# RD Sharma Solutions Class 9 Chapter 18 Surface Area and Volume of Cuboid and Cube

Read RD Sharma Solutions Class 9 Chapter 18 Surface Area and Volume of Cuboid and Cube below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

Exercise 18.1

Question 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.
Solution:
We have,
Length of a cuboid (l) = 80 cm
Breadth of a cuboid (b) = 40 cm
Height of a cuboid (h) = 20 cm

Total Surface Area cuboid = 2(lb + bh + hl)
Total Surface Area cuboid = 2[(80)(40)+(40)(20)+(20)(80)]
Total Surface Area cuboid = 2[3200+800+1600]
Total Surface Area cuboid = 2[5600]
Total Surface Area cuboid = 11200 cm2

Lateral Surface Area cuboid = 2[l + b]h
Lateral Surface Area cuboid = 2[80+40]20
Lateral Surface Area cuboid = 40[120]
Lateral Surface Area cuboid = 4800cm
Hence, the total surface area is 11200cm2 and Lateral Surface Area is 4800 cm2.

Question 2: Find the lateral surface area and total surface area of a cube of edge 10 cm.
Solution:
We have,
Side of a Cube = 10 cm

Lateral Surface Area of Cube = 4 (side)2
Lateral Surface Area of Cube = 4(10 x 10)
Lateral Surface Area of Cube = 400 cm2

Total Surface Area = (6 Side)2
Total Surface Area = 6(102)
Total Surface Area = 600 cm2
Hence, the total surface area of cube is 600cm2 and Lateral Surface area of cube is 400cm2.

Question 3: Find the ratio of the total surface area and lateral surface area of a cube.
Solution:
We know that,
Total Surface Area of the Cube (TSA) = 6 Side2
Lateral surface area of the Cube (LSA) = 4 Side2
Ratio of TSA and LSA = (6(Side)2)/(4(Side)2
Ratio of TSA and LSA = 3/2
Ratio of TSA and LSA = 3:2
Hence, the ratio of TSA and LSA of cube is 3:2.

Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?
Solution:
We have,
Length of the wooden block (l) = 80cm
Breadth of the wooden block (b) = 40cm
Height of the wooden block (h) = 20cm

Surface Area of the wooden box = 2[lb + bh + hl]
Surface Area of the wooden box = 2[(80×40)+(40×20)+(20×80)]
Surface Area of the wooden box = 2[5600]
Surface Area of the wooden box is 11200 cm2

The Area of each sheet of the paper = 40 × 40 cm
The Area of each sheet of the paper = 1600 cm2

Number of sheets required = (area of the box)/(area of one sheet)
Number of sheets required = 11200/1600
Number of sheets required = 7
Hence, Marry would require 7 sheets.

Question 5: The length, breadth, and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 m2.
Solution:
We have,
Length = l = 5 m
Breadth = b = 4 m
Height = h = 3 m

Total Area to be washed = lb + 2(l + b)h
Total area to be white washed = (5 × 4) + 2(5 + 4)3
Total area to be white washed = 20 + 2(9)3
Total area to be white washed = 20 + 54
Total area to be white washed is 74m2

It is given that, the Cost of white washing 1m2 is Rs. 7.50
The cost of white washing 74m2 = 74 × 7.50
The cost of white washing 74m2 = Rs. 555
Hence, the cost of white washing 74m2 is Rs.555.

Question 6: Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let us assumed that,
Breadth of the cuboid = a
Length of the new cuboid = 3a
Height of the new cuboid = a

Total surface area of the cuboid = 2(lb + bh + hl)
Total surface area of the cuboid = 2(3a x a + a x a + a x 3a)
Total surface area of the cuboid = 14a2

Total Surface area of three cubes = 3 × (6 side2)
Total Surface area of three cubes = 3 × 6a2
Total Surface area of three cubes = 18a2

Ratio of a total surface area of the cuboid to total surface areas of the three cubes = 14a2/18a2
Ratio of a total surface area of the cuboid to total surface areas of the three cubes = 7/9
Ratio of a total surface area of the cuboid to total surface areas of the three cubes = 7:9
Hence, Ratio of a total surface area of the cuboid to total surface areas of the three cubes is 7:9.

Question 7: A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.
Solution:
Edge of the cube = 4 cm (Given)
Volume of the cube = Side3 = 43 = 64
Volume of the cube is 64cm3
Again,
Edge of the cube = 1 cm3
So, Total number of small cubes = 64cm3/1cm3 = 64
And, total surface area of all the cubes = 64 x 6 x 1 = 384 cm2

Question 8: The length of a hall is 18 m and the width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.
Solution:
We have,
Length of hall = 18m
Width of hall = 12m
It is given that,
Area of the floor and the flat roof = Sum of the areas of four walls

Area of the floor and the flat roof = 2lb
Area of the floor and the flat roof = 2 x 18 x 12
Area of the floor and the flat roof = 432 sq/ft

Sum of the areas of four walls = (2 × 18h + 2 × 12h) sq/ft
432 = 2 × 18h + 2 × 12h
432 = 2 (18h + 12h)
432/2 = 18h + 12h
216 = 30h
216/30 = h
h = 7.2
Hence, the height of the hall is 7.2m.

Question 9: Hameed has built a cubical water tank with lid for his house, with each other edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles if the cost of tiles is Rs. 360 per dozen.
Solution:
Edge of the cubical tank = 1.5m or 150 cm
Surface area of the cubical tank (5 faces) = 5 × Area of one Face
Surface area of the cubical tank = 5 × (150 × 150)cm

It is given that,
Find area of each square tile
Side of tile = 25 cm
Area of one tile = 25 x 25 cm2

Number of tiles required = (Surface Area of Tank)/(Area of each Tile)
Number of tiles required = ((5×150×150))/(25×25)
Number of tiles required = 180

Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs. 360
Cost of one tile = Rs.360/12
Cost of one tile = Rs.30

The cost of 180 tiles = 180 × 30
The cost of 180 tiles = Rs. 5400

Question 10: Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.
Solution:
Let 'a' be the edge of the cube
The surface area of the cube = 6a2
S1 = 6a2
We get a new edge after increasing the edge by 50%
The new edge = a + (50%)a
The new edge = a + (50/100 a)
The new edge = a + 1/2 a
The new edge = a + (2a + 1a)/2
The new edge = 3/2 a
The new surface area = 6 × (3/2 a)2
The new surface area = 6 × 9/4 a2
The new surface area = 3 × 9/2 a2
The new surface area = 27/2 a2

Increase in the Surface Area
Increase in the Surface Area =27/2 a2- 6a2
Increase in the Surface Area = (27a-12a2)/2
Increase in the Surface Area =15a2/2

Increase in the surface area
Increase in the surface area =(15/2 a2)/(6a2 )×100
Increase in the surface area =(15a2)/(12a2 )×100
Increase in the surface area = 125%
Hence, percentage increase in the surface area of a cube is 125%.

Question: 11  The dimensions of a rectangular box are in the ratio of 2: 3: 4 and the difference between the cost of covering it with a sheet of paper at the rates of Rs. 8 and Rs. 9.50 per m2 is Rs. 1248. Find the dimensions of the box.
Solution:
Let Length rectangular box (l) = 2x
Let Breadth rectangular box (b) = 3x
Let Height rectangular box (h) = 4x

Total Surface area = 2(lb + bh + hl)
Total Surface area = 2((2x)(3x) + (3x)(4x) + (4x)(2x))
Total Surface area = 2(6x2 + 12x2 + 8x2)
Total Surface area = 52×2

If the cost is at Rs.8 per m2
The total cost = 52x
The total cost = 8 × 52×2
The total cost = Rs. 416

If the cost at Rs. 9.5 per m2,
Total cost = 52x2 × 2m
Total cost = 9.5 × 52x2 × 2
Total cost = Rs. 494x2

Difference in cost = Rs. 494x2 - Rs. 416x2
1248 = Rs. 78x2
1248/78 = x2
16 = x2
x2 = 16
x = 4
Hence, the Length rectangular box is = 2 × 4 = 8m
Breadth rectangular box (b) = 3 × 4 = 12m
Height rectangular box (h) = 4 × 4 = 16m

Question 12:  A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per meter sheet, a sheet being 2 m wide.
Solution:
Length of the closed iron tank = 12 m
Breadth of the closed iron tank = 9 m
Height of the closed iron tank = 4 m
Total surface area of the tank = 2[lb + bh + hl]
Total surface area of the tank = 2[(12 × 9) + (9 × 4) + (12 × 4)]
Total surface area of the tank = 2[108 + 36 + 48]
Total surface area of the tank = 384 m2

The Length of the Iron Sheet =  (Area of the Sheet)/(Width of the Iron Sheet)
The Length of the Iron Sheet =  384/2
The Length of the Iron Sheet = 192m.

Cost of the Iron Sheet = Length of the Iron Sheet × Cost rate
Cost of the Iron Sheet = 192 × 5
Cost of the Iron Sheet = Rs. 960
Hence, the cost of iron sheet is Rs. 960.

Question 13:  Ravish wanted to make a temporary shelter for his car by making a box-like structure with the tarpaulin that covers all the four sides and the top of the car (with the front face of a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5m with base dimensions 4m × 3m?
Solution:
We have,
Length of the Shelter = 4m
Breadth of the Shelter = 3m
Height of the Shelter = 2.5m

The Area of tarpaulin required = 2h(l + b) + lb
The Area of tarpaulin required = 2 × 2.5 (4 + 3) + 4×3
The Area of tarpaulin required = 5(7) + 12
The Area of tarpaulin required = 47m2
Hence, the area of tarpaulin is 47m2.

Question 14:  An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface of Rs. 50 per sq. metre.
Solution:
We have,
External Length of the box = 148 cm
External Breadth of the box = 116 cm
External Height of the box = 83 cm

Inner Length of the box = 148 - (2 × 3) = 142 cm
Inner Breadth of the box = 116 - (2 × 3) = 110 cm
Inner Height of the box = 83 - 3 = 80

Surface area of the Inner region = 2h(l + b) + lb
Surface area of the Inner region = 2 × 80(142 + 110) + (142 × 110)
Surface area of the Inner region = 160 × 252 + (15620)
Surface area of the Inner region = 40320 + 15620
Surface area of the Inner region = 55940 cm2
Surface area of the Inner region = 5.2904 m2
The cost of Painting the Surface Area of the Inner region = 5.2904 × 50
The cost of Painting the Surface Area of the Inner region = Rs. 279.70
Hence, the cost of painting the inner region is Rs. 279.70

Question 15:  The cost of preparing the walls of a room 12m long at the rate of Rs. 1.35 per square meter is Rs. 340.20 and the cost of matting the floor at 85paise per square meter is Rs. 91.80. Find the height of the room.
Solution:
We have,
Length of the room = 12m
Let the height of the room = h
Area of 4 walls = (Total Cost of preparing the wall)/(Rate of preparing the wall)
Area of 4 walls = (Rs.340.20)/(Rs.1.35)
Area of 4 walls = (Rs.340.20)/(Rs.1.35)
Area of 4 walls = 252m2

Area of 4 walls = 2(l + b) × h
252 = 2(l + b) × h
252/2 = (12 + b) × h
126 = (12 + b) × h______________(1)

Area of the floor = (Total Cost of floor)/(Rate of floor)
Area of the floor = (Rs.91.80)/(Rs.0.85)
Area of the floor = (Rs.91.80)/(Rs.0.85)
Area of the floor = 108m2

Area of the Floor = l × b
108 = 12 × b
180/12 = b
9m = b
Put the value of ‘b’ equation (1)
126 = (12 + 9) × h
126 = (21) × h
126/21 = h
6 = h
h = 6m
Hence, the height of the room is 6m.

Question 16:  The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square meter.
Solution:
We have,
Length of the room = 12.5 m
Breadth of the room = 9 m
Height of the room = 7 m
Total surface area of the four walls = 2(l + b) × h
Total surface area of the four walls = 2(12.5 + 9) × 7
Total surface area of the four walls = 301m2

Area of 2 doors = 2(2.5 × 1.2)
Area of 2 doors = 6 m2

Area of 4 windows = 4 (1.5 × 1)
Area of 2 doors = 6 m2

Area to be painted on 4 walls = 301 - (6 + 6)
Area to be painted on 4 walls = 301 - 12
Area to be painted on 4 walls = 289 m2

Cost of painting = 289 × 3.50
Cost of painting = Rs. 1011.5
Hence, the cost of painting is Rs. 1011.5.

Question 17:  The length and breadth of a hall are in the ratio 4: 3 and its height is 5.5 meters. The cost of decorating its walls (including doors and windows) at Rs. 6.60 per square meter is Rs. 5082. Find the length and breadth of the room.
Solution:
It is given that, Height is 5.5m
Let us assumed that,
Length of the hall = 4a
Breadth of the hall = 3a

Area of 4 wall = (Total cost of decoration of wall)/(Rate of decoration  )
Area of 4 wall = (Rs.5082)/(Rs.6.6)
Area of 4 wall = 770m2

Area of four walls = 2(l + b) × h
770 = 2(4a + 3a) × 5.5
770/5.5 = 2(7a)
140 = 14a
140/14 = a
10 = a
a = 10

Length of the hall = 4a
Length of the hall = 4 × 10
Length of the hall = 40m

Breadth of the hall = 3a
Breadth of the hall = 3 × 10
Breadth of the hall = 30m
Hence, the length and the breadth of hall are 40m and 30m.

Question 18:  A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85cm (See figure 18.5). The thickness of the plank is 5cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.

Solution:
External dimension
Length of book shelf = 85 cm
Breadth of book shelf = 25 cm
Height of book shelf = 110 cm

External surface area of shelf while leaving front face of shelf = 2(lb + bh) + lh
External surface area of shelf while leaving front face of shelf = {2(85 × 25 + 25 × 110) + 85 × 110}
External surface area of shelf while leaving front face of shelf = {2(2125 + 2750) + 9350}
External surface area of shelf while leaving front face of shelf = {2(4875) + 9350}
External surface area of shelf while leaving front face of shelf = {9750 + 9350}
External surface area of shelf while leaving front face of shelf = 19100 cm2

Area of Front face = [85 × 110 - 75 × 100 + 2(75 × 5)] cm2
Area of Front face = 1850 + 750 cm2
Area of Front face = 2600 cm2

Area to be polished = 19100 + 2600 cm2
Area to be polished = 21700 cm2

Cost of polishing 1cm2 area = Rs. 0.20
Cost of polishing 21700 cm2 area = 21700 * 0.20
Cost of polishing 21700 cm2 area = Rs. 4340

Length (l), Breadth (b), height (h) of each row of book shelf is
Length (l) = 75 cm,
Height (h) = (110-20)/3= 90/3 = 30 cm

Area to be painted in 1 row = 2(l + h)b + lh
Area to be painted in 1 row = [2(75 + 30) × 20 + 75 × 30)]cm2
Area to be painted in 1 row = [2(105) × 20 + 75 × 30)]cm2
Area to be painted in 1 row = [210 × 20 + 75 × 30)]cm2
Area to be painted in 1 row = (4200 + 2250) cm2
Area to be painted in 1 row = 6450 cm2

Area to be painted in 3 rows = 3 × 6450
Area to be painted in 3 rows = Rs. 19350 cm2

Cost of painting 1cm2 area = Rs. 0.10
Cost of painting 19350cm2 area = 19350 * 0.10
Cost of painting 19350cm2 area = Rs.1935
Total expense = 4340 + 1935
Total expense = Rs.6275
Hence, total expense required for polishing and painting the surface of the bookshelf is Rs. 6275

Question 19:  The paint in a certain container is sufficient to paint on an area equal to 9.375 m2, how many bricks of dimension 22.5cm × 10cm × 7.5cm can be painted out of this container?
Solution:
Area = 9.375 m2
Area = 9.375 × 100 × 100
Area = 93750 cm

Length of a single brick (l) = 22.5 cm
Breadth of a single brick (b) = 10 cm
Height of a single brick (h) = 7.5 cm
Surface area of a brick = 2 (lb + bh + hl)
Surface area of a brick = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)
Surface area of a brick = 2(225 + 75 + 168.75)
Surface area of a brick = 937.50 cm2
Number of bricks that can be painted = (Area )/(Surface area of a brick)
Number of bricks that can be painted = 93750/937.5
Number of bricks that can be painted = 100
Hence, 100 bricks can be painted out of the container.

Exercise 18.2

Question 1: A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold?
Solution:
Length of cuboidal water tank (l) = 6m
Breadth of cuboidal water tank (b) = 5m
Height of cuboidal water tank (h) = 4.5m

Volume of the cuboidal water tank = l × b × h
Volume of the cuboidal water tank = 6 × 5 × 4.5
Volume of the cuboidal water tank = 135m3

1m3 = 1000 liters
135m3 = (135×1000)liters
135m3 = 135000 liters
Hence, water hold by tank 1,35,000 liters.

Question 2: A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
Solution:
Length of cuboidal vessel (l) = 10 m
Breadth of cuboidal vessel (b) = 8 m
Volume of the vessel = 380m3

Volume of cuboidal vessel = l × b × h
380m3 = l × b × h
380m3 = 10 × 8 × h
h =( 380m^3)/(10m × 8m)
h = 4.75m
Hence, the height of the vessel is 4.75m.

Question 3: Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m3.
Solution:
Length of cuboidal pit (l) = 8 m
Breadth of cuboidal pit (b) = 6 m
Height of cuboidal pit (h) = 3 m

Volume of the cuboidal water tank = l × b × h
Volume of the cuboidal water tank = 8 × 6 × 3
Volume of the cuboidal water tank = 144
Volume of the Cuboidal pit is 144 m3

Cost of digging 1m3 = Rs. 30
Cost of digging 144m3 = 144 x 30
Cost of digging 144m3 = Rs. 4320
Hence, the cost of digging 144m3 is Rs. 4320

Question 4:  If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that
1/V=2/S (1/a+1/b+1/c)
Solution:
Length of cube (l) = a
Breadth of cube (b) = b
Height of cube (h) = c
Volume of the cube = l × b × h
Volume of the cube = a × b × c
Volume of the cube = abc

Surface area of the cube = 2(lb + bh + hl)
Surface area of the cube = 2(ab + bc + ca)

It is given that,
1/V=2/S (1/a+1/b+1/c)
1/V=2/S ((bc+ac+ab)/abc) ________(i)
We know that S = 2(ab + bc + ca)
Put the value of S in equation (i)
1/V=2/(2(ab+bc+ca)) ((bc+ac+ab)/abc)
1/V=2/(2(ab+bc+ca)) ((bc+ac+ab)/abc)
1/V=1/abc
Hence Proved

Question 5: The areas of three adjacent faces of a cuboid are x, y and z. If the volume is V, Prove that V2 = xyz.
Solution:
Let us assumed that,
Length of a cuboid = a
Breadth of a cuboid = b
Height of a cuboid = c
It is given that, the areas of three adjacent faces of a cuboid are x, y and z
x = ab
y = bc
z = ca
xyz = ab × bc × ca
xyz = a2b2c2
xyz = (abc)2_______(1)
Volume of a cuboid (V) = abc
Put the value of V in equation (1)
V2 = xyz
Hence proved

Question 6: If the areas of three adjacent face of a cuboid are 8cm2, 18cm2 and 25cm2. Find the volume of the cuboid.
Solution:
Let the areas of three adjacent faces of a cuboid is a, b, c.
a = 8 cm2 = lb
b = 18 cm2 = bh
c = 25 cm2 = lh
l = length of cuboid
h = height of cuboid
abc = 8 × 18 × 25
abc = 3600
Volume of cuboid = lbh
abc = lb × bh × lh
abc = (lbh)2
abc = V2
Put the value of abc in above equation.
V2 = 3600
V = 60
Hence, Volume of the cuboid is 60 cm3.

Question 7: The breadth of a room is twice its height, one half of its length and the volume of the room is 512dm3. Find its dimensions.
Solution:
Let us assume that,
Length of the room = l
Breadth of the room = b
Height of the room = h
It is given that, the breadth of a room is twice its height, one half of its length
b = 2h and b = 1/2
1/2 = 2h
l = 4h
b = 2h
Volume of the room = l × b × h
512 = 4h × 2h × h
512 = 8h3
512/8 = h3
64 = h3
h = 4
Length of the room (l) = 4h
Length of the room (l) = 4 × 4
Length of the room (l) = 16dm

Breadth of the room (b) = 2h
Breadth of the room (b) = 2 × 4
Breadth of the room (b) = 8dm

Height of the room (h) = 4dm
Hence, the length, breadth and height of the room is 16dm, 8dm and 4dm.

Question 8: 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?
Solution:
Water flow of a river = 2 km/hour
Water flow of a river in meter per minutes = 2000/60m/min
Water flow of a river in meter per minutes = 100/3m/min
1 km = 1000m
1 hour = 60mins
Depth of the river (h) = 3m
Width of the river (b) = 40m
Volume of the water flowing in 1 min = 100/3 × 40 × 3
Volume of the water flowing in 1 min = 4000 m3
Water flowing in litres = 40,00,000 litres
Hence, 40,00,000 litres of water will fall in the sea in 1minute.

Question 9: Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km every hour. What much area will it irrigate in 30 minutes if 8 cm of standing water is desired?
Solution:
Water in the canal forms a cuboid of Width (b) and Height (h).
Width of the cuboid (b) = 30dm = 3m
Height of the cuboid (h) = 12dm = 1.2m
Length of the cuboid (l) = 100 × 30/60  (Distance travelled in 30 min with a speed of 100 km per hour)
Length of the cuboid (l) = 50000m

Volume of water used for irrigation = l × b × h
Volume of water used for irrigation = 50000 × 3 × 1.2
Volume of water used for irrigation = 1,80,000m3
It is given that, Water deposited in the field forms a cuboid with a base area equal to the field's area and a height equal to the field's height = 8/100m

Area of field × 8/100 = 1,80,000m
Area of field = 1,80,000 × 100/8
Area of field = 22,50,000m
Hence, the area of field is 22,50,000m2

Question 10: Three metal cubes with edges 6cm, 8cm, 10cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.
Solution:
Side of first Cube = 6cm
Side of second Cube = 8cm
Side of third Cube = 10cm

Volume of cube = x3
x3 = (63 + 83 + 103)
x3 = 1728
x = 12
Volume of the new cube = x3
Volume of the new cube = (12cm)3
Volume of the new cube = 1728cm3

Surface area of the new cube = 6(side)2
Surface area of the new cube = 6(12)2
Surface area of the new cube = 6(144)
Surface area of the new cube = 864cm2

Diagonal of the newly formed cube = √3Side
Diagonal of the newly formed cube = 12√3cm
Hence, the Volume, Surface area and Diagonal of new cube is 1728cm3, 864cm2 and 12√3cm.

Question 11: Two cubes, each of volume 512 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
It is given that, the volume of the cube = 512cm3
Volume cube = (side)3
512=(Side)3
8=Side
Side=8
Side of a cube is 8 cm.
Length of new cuboid (l) = 8 + 8 = 16 cm,
Breadth of new cuboid (b) = 8 cm
Height of new cuboid (h) = 8 cm

Surface area of cuboid = 2(lb + bh + hl)
Surface area of cuboid = 2(16 × 8 + 8 × 8 + 16 × 8)
Surface area of cuboid = 2(128 + 64 + 128)
Surface area of cuboid = 2(320)
Surface area of cuboid = 640cm2
Hence, the surface area of a cube is 640 cm2.

Question 12: Half cubic meter of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet.
Solution:
It is given that, volume of gold-sheet = 1/2 m3 = 0.5 m3
Area of the gold-sheet = 1 hectare
Area of the gold-sheet = 10000m2
Thickness of gold sheet = (Volume of solid)/(Area of gold sheet)
Thickness of gold sheet = (1 m3)/(2 × 10000 m2 )
Thickness of gold sheet = 1/20000 m
Thickness of gold sheet in centimetres.
Thickness of gold sheet = 1/20000×100cm
Thickness of gold sheet = 1/200 cm
Hence, the thickness of the silver sheet is 1/200cm.

Question 13: A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.
Solution:
It is given that,
Volume of the large cube = (Side)3
Volume of the large cube = (12)3
Volume of the large cube = 1728cm3
The length of the third side:-
Volume of cube = (Side)3
(12)3 = (6)3 + (8)3 + (x)3
1728 = 216 + 512 + (x)3
1728 = 728 + (x)3
1728 - 728 = (x)3
1000 = (x)3
x = 10
Hence, the length of the third side is 10 cm.

Question 14:  The dimensions of a cinema hall are 100 m, 50 m, 18 m. How many persons can sit in the hall, if each person requires 150m3 of air?
Solution:
It is given that,
Length of cinema hall (l) = 100m
Breadth of cinema hall (b) = 50m
Height of cinema hall (h) = 18m

Volume of cuboid = l × b × h
Volume of cinema hall = 100m × 50m × 18m
Volume of air required by each person = 150 m3

Number of persons can sit in hall = (Volume of Cinema hall)/(Volume of air)
Number of persons can sit in hall = (100 × 50 × 18)/150
Number of persons can sit in hall = (100 × 50 × 18)/150
Number of persons can sit in hall = 600
Hence, 600 members can sit in the hall.

Question 15: Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block.
Solution:
It is given that,
Width of marble = 28cm
Thick of marble = 5cm
Let us assumed that,
Length of marble = l

Volume of the marble block = l × b × h
Volume of the marble block = l × 28 × 5
Volume of the marble block = l × 140

Weight of the marble square = 140 × l × 0.25kg
Weight of the marble = 112kgs
112kgh = 140 × l × 0.25
112/(140 × 0.25) = l
112/35 = l
112/35 = l
3.2 = l
l = 3.2
Hence, the length of marble is 3.2cm

Question 16:  A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a fluid can be placed in it? Also, find the volume of the wood used in it.
Solution:
External Length of the cuboid (l) = 25 cm
External Breadth of the cuboid (b) = 18 cm
External Height of the cuboid (h) = 15 cm

External volume of cuboid = l × b × h
External volume of cuboid = 25 × 18 × 15
External volume of cuboid = 6750cm3

Internal dimensions of cuboid is as follows
Internal Length of cuboid (l) = 25 - (2 × 2) = 21 cm
Internal Breadth of cuboid (b) = 18 - (2 × 2) = 14 cm
Internal Height of cuboid (h) = 15 - (2 × 2) = 11cm

Internal volume cuboid = l × b × h
Internal volume cuboid = 21 * 14 * 11 cm3
Internal volume cuboid = 3234 cm3
Volume of the wood utilized = External volume – Internal volume
Volume of the wood utilized = 3516cm3
Hence, the volume of the wood used 3516cm3

Question 17: The external dimensions of a closed wooden box are 48cm, 36cm, 30cm. The box is made of 1.5cm thick wood. How many bricks of size 6 cm × 3 cm × 0.75 cm can be put in this box?
Solution:
External Length of wooden box (l) = 48 cm
External Breadth of wooden box (b) = 36 cm
External Height of wooden box (h) = 30 cm

Internal Length of the wooden box (l) = 48 - (2 × 1.5) = 45 cm
Internal Breadth of the wooden box (b) = 36 - (2 × 1.5) = 33 cm
Internal Height of the wooden box (h) = 30 - (2 × 1.5) = 27 cm

Internal volume of the wooden box = l × b × h
Internal volume of the wooden box = 45 × 33 × 27cm3
Internal volume of the wooden box = 40095cm3

Volume of the brick = l × b × h
Volume of the brick = 6 × 3 × 0.75
Volume of the brick = 13.5 cm3
Number of bricks = 40095/13.5
Number of bricks = 2970 bricks
Hence, Bricks can be kept inside the wooden box are 2970.

Question 18:  How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 gms. Find the weight of the empty box in kg.
Solution:
Outer Length of iron (l) = 36 cm
Outer Breadth of iron (b) = 25 cm
Outer Height of iron (h) = 16.5 cm

Outer Volume of iron = l × b × h
Outer Volume of iron = 36 × 25 × 16.5
Outer Volume of iron = 36 × 25 × 16.5
Outer Volume of iron = 14,850

Inner Length of iron (l) = 36 - (2 × 1.5) = 33 cm
Inner Breadth of iron (b) = 25 - (2 × 1.5) = 22 cm
Inner Height of iron (h) = 16.5 - 1.5 = 15 cm

Inner Volume of iron = l × b × h
Inner Volume of iron = 36 × 25 × 16.5
Inner Volume of iron = 36 × 25 × 16.5
Inner Volume of iron = 14,850

Volume of Iron = Outer volume – Inner volume
Volume of Iron = (36 × 25 × 16.5) – (33 × 22 × 15)
Volume of Iron = 3960 cm3
Weight of Iron = 3960 × 15 = 59400grams
Weight of Iron = 59.4 kgs
Hence, the weight of Iron is 59.4 kgs

Question 19:  A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.
Solution:
Volume of the cube = (Side)
Volume of the cube = (9)
Volume of the cube = 729 cm3

Area of the base = l × b
Area of the base = 15 × 12
Area of the base = 180 cm2

Raise water level in the vessel = (Volume of the cube)/(Area of base of rectanglar vessel)
Raise water level in the vessel = 729/180
Raise water level in the vessel = 729/180
Raise water level in the vessel = 4.05 cm
Hence, the raise water level in the vessel is 4.05 cm

Question 20:  A rectangular container, whose base is a square of side 5cm, stands on a horizontal table, and holds water up to 1cm from the top. When a solid cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.
Solution:
Volume of water inside the tank = l × b × h
Volume of water inside the tank = 5 × 5 × 1
Volume of water inside the tank = 25cm3

Volume of water that overflowed = 2cm3

Volume of the cube = Volume of water inside the tank + Volume of water that overflowed
Volume of the cube 3 = 25 + 2
Volume of the cube 3 = 27
Volume of the cube = 27cm3
Edge of the cube = 3cm
Hence, the volume of the cube and the length of its edge is 27cm3 and 3cm.

Question 21: A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7m deep and the earth taken out is spread evenly on the field. By how many meters is the level of the field raised? Give the answer to the second place of decimal.
Solution:
Volume of the earth dug out = 50 × 40 × 7
Volume of the earth dug out = 14000m3
Let 'h' be the rise in the height of the field
Volume of the cuboidal field = Volume of the earth dug out
Volume of the cuboidal field = 200 × 150 × h
14000 = 200 × 150 × h
14000/(200×150)=h

0.47 = h
h = 0.47

Question 22:  A field is in the form of a rectangular length 18m and width 15m. A pit 7.5m long, 6m broad and 0.8m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.
Solution:
Length of Rectangular field = 18m
Width of the Rectangular field = 15m
Area of the field on which the earth taken out is to be spread = 18 × 15 - 7.5 × 6
Area of the field on which the earth taken out is to be spread = 225 m2

Length of the Pit = 7.5m
Width of the Pit = 6m
Depth of the Pit = 0.8m
Volume of earth taken out from the pit = 7.5 × 6 × 0.8
Volume of earth taken out from the pit = 36m3

Area of the field × h = Volume of the earth taken out from the pit
225 × h = 36
h = 36/225
h = 0.16 m
h = 0.16 × 100
h = 16 cm
Hence, the height of rectangular field is 16cm

Question 23: A rectangular tank is 80 m long and 25 m broad. Water flows into it through a pipe whose cross-section is 25 cm2, at the rate of 16 km per hour. How much the level of the water rises in the tank in 45 minutes?
Solution:
Length of the rectangular tank = 80m = 8000cm
Breadth of the rectangular tank = 25m = 2500cm
Volume of water in rectangular tank = 8000 × 2500 × h

Cross-sectional area of the pipe = 25 cm2
It is given that, the water flows forms a cuboid of base area 25 cm2 and length equal to the distance travelled in 45 minutes with the speed 16 km/hour.
Length = 16000 × 100 × 45/60cm
Volume of water coming out pipe in 45 minutes = 25 × 16000 × 100 × 45/60
Volume of water in the tank = Volume of water coming out of the pipe in 45 minutes
8000 × 2500 × h = 16000 × 100 × 45/60 × 25
h =  (16000 × 100 × 45 × 25)/(60 × 8000 × 2500)
h = 1.5cm
Hence, the level of the water rises in the tank in 45 minutes is 1.5cm.

Question 24:  Water in a rectangular reservoir having base 80m by 60m is 6.5m deep. In what time can the water be pumped by a pipe of which the cross-section is a square of side 20 cm if the water runs through the pipe at the rate of 15km/hr.
Solution:
Water Flow = 15 km/hr
Water Flow = 15000 m/hr
Volume of water coming out of the pipe in one hour = 20/100 × 20/100 × 15000
Volume of water coming out of the pipe in one hour = 600m3

Length of the tank = 80m
Breadth of the tank = 60m
Height of the tank = 6.5m
Volume of the tank = l × b × h
Volume of the tank = 80 × 60 × 6.5
Volume of the tank = 31200 m3

Time taken to empty the tank = (Volume of tank)/(Volume of water coming out from the pipe)
Time taken to empty the tank = 31200/600
Time taken to empty the tank = 52 hours
Hence, the time taken to empty the tank is 52 hours.

Question 25:  A village having a population of 4000 requires 150liters of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Solution:
Length of the cuboidal tank (l) = 20 m
Breadth of the cuboidal tank (b) = 15 m
Height of the cuboidal tank (h) = 6 m
Volume of the tank = l × b × h
Volume of the tank = 20 × 15 × 6
Volume of the tank = 1800 m3
Volume of the tank = 1800000litres
The amount of water consumed by the village residents in a single day = 4000 × 150 litres
The amount of water consumed by the village residents in a single day = 600000 litres

In n days, the entire village drank water = Capacity of the tank
n × 600000 = 1800000
n = 1800000/600000
n = 3
Hence, the water will last for 3 days in the tank.

Question 26:  A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in Fig. 18.12. If the edge of each cube is 3cm, find the volume of the structure built by the child.

Solution:
Volume of each cube = (edge)3

Volume of each cube = (3)3
Volume of each cube = 27 cm3

Number of cubes in the structure = 15
Volume of the structure = 27 × 15
Volume of the structure = 405 cm3

Question 27:  A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Solution:
Length of Godown (lg) = 40 m
Breadth of Godown (bg) = 25 m
Height of Godown (hg) = 10 m
Volume of the godown = lg × bg × hg
Volume of the godown = 40 × 25 × 10
Volume of the godown = 10000 m3

Wooden crate length (lw) = 1.5 m
Wooden crate breadth (bw) = 1.25 m
Wooden crate height (hw) = 0.5 m
Volume of the wooden crate = lw × bw × h
Volume of the wooden crate = 1.5 × 1.25 × 0.5
Volume of the wooden crate = 0.9375 m3

The number of wooden crates in the godown is denoted by the letter 'n.'
Volume of 'n' wooden crates = Volume of godown
0.9375 n = 10000
n = 10000/0.9375 = 10666.66
Hence, In the godown, the maximum number of wooden crates that can be held is 10666.66.

Question 28 : A wall of length 10 m was to be built across an open ground. The height of the wall is 4m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required?
Solution:
Length of the wall = 10m × 100 = 1000 cm
Thickness of the wall = 24 cm
Height of the wall = 4m × 100 = 400 cm
Volume of the wall = l × b × h
Volume of the wall = 1000 × 24 × 400
Volume of the wall = 9600000

Length of Brick = 24 cm
Breadth of Brick = 12 cm
Height of Brick = 8 cm
Volume of each brick = l × b × h
Volume of each brick = 24 × 12 × 8
Volume of each brick = 2304 cm3

Number of bricks = (Volume of the wall)/(Volume of each brick)
Number of bricks = 9600000/2304
Number of bricks = 4166.67 bricks
Hence, 4167 bricks requires wall.

Exercise VSAQs

Question 1: If two cubes each of side 6 cm are joined face to face, then find the volume of the resulting cuboid.
Solution:
Length of a cuboid = 6 cm + 6 cm = 12 cm
Breadth of a cuboid = 6 cm
Height of a cuboid = 6 cm
Volume of cuboid = l × b × h
Volume of cuboid = 12 × 6 × 6
Volume of cuboid = 432 cm3

Question 2: Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are melted down into a single cube whose diagonal is 12√3cm. Find the edges of three cubes.
Solution:
Let us assumed that, edges = 3x, 4x and 5x
It is given that, diagonal of new cube formed = 12√3 cm
Volume of new cube = (3x)3 + (4x)3 + (5x)3
Volume of new cube = 27x3 + 64x3 + 125x3
Volume of new cube = 216x3
New diagonal of a cube = √3a
12√3 = √3a
12 = a
a = 12
Side of new cube is 12 cm.
Volume of cube = (12)3
216x3 =(12)3
x3 = (12×12×12)/216
x3 = 8
x = ∛8
x = 2
Hence,
Length = 3x = 3 × 2 = 6 cm
Breadth = 4x = 4 × 2 = 8cm
Height = 5x = 5 × 2 = 10 cm

Question 3: If the perimeter of each face of a cube is 32 cm, find its lateral surface area. Note that four faces which meet the base of a cube are called its lateral faces.
Solution:
It is given that,
Perimeter of each face of a cube = 32 cm
Perimeter of each face of a cube = 4(Side)
4(Side) = 32
Side = 32/4
Side = 8
Lateral surface area = 4(Side)
Lateral surface area = 4 × (8)
Lateral surface area = 256 cm2
Hence, the Lateral surface area of cube is 256cm2