# Frank Solutions for Chapter 25 Surface Area and Volume of Solids Class 9 Mathematics ICSE

### Exercise 25.1

1. The volume of a cube is 1331 cm3. Find its total surface area.

Let the side of the cube = a cm

Hence,

a= 1331 cm3

Taking cube roots on both sides, we get,

a = 11 cm

Surface area of a cube = 6a2

Surface area of a cube = 6×11

Surface area of a cube = 6×121

We get,

Surface area of a cube = 726 cm2

Therefore, the surface area of the cube is 726 cm2

2. The total surface area of a cube is 864 cm2. Find its volume.

Let the side of the cube = a cm

Hence,

Total surface area of a cube = 6a2

864 = 6a2 …(Given)

On further calculation, we get,

a2 = 864/6

⇒ a2 = 144

Taking square root on both sides, we get,

a = √144

⇒ a = 12 cm

Volume of a cube = a3

Volume of a cube = 123

Volume of a cube = 1728 cm3

Therefore, volume of the cube = 1728 cm3

3. The length, breadth and height of a rectangular solid are in the ratio 6: 4: 3. If the total surface area is 1728 cm2. Find its dimensions.

Let a, b and c be the length, breadth and height of a rectangular solid respectively

Hence,

a: b: c = 6: 4: 3 …(Given)

Let the common multiple be x

Then,

a = 6x cm

b = 4x cm

c = 3x cm

The total surface area of the cuboid = 1728 cm2 …(Given)

2 (lb + bh + hl) = 1728

⇒ 2 (ab + bc + ca) = 1728

⇒ 2 [(6x. 4x) + (4x.3x) + (3x.6x)] = 1728

⇒ 2 [24x2 + 12x2 + 18x2] = 1728

⇒ 24x+ 12x2 + 18x2 = (1728/2)

⇒ 54x2 = 864

⇒ x2 = (864/54)

⇒ x2 = 16

⇒ x = 4

We get,

x = 4

Hence,

a = 6x = 6×4 = 24 cm

b = 4x = 4×4 = 16 cm

c = 3x = 3×4 = 12 cm

Therefore, the dimensions of the rectangular solid are 24 cm, 16 cm and 12 cm

4. Find the volume of a cube whose diagonal is √48 cm.

Given

Diagonal of a cube = √48 cm

We know that,

Diagonal of a cube = √3 x l

Hence,

√3 ×l = √48

⇒ l = (√48/√3)

⇒ l = (√48/3)

We get,

l = √16

⇒ l = 4 cm

Therefore, side (l) = 4 cm

Now,

Volume of cube = l3

Volume of cube = l×l×l

Volume of cube = 4× 4×4

Volume of cube = 64 cm3

Hence, volume of cube = 64 cm3

5. The length and breadth of a cuboid are 20 cm and 15 cm respectively. If its volume is 2400 cm3, find its height.

Given

Volume of a cuboid = 2400 cm3

Length of a cuboid = 20 cm

Breadth of a cuboid = 15 cm

We know,

Volume of a cuboid = l×b×h

2400 = 20×15×h

⇒ 2400 = 300h

⇒ h = 2400/300

We get,

h = 8 cm

Therefore, height of a cuboid is 8 cm

6. A cylinder has a diameter 20 cm. The area of the curved surface is 1100 cm2. Find the height and volume of the cylinder.

Given

Diameter of cylinder = 20 cm

Hence,

Radius (r) = (20/2) = 10 cm

Let ‘h’ be the height of the cylinder

Given

Area of curved surface = 1100 cm2

i.e, L.S.A of cylinder = 1100 cm2

2×π×r×h = 1100 [Since L.S.A of cylinder = 2×π×r×h]

⇒ 2×(22/7)×10×h = 1100

⇒(440/7) h = 1100

⇒ h = (1100×7)/440

⇒ h = (70/4)

We get,

h = 17.5 cm

Therefore, volume of cylinder = π × r2 ×h

= (22/7)× 102 ×17.5

On calculation, we get,

= 5500 cm3

7. Find the volume of a cylinder which has a base diameter 14 cm and height 24 cm.

Given

Diameter of base = 14 cm

Hence,

Radius (r) = 7 cm

Height (h) = 24 cm

Volume of cylinder = π ×r2 ×h

Volume of cylinder = (22/7) ×72 ×24

On calculation, we get,

Volume of cylinder = 3696 cm3

Therefore, volume of a cylinder is 3696 cm3

8. The total surface area of a cylinder is 264 m2. Find its volume if its height is 5 more than its radius.

Let r be the radius of the cylinder

Then,

Height (h) = 5 + r …(i)

Given, total surface area of cylinder = 264 m2

(2× π× r× h) + (2 ×π×r2) = 264

⇒ 2 πr (h + r) = 264

Substituting the value of h, we get,

2 πr (5 + r + r) = 264

⇒ r (5 + 2r) = (264/2 π)

On calculation, we get,

5r + 2r2 = 42

⇒ 2r2 + 5r – 42 = 0

⇒ 2r2 + 12r – 7r – 42 = 0

⇒ 2r (r + 6) – 7 (r + 6) = 0

⇒ (2r – 7) (r + 6) = 0

i.e,

(2r – 7) = 0

⇒ 2r = 7

⇒ r = (7/2)

⇒ r = 3.5 m

or

(r + 6) = 0

r = –6

Radius of a cylinder cannot be negative,

Hence, we take the value of r = 3.5 m

Therefore,

Height (h) = 5 + 3.5

Height (h) = 8.5 m

Volume of a cylinder = π × r2 ×h

Volume of a cylinder = (22/7)×(3.5)2×8.5

We get,

Volume of a cylinder = 327. 25 m

9. The curved surface area of a cylinder is 198 cm2and its base has diameter 21 cm. Find the height and the volume of the cylinder.

Given

L.S.A. of cylinder = 198 cm2

Diameter of base = 21 cm

Hence,

Radius (r) = 10.5 cm

Let h be the height of the cylinder

L.S.A. of cylinder = 198 cm2

2× π ×r× h = 198 (∵ L.S.A. of cylinder = 2 ×π ×r ×h)

⇒ 2×(22/7) ×10.5×h = 198

On further calculation, we get,

(462/7) h = 198

⇒ h = (198 ×7)/462

⇒ h = (1386/462)

We get,

h = 3 cm

Volume of a cylinder = π×r2 ×h

Volume of a cylinder = (22/7)×(10.5)2× 3

We get,

Volume of a cylinder = 1039.5 cm3

10. The volume of a solid cylinder is 7700 cm3. Find its height and total surface area if the diameter of its base is 35 cm.

Given

Volume of cylinder = 7700 cm3

Diameter of base = 35 cm

Hence,

Radius (r) = 17.5 cm

Let h be the height of the cylinder

Volume of cylinder = 7700

π× r2 ×h = 7700 (∵ volume of cylinder = π ×r2 ×h)

⇒ (22/7) ×(17.5)2 ×h = 7700

On calculating, we get,

962.5h = 7700

⇒ h = (7700/962.5)× (10/10)

⇒ h = (77000/9625)

We get,

h = 8 cm

Now,

T.S.A. of cylinder = (2× π ×r ×h) + (2× π× r2)

T.S.A. of cylinder = {2× (22/7)×17.5×8} + {2×(22/7)×(17.5)2}

T.S.A. of cylinder = 880 + 1925

We get,

T.S.A. of cylinder = 2805 cm2

11. The figure represents the cross section of a swimming pool 10m broad, 2 m deep at one end, 3 m deep at the other end. Calculate the volume of water it will hold when full, given that its length is 40 m.

The given figure is a trapezium, since two opposite sides are parallel

Length of a pool = 40 m

Height of the trapezium = 10 m

Area of cross section = Area of trapezium

Area of cross section = (1/2) ×(sum of parallel sides) x height

Area of cross section = (1/2)×(2 + 3)×10

Area of cross section = 50/2

We get,

Area of cross section = 25 m2

Volume of the pool = Area of cross section x length

Volume of the pool = 25×40

Volume of the pool = 1000 m3

Therefore, the volume of the pool is 1000 m3

12. The given figure is a cross section of a victory stand used in sports. All measurements are in centimeters. Assume all angles in the figure are right angles. If the width of the stand is 60 cm, find

(a) The space it occupies in cm3

(b) The total surface area in m2

(a) To find the volume, first find the area of the figure

We divide the figure into 3 different rectangles to find the area

Rectangle 1 (left):

Length = 50 cm

Width = 40 cm

Area = Length ×width

Area = 50 x 40

We get,

Area = 2000 cm2

Rectangle 2 (middle):

Length = (60 + 30) cm

= 90 cm

Width = 40 cm

Area = Length×width

Area = 90× 40

We get,

Area = 3600 cm2

Rectangle 3 (right):

Length = 60 cm

Width = 40 cm

Area = Length × width

Area = 60 × 40

Area = 2400 cm2

Total area = 2000 + 3600 + 2400

Total area = 8000 cm2

Volume = Total area × length

Volume = 8000 × 60

We get,

Volume = 480000 cm3

(b) Total surface area = 2×Area of cross section + Area of bottom face + Area of left face + Area of right face + Area of top face

Area of cross-section = 8000 cm2 [from (a)]

Width of a stand = 60 cm [Given]

Area of bottom face = 130×60

= 7800 cm2

Area of the left face = 40×60 + 50×60 + 50×60

= 8400 cm2

Area of right face = 60×60 + 40×60 + 30×60

= 7800 cm2

Area of the top face = 40×60

= 2400 cm2

Total surface area = 2×8000 + 7800 + 8400 + 7800 + 2400

We get,

= 42400 cm2

= 4.24 m2

Hence, the total surface area is 4.24 m2

13. Water flows at the rate of 1.5 meters per second through a pipe with area of cross section 2.5 cm2into a rectangular water tank of length 90 cm and breadth 50 cm. Find the rise in water level in the tank after 4 minutes.

Given

Rate of flow of water = 1.5 m/s

= 150 cm/s

Rate of volume of water flown = Rate of flow ×cross section area

= 150×2.5

We get,

= 375 cm3/s

Total volume of water flow = Rate of volume of water flown x Time

= 375×(4×60 seconds)

= 90000 cm3

Volume of water flown = volume of water in the tank

90000 = l×b ×h

⇒ 90000 = 90×50 ×h

⇒ h = 90000/4500

We get,

h = 20 cm

Therefore, the rise in the level of water is 20 cm

14. A swimming pool is 50 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 4.5 m respectively. If the bottom of the pool slopes uniformly, find the amount of water in kilolitres required to fill the pool (1 m3= 1000 litres).

Area of cross section = Area of trapezium

= (1/2)×(sum of parallel sides) x height

= (1/2) × (1.5 + 4.5) × 50

= (1/2) × 6 × 50

We get,

= 150 m2

Hence,

Volume of the pool = area of cross section × height

= 150 × 15

We get,

= 2250 m3

Since 1 m3 = 1 kilolitre

Therefore, volume of the pool is 2250 kilolitres

15. Rain falls on a rectangular roof 28 m by 9 m and the water flows into a tank of dimensions 90 m by 70 cm by 84 cm. How much rainfall will fill the tank completely?

Dimensions of the tank:

Length = 90 m

Breadth = 70 cm

= 0.7 m

Height = 84 cm

= 0.84 m

Amount of rainfall = Volume of tank/Area of roof

= (90 × 0.7 × 0.84)/(28 × 9)

On further calculation, we get,

= (52.92/252)

= 0.21 m

= 21 cm

Therefore, amount of rainfall is 21 cm

### Exercise 25.2

1. Find the volume of a cuboid whose diagonals is 3√29. cm when its length, breadth and height are in the ratio 2 : 3: 4.

2. The total surface area of a cube is 294 cm2. Find its volume.

3. The total surface area of a cuboid is 46 m2. If its height is 1 m and breadth 3 m, find its length and volume.

4. The length, breadth and height of a cuboid are in the ratio of 3 : 3 : 4. Find its volume in m3 if its diagonals is 5 √34 cm.

5. A square plate of side ‘x’ cm is 4 mm thick. If its volume is 144 cm3, find the value of ‘x’.

Volume of the square plate = Volume of cuboid

6. A room is 22 m long, 15 m broad and 6 m high. Find the area of its four walls and the cost of painting it including doors and windows at the rate of Rs 12 per m2.

7. A cuboid is 25 cm long, 15 cm broad and 9 cm high. Find the whole surface of a cube having its volume equal to that of the cuboid.

8. The square on the diagonal of a cube has an area of 441 cm2. Find the length of the side and total surface area of the cube.

9. Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the diagonal of this cube.

10. Three equal cubes of side 5 cm each are placed to form a cuboid. Find the volume and the total surface area of the cuboid.

11. Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.

12. A metal cube of side 4 cm is immersed in a water tank. The length and breadth of the tank are 8 cm and 4 cm respectively. Find the rise in level of the water.

13. A metal piece 6 cm long, 5 cm broad and x cm, high is dropped in a glass box containing water. The dimensions of the base of the glass box are 18 cm by 8 cm and the rise in water level is 0.5 cm. Find x.

14. A closed box is made of wood 5 mm thick. The external length, breadth and height of the box are 21 cm, 13 cm and 11 cm respectively. Find the volume of the wood used in making of the box.

15. A room is 5 m long, 2 m broad and 4 m high. Calculate the number of persons it can accommodate if each person needs a 0.16 m3 of air.

16. 375 persons can be accommodated in a room whose dimensions are in the ratio of 6 : 4 : 1. Calculate the area of the four walls of the room if the each person consumes 64 m3 of air.

17. A class room is 7 m long, wide and 4 m high. It has two doors each of 3 m × 1.4 m and six windows each of 2 m × 1 m. The interior walls are to be coloured at the rate of Rs 15 per m2. Find the cost of colouring the walls.

18. The cost of papering the four walls of a room at Rs 1 per m2 is Rs. 210. The height of the room is 5 m. Find the length and breadth of the room if they are in the ratio 5.2.

19. Find the volume of wood used in making a closed box 22 cm by 18 cm by 14 cm, using a 1 cm thick wood. Also, find the cost of wood. Also, find the cost of wood required to make the box at the rate of Rs. 5 per cm3. How many cubes of side 2 cm can be placed in the box?

20. The length of a cold storage is double its breadth. Its height is 3 m. the area of its four walls including doors is 108 m2. Find its volume.

21. A metallic sheet is of the rectangular shape with dimensions 48 cm × 36 cm. From each one of its corners, a square of 8 cm is cutoff. An open box is made of the remaining sheet. Find the volume of the box.

22. The area of a playground is 4800 m2. Find the cost of covering it with gravel 2.5 cm deep, if the gravel cost Rs 7.25 per m3.

23. A rectangular container has base with dimensions 6 cm × 9 cm. A cube of edge 3 cm is placed in the container and then sufficient water is filled into it so that the cube is just submerged. Find the fall in the level of the water in the container, when the cube is removed.

24. The base of rectangular container is a square of side 12 cm. this container holds water up to 2 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 224 cm3 of water overflows. Find the volume and surface area of the cube.

### Exercise  25.2

1. Find the lateral surface area, total surface area and the volume of the following cylinders:

(i) Radius = 4.2 cm, Height = 12 cm

(ii) Diameter = 10 m, Height = 7 m

(i)

(ii)

2. A cylinder has a diameter 20 cm. The area of the curved surface is 1100 cm2. Find the height and volume of the cylinder.

3. Find the volume of a cylinder which has a base diameter 14 cm and height 24 cm.

4. The total surface area of a cylinder is 264 m2. Find its volume if its height is 5 more than its radius.

5. The curved surface area of a cylinder is 198 cm2 and its base has diameter 21 cm. Find the height and the volume of the cylinder.

6. The volume of a solid cylinder is 7700 cm3. Find its height and total surface area if the diameter of its base is 35 cm.

7. The total surface area of a cylinder is 3872 cm2. Find its height and volume if the circumference of the base is 88 cm.

8. The radius of the base of a right circular cylinder is doubled and the height is halved. What is the ratio of volume of the new cylinder to that of the original cylinder?

9. The radius of the base of a right circular cylinder is tripled and the height is doubled. What is the ratio of volume of the new cylinder to that of the original cylinder?

10. The volume of a cylinder of height 8 cm is 392
π cm3. Find its lateral surface area and its total surface area.

11. Find the length of 22 kg copper wire of diameter 0.8 cm, if the weight of 1 cm3 copper is 4.2 g.

12. Find the length of a solid cylinder of diameter 4 cm when recast into a hollow cylinder of outer diameter 10 cm, thickness 0.25 cm and length 21 cm? Give your answer correct to two decimal places.

13. A hollow garden roller, 1 m wide with outside diameter of 30 cm, is made of 2 cm thick iron. Find the volume of the iron. If the roller rolls without sliding for 6 seconds at the rate of 8 complete rounds per second, find the distance travelled and the area covered by the roller in 6 seconds.

14. A rectangular metal sheet 36 cm × 20 cm can be formed into a right circular cylinder, either by rolling along its length or by rolling along its breadth. Find the ratio of the volumes of the two cylinders thus formed.

15. Find the ratio of the volumes of the two cylinders formed by rolling an iron sheet 2.2 m × 1.5 m either along its length or by rolling along its breadth.

16. A rectangular strip 36 cm × 22 cm is rotated about the longer side. Find the volume and the total surface area of the cylinder formed.

17. The height of a right circular cylinder is 4.2 cm. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder correct to 1 decimal place.

18. How many cubic metres of earth must be dug out to sink a well 42 m deep and 14 m in diameter? Find the cost of plastering the inside walls of the well at Rs 15 per m2.

19. A well with 14 m diameter is dug 21 m deep. The earth taken out of it has been evenly spread all around it to a width of 14 m of form an embankment. Find the height of the embankment.

20. A well with 6 m diameter is dug. The earth taken out of it is spread uniformly all around it to a width of 2 m to form an embankment of height 2.25 m. Find the depth of the will.

21. A cylindrical container with internal diameter of its base 20 cm, contains water upto a height 14 cm. Find the area of the wet surface of the cylinder.

22. The radius of a solid cylinder decreases by 10% and its height increases by 20%. Find the change in percentage of its volume and curved surface area.

23. From a tap of inner radius 0.80 cm, water flows at the rate of 7 m/s. Find the volume in litres of water delivered by the pipe in 75 minutes.

24. A cylindrical water tank has a diameter 4 m and is 6 m high. Water is flowing into it from a cylindrical pipe of diameter 4 cm at the rate of 10 m/s. In How much time the tank will be filled?

25. The difference between the outer and inner curved surface area of a hollow cylinder is 264 cm2. It its height is 14 cm and the volume of the material in it is 1980 cm3, find its total surface area.

26. The sum of the height and the radius of a cylinder is 28 cm and its total surface area is 616 cm2, find the volume of the cylinder.

27. A cylindrical tube, open at both ends, is made of metal. The bore (internal diameter) of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal. Also, find the weight of the tube if 1 cm3 of the metal weights 1.42 g.

### Exercise 25.3

1. The cross section of a piece of metal 2 m in length is shown.

(a) Calculate the area of cross section.

(b) Calculate the volume of the piece of metal.

(a)

(b)

2. The area of cross section of a pipe is 5.4 square cm and water is pumped out of it at the rate of 27 km per hour. Find in litres, the volume of water which flows out of the pipe in 2 minutes.

3. The figure shows the cross section of 0.2 m a concrete wall to be constructed. It is 0.2 m wide at the top, 0.2 m wide at the bottom and its height is 4.0 m, and its length is 40 m.

(a) Calculate the cross sectional area

(b) Calculate the volume of the concrete in the wall.

(c) If the whole wall is to be painted, find the cost of painting it at 2.50 per sq. m.

(a) Complete the diagram as shown:

(b)

(c)

4. The cross section of tunnel perpendicular to its length is a trapezium ABCD as shown in the figure. AM = BN; AB = 4.4 m CD = 3 m. The height of a tunnel is 2.4 m. The tunnel is 5.4 m long.

(a) Calculate the cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs 5 per m2.

(b) Calculate the cost of flooring at the rate of Rs. 2.5 per m2.

(a)

(b)

5. ABCDE is the end view of a factory shed which is 50 m long. The roofing of the shed consists of asbestos sheets as shown in the figure. The two ends of the shed are completely closed by brick walls.

(a) Calculate the total volume content of the shed.

(b) If the cost of asbestos sheet roofing is Rs. 20 per m2, find the cost of roofing.

(c) Find the total surface area (including roofing) of the shed.

(a)

(b)

(c)

6. The cross section of a swimming pool is a trapezium whose shallow and deep ends are 1 m and 3 m respectively. If the length of the pool is 50 m and its width is 1.5 m, calculate the volume of water it holds.

7. A hose-pipe of cross section area 3 cm2 delivers 1800 litres of water in 10 minutes. Find the speed of water in km/h through the pipe.

8. The cross-section of a canal is a trapezium with the base of length of 3 m and the top length of 5 m. It is 2 m deep and 400 m long. Calculate the volume of water it holds.