 l = 1 kl
∴ 96.5 m3 = 96.5 kl
Thus, the required volume = 96.25 kl

6. The capacity of closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Capacity of the cylindrical vessel = 15.4 l
= 15.4×1000 cm ⇒ Volume of the vessel = (15.4/1000)m3
Height of the vessel = 1m Let ‘r’ metres be the radius of the base of the vessel
∴ Volume = πr2h
⇒ πr2h= 15.4/1000 Thus, the required sheet = 0.4708 m2

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Since, 10 mm = 1cm
∴ 1 mm = (1/10) cm
For graphite cylinder  Thus, the required volume of the graphite = 0.11 cm
For the pencil Diameter of the pencil = 7 mm = (7/10)cm
∴ Radius of the pencil (R) = (7/20) cm
Height of the pencil (h) = 14 cm
Volume of the pencil = πr2h Volume of the wood Volume of the wood = [Volume of the pencil] – [Volume of the graphite]
= 5.39 cm3 – 0.11 cm3 = 5.28 cm
Thus, the required volume of the wood is 5.28 cm3.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

The bowl is cylindrical.
Diameter of the base = 7 cm
⇒ Radius of the base (r) = (7/3) cm
Height (h) = 4 cm
Volume of soup = πr2h = 38500 / 100 liters
Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

Page No: 233

Exercise 13.7

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

(i) Radius (r) = 6 cm
Height (h) = 7 cm
Volume of the cone = 1/3 πr2h
= (1/3 × 22/7 × 6 × 6 × 7) cm3
= 264 cm3
(ii) Radius (r) = 3.5 cm
Height (h) = 12 cm
Volume of the cone = 1/3 πr2h
= (1/3 × 22/7 × 3.5 × 3.5 × 12) cm3
= 154 cm3

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

(i) Radius (r) = 7 cm
Slant height (l) = 25 cm
Let h be the height of the conical vessel.
∴ h = √l- r2
⇒ h = √25- 72
⇒ h = √576
⇒ h = 24 cm
Volume of the cone = 1/3 πr2h
= (1/3 × 22/7 × 7 × 7 × 24) cm3
= 1232 cm3
Capacity of the vessel = (1232/1000) l = 1.232 l

(i) Height (h) = 12 cm
Slant height (l) = 13 cm
Let r be the radius of the conical vessel.
∴ r = √lh2
⇒ r = √13- 122
⇒ r = √25
⇒ r = 5 cm
Volume of the cone = 1/3 πr2h
= (1/3 × 22/7 × 5 × 5 × 12) cm3
= (2200/7) cm3
Capacity of the vessel = (2200/7000) l = 11/35 l

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)

Height (h) = 15 cm
Volume = 1570 cm3
Let the radius of the base of cone be r cm
∴ Volume = 1570 cm3
⇒ 1/3 πr2h = 1570
⇒ 13 × 3.14 × r× 15 = 1570
⇒ r= 1570/(3.14×5) = 100
⇒ r = 10

4. If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.

Height (h) = 9 cm
Volume = 48π cm3
Let the radius of the base of the cone be r cm
∴ Volume = 48π cm3
⇒ 1/3 πr2h = 48π
⇒ 13 × r× 9 = 48
⇒ 3r= 48
⇒ r= 48/3 = 16
⇒ r = 4

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Diameter of the top of the conical pit = 3.5 m
Radius (r) = (3.5/2) m = 1.75 m
Depth of the pit (h) = 12 m
Volume = 1/3 πr2h
= (13 × 22/7 × 1.75 × 1.75 × 12) m3
= 38.5 m3
1 m= 1 kilolitre

Capacity of pit = 38.5 kilolitres.

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

(i) Diameter of the base of the cone = 28 cm
Radius (r) = 28/2 cm = 14 cm
Let the height of the cone be h cm
Volume of the cone = 13πr2h = 9856 cm3
⇒ 1/3 πr2h = 9856
⇒ 1/3 × 22/7 × 14 × 14 × h = 9856
⇒ h = (9856×3)/(22/7 × 14 × 14)
⇒ h = 48 cm

(ii) Radius (r) = 14 m
Height (h) = 48 cm
Let l be the slant height of the cone
l2 = h+ r2
⇒ l2 = 48+ 142
⇒ l2 = 2304+196
⇒ l= 2500
⇒ ℓ = √2500 = 50 cm

(iii) Radius (r) = 14 m
Slant height (l) = 50 cm
Curved surface area = πrl
= (22/7 × 14 × 50) cm2
= 2200 cm2

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.

On revolving the  ⃤  ABC along the side 12 cm, a right circular cone of height(h) 12 cm, radius(r) 5 cm and slant height(l) 13 cm will be formed.
Volume of solid so obtained = 1/3 πr2h
= (1/3 × π × 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.

On revolving the  ⃤  ABC along the side 12 cm, a cone of radius(r) 12 cm, height(h) 5 cm, and slant height(l) 13 cm will be formed.
Volume of solid so obtained =1/3 πr2h
= (1/3 × π × 12 × 12 × 5) cm3
= 240π cm3
Ratio of the volumes = 100π/240π = 5/12 = 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.

Diameter of the base of the cone = 10.5 m
Radius (r) = 10.5/2 m = 5.25 m
Height of the cone = 3 m
Volume of the heap = 1/3 πr2h
= (1/3 × 22/7 × 5.25 × 5.25 × 3) m3
= 86.625 m3
Also,
l2 = h+ r2
⇒ l2 = 3+ (5.25)2
⇒ l2 = 9 + 27.5625
⇒ l= 36.5625
⇒ l = √36.5625 = 6.05 m
Area of canvas = Curve surface area
= πrl = (22/7 × 5.25 × 6.05) m2
= 99.825 m(approx)

Page No: 236

Exercise 13.8

1. Find the volume of a sphere whose radius is
(i) 7 cm        (ii) 0.63 m

(i) Radius of the sphere(r) = 7 cm
Therefore, Volume of the sphere = 4/3 πr3
= (4/3 × 22/7 × 7 × 7 × 7) cm3
= 4312/3 cm3

(ii) Radius of the sphere(r) = 0.63 m
Volume of the sphere = 4/3 πr3
= (4/3 × 22/7 × 0.63 × 0.63 × 0.63) m3
= 1.05 m3

2. Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm                     (ii) 0.21 m

(i) Diameter of the spherical ball = 28 cm
Radius = 28/2 cm = 14 cm
Amount of water displaced by the spherical ball = Volume
= 4/3 πr3
= (4/3 × 22/7 × 14 × 14 × 14) cm3
= 34496/3 cm3

(ii) Diameter of the spherical ball = 0.21 m
Radius (r) = 0.21/2 m = 0.105 m
Amount of water displaced by the spherical ball = Volume
= 4/3 πr3
= (43×227×0.105×0.105×0.105) m3
= 0.004851 m3

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?

Diameter of the ball = 4.2 cm
Radius = (4.2/2) cm = 2.1 cm
Volume of the ball = 4/3 πr3
= (4/3 × 22/7 × 2.1 × 2.1 × 2.1) cm3
= 38.808 cm3
Density of the metal is 8.9g per cm3
Mass of the ball = (38.808 × 8.9) g = 345.3912 g

4. The diameter of the moon is approximately one-fourth of the diameter of the earth.What fraction of the volume of the earth is the volume of the moon?

Let the diameter of the moon be r
Radius of the moon = r/2
A/q,
Diameter of the earth = 4r
Volume of the moon = v = 4/3 π(r/2)3
= 4/3 πr× 1/8
⇒ 8v = 4/3 πr--- (i)
Volume of the earth = r3 = 4/3 π(2r)3
= 4/3 πr3× 8
⇒ V/8 = 4/3 πr3 --- (ii)
From (i) and (ii), we have
8v = V/8
⇒ v = 1/64 V
Thus, the volume of the moon is 1/64 of the volume of the earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Diameter of a hemispherical bowl = 10.5 cm
Radius(r) = (10.5/2) cm = 5.25cm
Volume of the bowl = 2/3 πr3
= (2/3 × 22/7 × 5.25 × 5.25 × 5.25) cm3
= 303.1875 cm3
Litres of milk bowl can hold = (303.1875/1000) litres
= 0.3031875 litres (approx.)

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Internal radius = r = 1m
External radius = R = (1 + 0.1) cm = 1.01 cm
Volume of iron used = External volume – Internal volume
= 2/3 πR- 2/3 πr3
= 2/3 π(R- r3)
= 2/3 × 22/7 × [(1.01)3−(1)3] m3
= 44/21 × (1.030301 - 1) m3
= (44/21 × 0.030301) m3
= 0.06348 m3(approx)

7. Find the volume of a sphere whose surface area is 154 cm2.

Let r cm be the radius of the sphere
So, surface area = 154cm2
⇒ 4πr= 154
⇒ 4 × 22/7 × r= 154
⇒ r= (154×7)/(4×22) = 12.25
⇒ r = 3.5 cm
Volume = 4/3 πr3
= (4/3 × 22/7 × 3.5 × 3.5 × 3.5) cm3
= 539/3 cm3

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹498.96. If the cost of white-washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,      (ii) volume of the air inside the dome.

(i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
=  (498.96/2.00) m= 249.48 m

(ii) Let r be the radius of the dome.
Surface area = 2πr2
⇒ 2 × 22/7 × r= 249.48
⇒ r= (249.48×7)/(2×22) = 39.69
⇒ r2=  39.69
⇒ r = 6.3m
Volume of the air inside the dome = Volume of the dome
= 2/3 πr3
= (2/3 × 22/7 × 6.3 × 6.3 × 6.3) m3
= 523.9 m(approx.)

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,      (ii) ratio of S and S′.

(i) Volume of 27 solid sphere of radius r = 27 × 4/3 πr3 --- (i)
Volume of the new sphere of radius r′ = 4/3 πr'3 --- (ii)
A/q,
4/3 πr'3= 27 × 4/3 πr3
⇒ r'= 27r3
⇒ r'3 = (3r)3
⇒ r′ = 3r
(ii) Required ratio = S/S′ = 4 πr2/4πr′= r2/(3r)2
= r2/9r2 = 1/9 = 1:9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Diameter of the spherical capsule = 3.5 mm
= 1.75mm
Medicine needed for its filling = Volume of spherical capsule
= 4/3 πr3
= (4/3 × 22/7 × 1.75 × 1.75 × 1.75) mm3
= 22.46 mm(approx.)

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

Chapter 13 NCERT Solutions available here is useful in preparing for the exams. We will learn to find the surface areas and volumes of cuboids and cylinders in details and extend this study to some other solids such as cones and spheres.

• Surface Area of a Cuboid and a Cube: Surface Area of a Cuboid = 2(lb + bh + hl) and Surface Area of a Cube = 6a2

• Surface Area of a Right Circular Cylinder: Curved Surface Area of a Cylinder = 2πrh and Total Surface Area of a Cylinder = 2πr(r + h).

• Surface Area of a Right Circular Cone: Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl and Total Surface Area of a Cone = πrl + πr2 = πr(l + r).

• Surface Area of a Sphere: A sphere is like the surface of a ball. The word solid sphere is used for the
solid whose surface is a sphere. Surface Area of a Sphere = 4 πr2, Curved Surface Area of a Hemisphere = 2πr2 and Total Surface Area of a Hemisphere = 3πr2.

• Volume of a Cuboid: The measure of this occupied space is called the Volume of the object. Volume of a Cuboid = base area × height = length × breadth × height and Volume of a Cube = edge × edge × edge = a3.

• Volume of a Cylinder: The volume of a cylinder can be obtained as : base area × height = area of circular base × height = πr2h So, Volume of a Cylinder = πr2h.

• Volume of a Right Circular Cone: Volume of a Cone = 1/3 πr2h where r is the base radius and h is the height of the cone.

• Volume of a Sphere: Volume of a Sphere = 4/3πr3 where r is the radius of the sphere.

• Volume of a Hemisphere = 2/3πr3 where r is the radius of the hemisphere.

Surface Areas and Volumes contains total nine exercises in which the last one is optional not important for examinations. Below, we have provided exercisewise Chapter 13 NCERT Solutions which you can check.

Studyrankers experts have prepared these NCERT Solutions for Class 9 Maths in which every question's answers are detailed step by step that will be give in depth study of concepts.

### NCERT Solutions for Class 9 Maths Chapters:

 Chapter 1 Number Systems Chapter 2 Polynomials Chapter 3 Coordinate Geometry Chapter 4 Linear Equations in Two Variables Chapter 5 Introduction to Euclid’s Geometry Chapter 6 Lines and Angles Chapter 7 Triangles Chapter 8 Quadrilaterals Chapter 9 Areas of Parallelograms and Triangles Chapter 10 Circles Chapter 11 Constructions Chapter 12 Heron's Formula Chapter 14 Statistics Chapter 15 Probability

FAQ on Chapter 13 Surface Areas and Volumes

#### Why we should solve NCERT Solutions for Chapter 13 Surface Areas and Volumes Class 9?

These Chapter 13 NCERT Solutions provided here will make you equipped with variety of concepts and help you in learning the concepts embedded in the question. Through these NCERT Solutions, one can easily complete their homework.

#### What is total surface area of a cone?

Total Surface Area of a Cone = πrl + r2 = πr(l + r).

#### What do you mean by Solids?

The bodies occupying space are called solids, such as a cuboid, a cube, a cylinder, a cone, a sphere, etc. These solids have plane or curved surfaces.

#### The floor area of a room is 100 m2 and its height is 8 m. Find its volume.

∵ Volume of a cuboid = [Base area] × Height
∴ Volume of the room = [Area of the floor] × height = 100 m2 × 8 m = 800 m3
Thus, the volume of the room = 800 m3.