NCERT Class 8 Maths Chapter 9 Mensuration Exercise 9.1, 9.2, 9.3 Solutions
In this page we have provided solutions of the Exercises 9.1, 9.2, 9.3 of NCERT Class 8 Maths Chapter 9 Mensuration. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Publishing Organisation |
NCERT |
Class |
8 |
Subject |
Mathematics |
Chapter |
9: Mensuration |
Exercise |
9.1, 9.2, 9.3 |
Chapter – 9
Exercise – 9.1
(1) The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
Ans:- Area of trapezium = ½ (Sum of Parallel Sides) × (Distances between Parallel Sides)
= [1/2 (1 + 1.2) (0.8)] m^{2} = 0.88 m^{2}
(2) The area of a trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Ans:- It is given that, area of trapezium = 34 cm^{2} and height = 4cm
Let the length of one Parallel side be a, we know that.
Area of trapezium = 1/2 (Sum of Parallel Sides) × (Distance between Parallel Sides)
34 cm2 = 1/2 (10 cm + a) × (4 cm)
34 cm = 2 (10 cm + a)
17 cm = 10 cm + a
a = 17 cm – 10 cm = 7 cm
Thus, the length of the other Parallel Sides is 7 cm.
(3) Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Ans:- Fence of the trapezium Shaped ABCD = 120 m
⇒ AB + BC + CD + DA = 120
⇒ AB + 48 + 17 + 40 = 120
⇒ AB + 105 = 120
⇒ AB = 120 – 105
⇒ AB = 15m
∴ Area of the field
= (BC + AD) × AB/2
= (48 + 40) × 16/2 = 660 m^{2}
(4) The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Ans:- It is given that,
Length of the diagonal, d = 24 m
Length of the Perpendiculars, h and h_{2}, from the opposite vertices
To the diagonal are h_{1 }= 8 m and h_{2} = 13 m
Area of the quadrilateral – 1/2 d (h_{1} + h_{2})
= 1/2 (24m) × (13m + 8cm)
= 1/2 (24m) (21m)
= 252 m^{2}
Thus, the area of the field is 252 m^{2}.
(5) The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Ans:- Area of the rhombus
= 1/2 × d_{1} × d_{2}
= 1/2 × 7.5 × 12
= 45m^{2}
(6) Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Ans:- Let the length of the other diagonal of the rhombus be x.
A rhombus is a Special case of a Parallelogram.
The area of a Parallelogram is given by the product of its base and height.
Thus, area of rhombus = 1/2 (Product of it diagonals)
⇒ 24 cm2 = 1/2 (8 cm × x)
⇒ x = (24 × 2/8) cm = 6 cm
Thus, the length of the diagonal of the rhombus is 6 cm.
(7) The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m^{2} is ₹ 4.
Ans:- Area of a tile
= 1/2 × d_{1} × d_{2} = 1/2 × 45 × 30
= 672 cm^{2}
∴ Area of the floor
= 675 cm^{2}
∴ Area of the floor
= 675 × 3,000 cm^{2} = 20, 25, 000 cm2
= 20, 25, 000/100 × 100 m^{2}
∵ 1m^{2} = 100 × 100 cm^{2}
The Cost of Polishing Per m^{2} = Rs 4
∴ Total Cost of Polishing the floor
= 202. 50 × 4
= Rs 80
(8) Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Ans:- Let the length of the field along the road be/ m. Hence, the length of the field along the river will be 2/m.
Area of trapezium = 1/2 (Sum of Parallel Sides) (Distance between the Parallel Sides)
⇒ 10500 m2 = 1/2 (1 + 21) × (100 m)
3l = (2 × 10500/100) m = 210 m
Thus, length of the field along the river = (2 × 70) m = 140 m.
(9) Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Ans:- Area of rectangular surface + 2 (Area of trapezoidal Surface)
= 11 × 5 + 2 [(5 + 11) × 4/2] m^{2}
= 55 + 64 m^{2} = 119 m^{2}
(10) There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?
Ans:- Jyoti’s way of finding area is as follows.
Area of Pentagon = 2 (Area of trapezium ABCF)
= [2 × 1/2 (15 + 30) (15/2)] m^{2}
= 337.5m^{2}
Kavita’s way of finding area is as follows.
Area of Pentagon = Area of △ ABE + Area of square BCDE
= [1/2 × 15 × (30 – 15) + (15)2] m^{2}
= (1/2 × 15 × 15 + 225) m^{2}
= (112.5 + 225) m^{2}
= 337.5 m^{2}
(11) Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.
Ans:- Area of the right Section of the frame
= (28 + 20) × 1/2 (24 – 16)/2 cm^{2}
= 48 × 4/2 cm2 = 96 cm^{2}
Similarly, area of the left Section of the frame = 96 cm^{2}
Area of the upper Section of the frame
= (24 + 16) × 1/2 (28 – 20)/2 cm^{2}
= 40 × 4/2 cm^{2}
Similarly, area of the lower section of the frame = 80 cm^{2}
Exercise – 9.2
(1) There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Ans:- We Know that,
Total Surface area of the Cuboid = 2 (lh + bh + lb)
Total Surface area of cuboid (a) = [2{(60) (40) + (40) + (50) + (50) + (60)}] cm^{2}
= [2 (2400 + 2000 + 3000)] cm^{2}
= (2 × 7400) cm^{2}
= 14800 cm^{2}
Total Surface area of Cube (b) = 6 (50 cm)^{2} = 15000 cm^{2}
Thus, the Cuboidal box (a) will require lesser amount of material.
(2) A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Ans:- Total Surface area of the Suitcase
= 2 (lb + bh + hl)
= 2 (80 × 48 + 48 × 24 + 24 × 80) cm^{2}
= 2 (3840 + 1152 + 1920) cm^{2}
= (6912) cm^{2}
= 13824 cm^{2}
∴ Length of tarpaulin required to cover 1 Suitcase
= Total Surface area of the Suitcase/Width of tarpaulin
= 13824/96 = 144 cm
∴ Length of tarpaulin required to cover 100 such Suitcases
= 144 × 100 cm = 14400 cm
= 144 m
Hence, 144 m of tarpaulin is required.
(3) Find the side of a cube whose surface area is 600 cm^{2}.
Ans:- Given that, Surface area of cube = 600 cm^{2}
Let the length of each side of cube be l.
Surface area of cube = 6(side)^{2}
600 cm^{2} = 6l^{2}
L^{2} = 100 cm^{2}
l = 10 cm
Thus, the side of the cube is 10 cm.
(4) Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.
Ans:- l = 1 m
b = 2 m
h = 1.5 m
Required area
= 2 (l × b + b × h + h × l) – l × b
= 2 (1 × 2 + 2 × 1.5 + 1.5 × 1) m^{2} – (1 × 2) m^{2}
= 13m^{2} – 2m^{2}
= 11 m^{2}
Hence, She Covered 11 m^{2} of Surface area.
(5) Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room?
Ans:- Given that
Length (l) = 15 m. breadth (b) = 10 m, height (h) = 7 m
Area of the hall to be painted = Area of the wall + Area of the celling
= 2h ( I + b) + lb
= [2 (7) (15 + 10) + 15 × 10] m^{2}
= [14 (25) + 150] m^{2}
= 500 m^{2}
It is given that 100 m^{2} area can be painted from each can.
Number of cans required to Paint an area of 500 m^{2}.
= 500/100 = 5
Hence, 5 Cans are required to paint the walls and Celling of the Cuboidal hall.
(6) Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Ans:- Similarity between both the figures is that both have the Same heights.
The difference between the two figures is that one is a Cylinder and the other is a Cube.
Lateral Surface area of the Cube = 41^{2} = 4 (7 cm)^{2} = 196 cm^{2}
Lateral Surface area of the Cylinder = 2πrh = (2 × 22/7 × 7/2 × 7) cm^{2} = 154 m^{2}
Hence, the Cube has longer lateral Surface area.
(7) A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Ans:- r = 7m
h = 3 m
∴ Total Surface area
= 2 πr (r + h)
= 2 × 22/7 × 7 × (7 + 3)
= 440 m^{2}
Hence, 440 m^{2} of metal sheet is required.
(8) The lateral surface area of a hollow cylinder is 4224 cm^{2} . It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Ans:- A hollow Cylinder is cut along its height to from a rectangular Sheet.
Area of Cylinder = Area of rectangular Sheet
4224 cm^{2} = 33 cm × Length
Length = 4224 cm^{2}/33 cm = 128 cm
Thus, the length of the rectangular sheet is 128 cm.
Perimeter of the rectangular Sheet = 2 (Length + Width)
= [2 (128 + 33)] cm
= (2 × 161) cm
= 322 m
(9) A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Ans:- In one revolution, the roller will cover an area equal to its lateral Surface area.
Thus, in 1 revolution, area of the road Covered = 2 πrh.
= 2 × 22/7 × 42 cm × 1m
= 2 × 22/7 × 42/100 m × 1 m
= 264/100 m^{2}
in 750 revolutions, area of the road Covered.
= (750 × 264/100) m^{2}
= 1980 m^{2}
(10) A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.
Ans:- Height of the label = 20cm – 2cm = 16cm
Radius of the Label = (14/2) cm = 7 cm
Label is in the form of a Cylinder having its height as 7 cm and 16 cm.
Area of the Label = 2 π (Radius) (Height)
= (2 × 22/7 × 7 × 16) cm^{2} = 704 cm^{2}
Exercise – 9.3
(1) Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
Ans:- Volume
(b) Number of cement bags required to plaster it.
Ans:- Surface area
(c) To find the number of smaller tanks that can be filled with water from it.
Ans:- Volume
(2) Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Ans:- The heights and diameters of these Cylinders A and B are interchanged.
We Know that,
Volume of Cylinder = πr^{2}h
If measure of r and h are same, then the Cylinder with greater radius will have greater area.
Radius of Cylinder A = 7/2 cm
Radius of Cylinder B = (14/2) cm = 7 cm
As the radius of Cylinder B is greater, therefor, the volume of cylinder B will be greater. Let us verify it by calculating the volume of both the Cylinders.
Volume of Cylinder A = πr^{2}h
= (22/7 × 7/2 × 7/2 × 14) cm^{2}
= 539 cm^{3}
Volume of Cylinder B = πr^{2}h
= (22/7 × 7 × 7 × 7) cm^{2}
= 1078 cm^{2}
Volume of Cylinder B is greater
Surface area of Cylinder A = 2πr (r + h)
= [2 × 22/7 × 7/2 (7/2 + 14)] cm^{2}
= [22 × (7 + 28)/2] cm^{2}
= (22 × 35/2) cm^{2}
= 385 cm^{2}
Surface area of Cylinder B = 2πr (r + h)
= [2 × 22/7 × 7 × (7 + 7)] cm^{2}
= (44 × 14) cm^{2}
= 616 cm^{2}
Thus, the Surface area of Cylinder B is also greater than the Surface area of Cylinder A.
(3) Find the height of a cuboid whose base area is 180 cm^{2} and volume is 900 cm^{3}?
Ans:- Height of the Cuboid
= Volume of the Cuboid/Base area of the Cuboid
= 900/180
= 5 cm
(4) A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Ans:- Volume of cuboid = 60 cm × 54 cm × 30 cm = 97200 cm^{3}
Side of the Cube = 6 cm
Volume of the Cube = (6)^{3} cm^{3 }= 216 cm^{3}
Required number of cubes = Volume of Cuboid/Volume of the Cube
= 97200/216 = 450
Thus, 450 Cubes can be placed in the given Cuboid.
(5) Find the height of the cylinder whose volume is 1.54 m^{3} and diameter of the base is 140 cm?
Ans:- ∵ Diameter of the base = 140 cm
∴ Radius of the base (r)
= 140/2 cm = 70 cm
∴ Area of the base = πr^{2}
= 22/7 × 70 × 70
= 15400 cm^{2}
Volume of the Cylinder
= 1.54 m^{3}
= 1.54 × 100 × 100 × 100 cm^{3}
= 1540000 cm^{3}
∴ Height of the Cylinder
= Volume of the Cylinder/Area of the base of the Cylinder
= 1540000/15400
= 100 cm
= 1 m
Hence, the height of the Cylinder is 1 m.
(6) A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank?
Ans:- Radius of Cylinder = 1.5 m
Length of Cylinder = 7m
Volume of Cylinder = πr^{2}h
= (22/7 × 1.5 × 1.5 × 7) m^{2}
= 49.5 m^{2}
Required quantity = (49.5 × 1000) L = 49500 L
Therefore, 49500 L of milk can be stored in the tank.
(7) If each edge of a cube is doubled,
(i) how many times will its surface area increase?
Ans:- Let the original edge of the cube be a cm.
Then, its new edge = 2a cm
Original Surface area of the cube = 6a^{2} cm^{2}
New. Surface are of the Cube
= 6 (2a)^{2} cm^{2}
= 24a^{2} cm^{2}
= 4 (6a^{2} cm^{2})
= 4 Original Surface area, Hence its Surface area will increase 4 times.
(ii) how many times will its volume increase?
Ans:- Original Volume of the Cube
= a^{3 }cm^{3}
New volume of the Cube
= (2a)^{3} cm^{3}
= 8a^{3} cm^{3}
= 8 Original Volume of the Cube.
Hence, its volume will increase 8 times.
(8) Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 m^{3}, find the number of hours it will take to fill the reservoir.
Ans:- Volume of Cuboidal reservoir = 108m^{3} = (108 × 1000) L = 108000L
It is given that water is being poured at the rate of 60 L Per minute.
That is, (60 × 60) L = 3600 L per hour
Required number of hours = 108000/3600 = 30 hours.
Thus, it will take 30 hours to fill the reservoir.
Chapter 8: Algebraic Expressions and Identities