NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 8 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 8 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 8 Mathematics are an important part of exams for Class 8 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 8 Mathematics and also download more latest study material for all subjects. Chapter 11 Mensuration is an important topic in Class 8, please refer to answers provided below to help you score better in exams
Chapter 11 Mensuration Class 8 Mathematics NCERT Solutions
Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 11 Mensuration in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks
Chapter 11 Mensuration NCERT Solutions Class 8 Mathematics
Exercise 11.1
Q.1) A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Sol.1) Given: The side of a square = 60 m
And the length of rectangular field = 80 m
According to question,
Perimeter of rectangular field = Perimeter of square field
⇒ 2(𝑙 + 𝑏) = 4 × 𝑠𝑖𝑑𝑒
⇒ 2(80 + 𝑏) = 4 × 60
⇒ 160 + 2𝑏 = 240
⇒ 2𝑏 = 240 − 160
⇒ 𝑏 = 80/2
⇒ 𝑏 = 40 𝑚
Now Area of Square field = (𝑠𝑖𝑑𝑒)2 = (60)2
= 3600 𝑚2
And Area of Rectangular field = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
= 80 × 40 = 3200 𝑚2
Hence, area of square field is larger.
Q.2) Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house.
Find the total cost of developing a garden around the house at the rate of 𝑅𝑠. 55 𝑝𝑒𝑟 𝑚2?
Sol.2) Side of a square plot = 25 𝑚
∴ Area of square plot (𝑠𝑖𝑑𝑒)2 = (25)2 = 625 𝑚2
Length of the house = 20 m and Breadth of the house = 15 𝑚
∴ Area of the house = length x breadth
= 20 × 15 = 300 𝑚2
Area of garden = Area of square plot – Area of house
= 625 – 300 = 325 𝑚2
Cost of developing the garden 𝑝𝑒𝑟 𝑠𝑞. 𝑚 = 𝑅𝑠. 55
∴ Cost of developing the garden 325 𝑠𝑞. 𝑚 = 𝑅𝑠. 55 × 325 = 𝑅𝑠. 17,875
Hence total cost of developing a garden around is 𝑅𝑠. 17,875.
Q.3) The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5 meters]
Sol.3) Given: Total length = 20 m
Diameter of semi-circle = 7 m
∴ Radius of semi-circle = 7/2 = 3.5𝑚
Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 𝑚
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = 𝑙 × 𝑏
= 13 × 7 = 91𝑚2
Area of two semi circles = 2 × (1/2)𝜋𝑟2
= 2 × 1/2 × 22/7 × 3.5 × 3.5 = 38.5 𝑚2
Area of garden = 91 + 38.5 = 129.5 𝑚2
Now Perimeter of two semi circles = 2 × π𝑟
= 2 × 22/7 × 3.5 = 22 𝑚
And Perimeter of garden = 22 + 13 + 13 = 48 m
Q.4) A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080𝑚2? [If required you can split the tiles in whatever way you want to fill up the corners]
Sol.4) Given: Base of flooring tile = 24 cm = 0.24 m
Corresponding height of a flooring tile = 10 cm = 0.10 m
Now Area of flooring tile = Base x Altitude
= 0.24 × 0.10 = 0.024 𝑚2
∴ Number of tiles required to cover the floor = Area of floor Area of one tile
= 1080/0.024 = 45000 𝑡𝑖𝑙𝑒𝑠
Hence 45000 tiles are required to cover the floor.
Q.5) An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression 𝑐 = 2𝜋𝑟 where 𝑟 is the radius of the circle.
Sol.5) (a) Radius = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 2.8/2 = 1.4𝑐𝑚
Circumference of semi-circle = π𝑟 = 22/7 × 1.4 = 4.4 𝑐𝑚
Total distance covered by the ant = Circumference of semi-circle + Diameter
= 4.4 + 2.8 = 7.2 𝑐𝑚
(b) Diameter of semi-circle = 2.8 cm
∴ Radius= 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 2.8/2 = 1.4𝑐𝑚
Circumference of semi-circle = π𝑟 = 22/7 × 1.4 = 4.4 𝑐𝑚
Total distance covered by the ant = 1.5 + 2.8 + 1.5 + 4.4 = 10.2 𝑐𝑚
(c) Diameter of semi circle = 2.8 cm
∴ Radius = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 = 2.8/2 = 1.4𝑐𝑚
Circumference of semi-circle = πr = 22/7 × 1.4 = 4.4 𝑐𝑚
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 𝑐𝑚
Hence for figure (b) food piece, the ant would take a longer round.
Exercise 11.2
Q.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.
Sol.1) Here one parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and height (h) = 0.8 m
∴ Area of top surface of the table = 1/2 (𝑎 + 𝑏) × ℎ
= 1/2 (1 + 1.2) × 0.8
= 1/2 × 2.2 × 0.8 = 0.88𝑚2
Hence, the surface area of the table is 0.88𝑚2.
Q.2) The area of a trapezium is 34𝑐𝑚2 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.
Sol.2) Let the length of the other parallel side be 𝑏.
Length of one parallel side (a) = 10 am and height (h) = 4 cm
Area of trapezium = 1/2 (𝑎 + 𝑏) × ℎ
⇒ 34 = 1/2 (10 + 𝑏) × 4
⇒ 34 = (10 + 𝑏) × 2
⇒ 34 = 20 + 2𝑏
⇒ 34 − 20 = 2𝑏
⇒ 14/2 = 𝑏
⇒ 7 = 𝑏 ⇒ 𝑏 = 7
Hence, the another required parallel side is 7 𝑐𝑚.
Q.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.
Sol.3) Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m
Perimeter of trapezium ABCD = AB + BC + CD + DA
⇒ 120 = AB + 48 + 17 + 40
⇒ 120 = AB = 105
⇒ AB = 120 – 105 = 15 m
Now Area of the field = 1/2 × (𝐵𝐶 + 𝐴𝐷) × 𝐴𝐵
= 1/2 × (48 + 40) × 15 = 1/2
× 88 × 15 = 660𝑚2
Hence, area of the field ABCD is 660𝑚2.
Q.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.
Sol.4) Here 1 h = 13 m, 2 h = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of Δ𝐴𝐵𝐶 + Area of Δ𝐴𝐷𝐶
Hence, required area of the field is 252𝑚2.
Q.5) The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Sol.5) Given: 𝑑1 = 7.5 cm and 𝑑2 = 12 𝑐𝑚
We know that, Area of rhombus = 1/2 × 𝑑1𝑑2 = 1/2 × 7.5 × 12 = 45𝑐𝑚2
Hence, area of rhombus is 45𝑐𝑚2.
Q.6) Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Sol.6) Since rhombus is also a kind of parallelogram.
∴ Area of rhombus = Base x Altitude
= 6 × 4 = 24 𝑐𝑚2
Also Area of rhombus = (1/2) 𝑑1𝑑2
⇒ 24 = 1/2 × 8 × 𝑑2
⇒ 24 = 4𝑑2
⇒ 𝑑2 = 24/2
= 6𝑐𝑚
Hence, the length of the other diagonal is 6 cm.
Q.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 𝑚2 is Rs. 4.
Sol.7) Here, 𝑑1 = 45 cm and 𝑑2 = 30 cm
Area of one tile = 1/2
𝑑1𝑑2 = 1/2 × 45 × 30 = 675 𝑐𝑚2
∴ Area of 3000 tiles = 675 × 3000 = 2025000 𝑐𝑚2
= 2025000/10000 = 202.50𝑐𝑚2
Cost of polishing the floor per sq. meter = 𝑅𝑠. 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 × 202.50 = 𝑅𝑠. 810
Hence, the total cost of polishing the floor is 𝑅𝑠. 810
Q.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 10500𝑚2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the
river.
Sol.8) Given: Perpendicular distance (h) = 100 m
Area of the trapezium shaped field = 10500𝑚2
Let side along the road be 𝑥 𝑚 and side along the river = 2𝑥 𝑚
∴ Area of the trapezium field = 1/2 (𝑎 + 𝑏) × ℎ
⇒ 10500 = 1/2 (𝑥 + 2𝑥) × 100
⇒ 10500 = 3𝑥 × 50
⇒ 3𝑥 = 10500/50
⇒ 𝑥 = 10500/50×3
⇒ 𝑥 = 70 𝑚
Hence, the side along the river = 2𝑥 = 2 × 70 = 140 𝑚.
Q.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.
Sol.9) Given: Octagon having eight equal sides, each 5 m.
Construction:
Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.
Now Area of two trapeziums = 2 × 1/2
(𝑎 + 𝑏) × ℎ = 2 × 1/2
(11 + 5) × 4 = 4 × 16 = 64𝑚2
Area of rectangle = length x breadth
= 11 × 5 = 55𝑚2
∴ Total area of octagon = 64 + 55 = 119 𝑚2
Q.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?
Sol.10) First way : By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= 1/2 (𝐴𝑃 + 𝐵𝐶) × 𝐶𝑃 + 1 2 (𝐸𝐷 + 𝐴𝑃) × 𝐷𝑃
= 1/2 (30 + 15 ) × 𝐶𝑃 + 1 2 (15 + 30) × 𝐷𝑃
= 1/2 (30 + 15) (𝐶𝑃 + 𝐷𝑃)
= 1/2 × 45 × 𝐶𝐷
= 1/2 × 45 × 15 = 337.5 𝑚2
Second way : By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
𝐴𝑀 = 30 – 15 = 15 𝑚
Area of pentagon = Area of Δ𝐴𝐵𝐸 + Area of square 𝐵𝐶𝐷𝐸
= 1 2 × 15 × 15 + 15 × 15 = 112.5 + 225.0 = 337.5 𝑚2
Hence, total area of pentagon shaped park = 337.5 𝑚2.
Q.11) Diagram of the adjacent picture frame has outer dimensions 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is same.
Sol.11) Here two of given figures (I) and (II) are similar in dimensions.
And also figures (III) and (IV) are similar in dimensions.
∴ Area of figure (I) = Area of trapezium
= 1/2 (𝑎 + 𝑏) × ℎ
= 1/2 (25 + 20) × 4
= 1/2 × 48 × 4 = 96 𝑐𝑚2
Also Area of figure (II) =96𝑐𝑚2
Now Area of figure (III) = Area of trapezium
= 1/2 (𝑎 + 𝑏) × ℎ
= 1/2 (24 + 16) × ℎ
= 1/2 × 40 × 4 = 80 𝑐𝑚2
Also Area of figure (IV) = 80 𝑐𝑚2
Exercise 11.3
Q.1) There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Sol.1) (a) Given: Length of cuboidal box (𝑙) = 60 𝑐𝑚
Breadth of cuboidal box (b) = 40 𝑐𝑚
Height of cuboidal box (h) = 50 𝑐𝑚
∴ Total surface area of cuboidal box = 2(𝑙𝑏 + 𝑏ℎ + ℎ𝑙)
= 2 (60 × 40 + 40 × 50 + 50 × 60)
= 2 (2400 + 2000 + 3000)
= 2 × 7400 = 14800 𝑐𝑚2
(b) Given: Length of cuboidal box (l) = 50 cm
Breadth of cuboidal box (b) = 50 cm
Height of cuboidal box (h) = 50 cm
∴ Total surface area of cuboidal box = 2(𝑙𝑏 + 𝑏ℎ + ℎ𝑙)
= 2 (50 × 50 + 50 × 50 + 50 × 50)
= 2 (2500 + 2500 + 2500)
= 2 × 7500 = 15000 𝑐𝑚2
Hence, the cuboidal box (a) requires the lesser amount of material to make, since surface area of box (a) is less than that of box (b).
Q.2) A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth.
How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Sol.2) Given: Length of suitcase box (l) = 80 cm,
Breadth of suitcase box (b) = 48 cm
And Height of cuboidal box (h) = 24 cm
∴ Total surface area of suitcase box = 2(𝑙𝑏 + 𝑏ℎ + ℎ𝑙)
= 2 (80 × 48 + 48 × 24 + 24 × 80)
= 2 (3840 + 1152 + 1920)
= 2 × 6912 = 13824 𝑐𝑚2
Area of Tarpaulin cloth = Surface area of suitcase
⇒ 𝑙 × 𝑏 = 13824
⇒ 𝑙 × 96 = 13824
⇒ 𝑙 = 13824/96 = 144𝑐𝑚
Required tarpaulin for 100 suitcases = 144 × 100 = 14400 𝑐𝑚 = 144 𝑚
Hence, the tarpaulin cloth required to cover 100 suitcases is 144 𝑚.
Q.3) Find the side of a cube whose surface area is 600 𝑐𝑚2.
Sol.3) Here Surface area of cube = 600𝑐𝑚2
⇒ 6𝑙2 = 600
⇒ 𝑙2 = 100
⇒ 𝑙 = 10 𝑐𝑚
Hence the side of cube is 10 𝑐𝑚.
Q.4) Rukshar 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?
Sol.4) Here, Length of cabinet (l) = 2 m,
Breadth of cabinet (b) = 1 m
And Height of cabinet (h) = 1.5 m
∴ Surface area of cabinet = 2(𝑙𝑏 + 𝑏ℎ + ℎ𝑙)
= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)
= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 𝑚2
Hence, the required surface area of cabinet is 11𝑚2.
Q.5) Daniel is paining 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 100𝑚2 of area is painted.
How many cans of paint will she need to paint the room?
Sol.5) Here, Length of wall (l) = 15 m, Breadth of wall (b) = 10 m
And Height of wall (h) = 7 m
∴ Total Surface area of classroom = 2(𝑙𝑏 + 𝑏ℎ + ℎ𝑙)
= 15 × 10 + 2 (10 × 7 + 7 × 15)
= 150 + 2 (70 + 105)
= 150 + 350 = 500 𝑚2
Now Required number of cans = 𝐴𝑟𝑒𝑎 𝑜𝑓 ℎ𝑎𝑙𝑙/𝐴𝑟𝑒𝑎 𝑜𝑓 1 𝑐𝑎𝑛
= 500/100 = 5 𝑐𝑎𝑛𝑠
Hence, 5 cans are required to paint the room.
Q.6) Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?
Sol.6) Given: Diameter of cylinder = 7 cm
∴ Radius of cylinder (r) = 7/2 𝑐𝑚
And Height of cylinder (h) = 7 cm
Lateral surface area of cylinder = 2πrh = 2 × 22/7 × 7/2 × 7 = 154𝑐𝑚2
Now lateral surface area of cube = 4𝑙2 = 4 × (7)2
= 4 × 49 = 196𝑐𝑚2
Hence, the cube has larger lateral surface area.
Q.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?
Sol.7) Given: Radius of cylindrical tank (r) = 7 m
Height of cylindrical tank (h) = 3 m
Total surface area of cylindrical tank = 2π𝑟 (ℎ + 𝑟)
= 2 × 22/7 × 7(3 + 7)
= 44 × 10 = 440 𝑚2
Hence, 440𝑚2 metal sheet is required.
Q.8) The lateral surface area of a hollow cylinder is 4224 𝑐𝑚2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Sol.8) Given: Lateral surface area of hollow cylinder = 4224 𝑐𝑚2
And Height of hollow cylinder = 33 cm
Curved surface area of hollow cylinder = 2πrh
⇒ 4224 = 22 2 33 7 x x xr
Perimeter of rectangular sheet = 2(𝑙 + 𝑏)
= 2 (128 + 33) = 2 × 161 = 322 𝑐𝑚
Hence, the perimeter of rectangular sheet is 322 𝑐𝑚.
Q.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 1 m.
Sol.9) Given: Diameter of road roller = 84 cm
∴ Radius of road roller (𝑟) = 𝑑/2 = 84/2 = 42𝑐𝑚
Length of road roller (h) = 1 𝑚 = 100 𝑐𝑚
Curved surface area of road roller = 2πrh
= 2(22/7) × 42 × 100 = 26400 𝑐𝑚2
∴ Area covered by road roller in 750 revolutions = 26400 × 750
= 1,98,00,000 𝑐𝑚2 = 1980 𝑚2 [ 1 𝑚2 = 10,000 𝑐𝑚2]
Hence, the area of the road is 1980 𝑚2.
Q.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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
Sol.10) Given: Diameter of cylindrical container = 14 cm
∴ Radius of cylindrical container (𝑟) = 𝑑/2 = 14/2 = 7𝑐𝑚
Height of cylindrical container = 20 cm
Height of the label (h) = 20 – 2 – 2 = 16 cm
Curved surface area of label = 2πrh = 2
22/7 × 7 × 16 = 704 c𝑚2
Hence, the area of the label of 704 𝑐𝑚2.
Exercise 11.4
Q.1) Given a cylindrical tank, in which situation will you find surface are and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Sol.1) We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.
(a) Volume (b) Surface area (c) Volume
Q.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.
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous).
Diameter of cylinder 𝐴 = 7 𝑐𝑚
⇒ Radius of cylinder 𝐴 = 7/2 𝑐𝑚
And Height of cylinder 𝐴 = 14 𝑐𝑚
∴ Volume of cylinder 𝐴 = π𝑟2 ℎ = 22/7 × 7/2 × 7/2 × 14 = 539 𝑐𝑚3
Now Diameter of cylinder 𝐵 = 14 𝑐𝑚
⇒ Radius of cylinder 𝐵 = 14/2 = 7𝑐𝑚
And Height of cylinder 𝐵 = 7 𝑐𝑚
∴ Volume of cylinder 𝐴 = π𝑟2 ℎ = 22/7 × 7 × 7 × 7 = 1078𝑐𝑚3
Total surface area of cylinder 𝐴 = 𝜋𝑟(2ℎ + 𝑟)
= 22 × (14 + 7) = 22 × 21 = 462 𝑐𝑚2
Yes, cylinder with greater volume also has greater surface area.
Q.3) Find the height of a cuboid whose base area is 180𝑐𝑚2 and volume is 900𝑐𝑚3?
Sol.3) Given: Base area of cuboid = 180 𝑐𝑚2 and
Volume of cuboid = 900 𝑐𝑚3
We know that, Volume of cuboid = 𝑙 × 𝑏 × ℎ
⇒ 900 = 180 × ℎ [Base area = 𝑙 × 𝑏 = 180 (given)]
⇒ ℎ = 900/180 = 5 𝑚
Hence, the height of cuboid is 5 𝑚.
Q.4) A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Sol.4) Given: Length of cuboid (l) = 60 𝑐𝑚,
Breadth of cuboid (b) = 54 𝑐𝑚 and
Height of cuboid (h) = 30 𝑐𝑚
We know that, Volume of cuboid = 𝑙 × 𝑏 × ℎ
= 60 × 54 × 30 𝑐𝑚3
And Volume of cube = (𝑠𝑖𝑑𝑒)3 = 6 × 6 × 6 𝑐𝑚3
∴2 Number of small cubes = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑢𝑏𝑜𝑖𝑑/𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑢𝑏𝑒 = 60×54×30/6×6×6 = 450
Hence, the required cubes are 450.
Q.5) Find the height of the cylinder whose volume if 1.54 𝑚3 and diameter of the base is 140 cm.
Sol.5) Given: Volume of cylinder = 1.54 𝑚3
and Diameter of cylinder = 140 cm
Q.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 litres that can be stored in the tank.
Sol.6) Given: Radius of cylindrical tank (r) = 1.5 m
And Height of cylindrical tank (h) = 7 m
Volume of cylindrical tank = π𝑟2 ℎ
= 22/7 × 1.5 × 1.5 × 7
= 49.5 𝑐𝑚3 = 49.5 × 1000 𝑙𝑖𝑡𝑟𝑒𝑠 [ 1 𝑚3 = 1000 𝑙𝑖𝑡𝑟𝑒𝑠]
= 49500 litres Hence, the required quantity of milk is 49500 𝑙𝑖𝑡𝑟𝑒𝑠.
Q.7) If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Sol.7) (i) Let the edge of cube be 𝑙.
Since, Surface area of the cube (𝐴) = 6𝑙2
When edge of cube is doubled, then
Surface area of the cube (𝐴′) = 6(2𝑙)2 = 6 × 4𝑙2
= 4 × 6𝑙2
𝐴′ = 4 × 𝐴
Hence, the surface area will increase four times.
(ii) Volume of cube (𝑉) = 𝑙3
When edge of cube is doubled, then
Volume of cube (𝑉’) = (2𝑙)3 = 8𝑙3
𝑉’ = 8 × 𝑉
Hence, the volume will increase 8 times
Q.8) Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 𝑚3, find the number of hours it will take to fill the reservoir.
Sol.8) Given: volume of reservoir = 108 𝑚3
Rate of pouring water into cuboidal reservoir = 60 litres/minute
| NCERT Solutions Class 8 Mathematics Chapter 1 Rational Numbers |
| NCERT Solutions Class 8 Mathematics Chapter 2 Linear Equations In One Variable |
| NCERT Solutions Class 8 Mathematics Chapter 3 Understanding Quadrilaterals |
| NCERT Solutions Class 8 Mathematics Chapter 4 Practical Geometry |
| NCERT Solutions Class 8 Mathematics Chapter 5 Data Handling |
| NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots |
| NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots |
| NCERT Solutions Class 8 Mathematics Chapter 8 Comparing Quantities |
| NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities |
| NCERT Solutions Class 8 Mathematics Chapter 10 Visualising Solid Shapes |
| NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration |
| NCERT Solutions Class 8 Mathematics Chapter 12 Exponents And Powers |
| NCERT Solutions Class 8 Mathematics Chapter 13 Direct and Inverse Proportions |
| NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation |
| NCERT Solutions Class 8 Mathematics Chapter 15 Introduction To Graphs |
| NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers |
NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration
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Chapter 11 Mensuration Class 8 Mathematics NCERT Solutions
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NCERT Solutions Chapter 11 Mensuration Class 8 Mathematics
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Chapter 11 Mensuration Class 8 NCERT Solution Mathematics
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Class 8 NCERT Solution Mathematics Chapter 11 Mensuration
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