NCERT Solutions for Class 8 Maths Chapter 11: Download free pdf
While learning CBSE Class 8 Chapter 11, it is important for students to refer to NCERT Class 8 Maths Chapter 11 (Mensuration) Solutions prepared by experts. For this reason, Embibe has curated solutions to every question in Class 8 NCERT Chapter 11 in accordance with the NCERT textbook. You can either directly view or download the NCERT Solutions for Class 8 Maths Chapter 11 in PDF format from this article. Having a soft copy of the solutions in offline mode is really useful for quick reference. Read on to download CBSE Class 8 Chapter 11 solutions PDF, and get important study tips and chapter summary.
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration PDF
NCERT Solutions for Class 8 Maths Chapter 11 exercise questions and answers are provided in a simple and precise manner which will help you to have a better understanding of the concepts and score well in the exam. Download NCERT Solutions for Class 8 Maths Chapter 11.1, 11.2, 11.3, 11.4, etc., from the table below:
Answers of NCERT Class 8 Maths Chapter 11 Exercise Questions
Exercise 11.1 Page No: 171
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Solution: Side of the square = 60 m (Given) The length of a rectangular field, l = 80 m (Given) As per the question, Perimeter of rectangular field = Perimeter of square field i.e., 2(l+b) = 4×Side (using formulas) 2(80+b) = 4×60 or 160+2b = 240 or b = 40 Therefore, breadth of the rectangle is 40 m. Now, Area of Square field = (side)2 = (60)2 = 3600 m2 And Area of Rectangular field = length×breadth = 80×40 = 3200 m2 Therefore, area of square field is larger.
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.
Solution: Side of the square plot = 25 m We know that area of the square plot is equal to the square of its side = (side)2 = (25)2 = 625 So, the area of a square plot is 625 m2 Length of the house = 20 m and
The breadth of the house = 15 m Area of the house = length×breadth = 20×15 = 300 m2 Area of garden = Area of the square plot – Area of the house = 625–300 = 325 m2 ∵ The cost of developing the garden per sq. m is Rs. 55 Cost of developing the garden 325 sq. m = Rs. 55×325 = Rs. 17,875 Therefore, the total cost of developing a garden around is Rs. 17,875.
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]
Solution:: It is given that: Total length = 20 m Diameter of the semi-circle = 7 m Radius of the semi-circle = 7/2 = 3.5 m Length of the rectangular field = 20-(3.5+3.5) = 20-7 = 13 m The breadth of the rectangular field = 7 m Area of the rectangular field = l×b = 13×7= 91m2 Area of the two given semi-circles = 2×(1/2)×π×r2 = 2×(1/2)×22/7×3.5×3.5 = 38.5 m2 Area of the garden = 91+38.5 = 129.5 m2 Perimeter of the two semi-circles = 2πr = 2×(22/7)×3.5 = 22 m And Perimeter of the garden = 22+13+13 = 48 m. This is the answer.
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 1080 m2? [If required you can split the tiles in whatever way you want to fill up the corners] Solution: Given that the base of the flooring tile = 24 cm = 0.24 m Corresponding height of the flooring tile= 10 cm = 0.10 m We know that area of flooring tile= Base×Altitude = 0.24×0.10 = 0.024 The area of flooring tile is 0.024m2 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 entire floor.
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, the circumference of a circle can be obtained by using the expression C = 2πr, where r is the radius of the circle.
Solution: (a) Radius = Diameter/2 = 2.8/2 cm = 1.4 cm Circumference of the semi-circle = πr = (22/7)×1.4 = 4.4 Circumference of the semi-circle is 4.4 cm Total distance covered by the ant= Circumference of the semi -circle+Diameter = 4.4+2.8 = 7.2 cm (b) Diameter of the semi-circle = 2.8 cm Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of the semi-circle = r = (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 a semi-circle = 2.8 cm Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of the semi-circle = π r = (22/7)×1.4 = 4.4 cm Total distance covered by the ant = 2+2+4.4 = 8.4 cm After analyzing the results that we got for the three figures, we concluded that for figure (b) food piece, the ant would take a longer round.
Exercise 11.2 Page No: 177
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 the perpendicular distance between them is 0.8 m.
Solution: One parallel side of the trapezium (a) = 1 m And second side (b) = 1.2 m and height (h) = 0.8 m The area of the top surface of the table= (½)×(a+b)h = (½)×(1+1.2)0.8 = (½)×2.2×0.8 = 0.88 The area of the top surface of the table is 0.88 m2.
2. The area of a trapezium is 34 cm2and 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.
Solution: Let the length of the other parallel side be b. Length of one parallel side of the quadrilateral, a = 10 cm height, (h) = 4 cm and Area of the trapezium is 34 cm2 Formula for the area of trapezium = (1/2)×(a+b)h 34 = ½(10+b)×4 34 = 2×(10+b) After simplifying the equation we get, b = 7 Hence, the other parallel side 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. Solution:
Given: BC = 48 m, CD = 17 m, AD = 40 m & perimeter = 120 m ∵ Perimeter of the trapezium ABCD = AB+BC+CD+DA 120 = AB+48+17+40 120 = AB = 105 AB = 120–105 = 15 m Now, the area of the field= (½)×(BC+AD)×AB = (½)×(48 +40)×15 = (½)×88×15 = 660 Therefore, the area of the field ABCD is 660m2.
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.
Assume, h1 = 13 m, h2 = 8 m and AC = 24 m Area of the quadrilateral ABCD = Area of the triangle ABC+Area of the triangle ADC = ½( bh1)+ ½(bh2) = ½ ×b(h1+h2)= (½)×24×(13+8) = (½)×24×21 = 252 Therefore, the required area of the field is 252 m2
5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Solution: It is given that d1 = 7.5 cm and d2 = 12 cm ,Area of rhombus = (½ )×d1×d2 = (½)×7.5×12 = 45 Hence, area of rhombus is 45 cm2 .
6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal. Solution: Since a rhombus is also a kind of a parallelogram. The formula for Area of rhombus = Base×Altitude Assigning these values, we have The area of rhombus = 6×4 = 24 The area of a rhombus is 24 cm2 Also, the formula for the area of rhombus = (½)×d1d2 After substituting these values in the equation, we get 24 = (½)×8×d2 d2 = 6 Therefore, the length of the other diagonal is 6 cm.
7. The floor of a building consists of 3000 tiles thatare rhombus-shaped and each of its diagonals is 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2is Rs. 4. Solution: The length of one diagonal, d1 = 45 cm & d2= 30 cm ∵ The area of one tile = (½)d1d2 = (½)×45×30 = 675 The area of one tile is 675 cm2 The area of 3000 tiles is = 675×3000 = 2025000 cm2 = 2025000/10000 = 202.50 m2 [∵ 1m2 = 10000 cm2] ∵ The cost of polishing the floor per sq. meter = 4 The cost of polishing the floor per 202.50 sq. meter = 4×202.50 = 810 Therefore, the total cost of polishing the floor is Rs. 810.
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 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution: The perpendicular distance between the bottom and top of the trapezium (h) = 100 m (Given) Area of the field = 10500 m2 (Given) Let the length of the side along the road be ‘x’ m and side along the river = 2x m The area of the trapezium field = (½)×(a+b)×h 10500 = (½)×(x+2x)×100 10500 = 3x×50 After simplification, we have x = 70, which means the side along the river is 70 m Therfore, the side along the river = 2x = 2( 70) = 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.
Solution: The octagon has eight equal sides, each 5 m. (given) Divide the octagon as shown in the below figure, into 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and a rectangle with length and breadth 11 m and 5 m respectively.
Now, the area of two trapeziums = 2 [(½)×(a+b)×h] = 2×(½)×(11+5 )×4 = 4×16 = 64 The area of the two trapeziums is 64 m2 Moreover, the area of the rectangle = length×breadth = 11×5 = 55 The area of the rectangle is 55 m2 The total area of octagon = 64+55 = 119 m2
10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it into two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area? Solution: First way: As per Jyoti’s diagram, The area of the pentagon = Area of trapezium ABCP + Area of trapezium AEDP = (½)(AP+BC)×CP+(1/2)×(ED+AP)×DP = (½)(30+15)×CP+(1/2)×(15+30)×DP = (½)×(30+15)×(CP+DP) = (½)×45×CD = (1/2)×45×15 =337.5 m2 The area of the pentagon is 337.5 m2 Second way: As per Kavita’s diagram
Draw a perpendicular AM to BE. AM = 30–15 = 15 m The area of pentagon = Area of triangle ABE+Area of square BCDE (from the above figure) = (½)×15×15+(15×15) = 112.5+225.0 = 337.5 Therefore, the total area of pentagon-shaped park = 337.5 m2
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 the same.
Solution: Divide the given figure into 4 parts, as shown below:
Here two of the given figures (I) and (II) are similar in dimensions. Moreover, figures (III) and (IV) are similar in dimensions. The area of the figure (I) = Area of trapezium = (½)×(a+b)×h = (½)×(28+20)×4 = (½)×48×4 = 96 The area of Figure (I) = 96 cm2 Moreover, area of figure (II) = 96 cm2 We know that area of figure (III) = Area of trapezium = (½)×(a+b)×h = (½)×(24+16)4 = (½)×40×4 = 80 The area of figure (III) is 80 cm2 and, the area of the figure (IV) = 80 cm2
1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Solution: (a) Given that length of cuboidal box (l) = 60 cm The breadth of the cuboidal box (b) = 40 cm The height of cuboidal box (h) = 50 cm Total surface area of the cuboidal box = 2×(lb+bh+hl) = 2×(60×40+40×50+50×60) = 2×(2400+2000+3000) = 14800 cm2 (b) The length of cubical box (l) = 50 cm The breadth of cubicalbox (b) = 50 cm The height of cubicalbox (h) = 50 cm The total surface area of the cubical box = 6(side)2 = 6(50×50) = 6×2500 = 15000 The surface area of the cubical box is 15000 cm2 From the result of (a) and (b), the cuboidal box requires a lesser amount of material to make.
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? Solution: Th length of suitcase box, l = 80 cm, The breadth of suitcase box, b= 48 cm And the height of the cuboidal box , h = 24 cm The total surface area of suitcase box = 2(lb+bh+hl) = 2(80×48+48×24+24×80) = 2 (3840+1152+1920) = 2×6912 = 13824 The total surface area of the suitcase box is 13824 cm2 The area of the tarpaulin cloth = Surface area of the suitcase l×b = 13824 l ×96 = 13824 l = 144 Tarpaulin required for 100 suitcases = 144×100 = 14400 cm = 144 m Therefore, tarpaulin cloth required to cover 100 suitcases is 144 m.
3. Find the side of a cube whose surface area is 600cm2. Solution: The surface area of cube = 600 cm2 (Given) The formula for surface area of a cube = 6(side)2 Substituting the values, we get 6(side)2 = 600 (side)2 = 100 Or side = ±10 Since the side cannot be negative, the measure of each side of a cube is 10 cm
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?
Solution: The length of the cabinet, l = 2 m, Breadth of cabinet, b = 1 m and Height of cabinet, h = 1.5 m The surface area of cabinet = lb+2(bh+hl ) = 2×1+2(1×1.5+1.5×2) = 2+2(1.5+3.0) = 2+9.0 = 11 The required surface area of the cabinet is 11m2.
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 m^2 of the area is painted. How many cans of paint will she need to paint the room? Solution: Length of wall, l = 15 m, Breadth of wall, b = 10 m and Height of wall, h = 7 m Total Surface area of the classroom = lb+2(bh+hl ) = 15×10+2(10×7+7×15) = 150+2(70+105) = 150+350 = 500 Now, the required number of cans = Area of hall/Area of one can = 500/100 = 5 Therefore, 5 cans are required to paint the room.
6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?
Solution: Diameter of the cylinder = 7 cm (Given) The radius of the cylinder, r = 7/2 cm Height of the cylinder, h = 7 cm The lateral surface area of the cylinder = 2πrh = 2×(22/7)×(7/2)×7 = 154 So, the lateral surface area of the cylinder is154 cm2 Now, the lateral surface area of the cube = 4 (side)2=4×72 = 4×49 = 196 The lateral surface area of the cube is 196 cm2 Hence, the cube has a larger lateral surface area.
7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How many sheets of metal is required? Solution:
The radius of the cylindrical tank, r = 7 m Height of the cylindrical tank, h = 3 m The total surface area of the cylindrical tank = 2πr(h+r) = 2×(22/7)×7(3+7) = 44×10 = 440 Therefore, a 440 m2 metal sheet is required.
8. The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet? Solution: The lateral surface area of the hollow cylinder = 4224 cm2 Height of the hollow cylinder, h = 33 cm and say r be the radius of the hollow cylinder The curved surface area of the hollow cylinder = 2πrh 4224 = 2×π×r×33 r = (4224)/(2π×33) r = 64/π Now, Length of the rectangular sheet, l = 2πr l = 2 π×(64/π) = 128 (using value of r) So the length of the rectangular sheet is 128 cm. Also, Perimeter of the rectangular sheet = 2(l+b) = 2(128+33) = 322 The perimeter of the rectangular sheet is 322 cm.
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.
Solution: Diameter of the road roller, d = 84 cm Radius of the road roller, r = d/2 = 84/2 = 42 cm Length of the road roller, h = 1 m = 100 cm Formula for finding Curved surface area of the road roller = 2πrh = 2×(22/7)×42×100 = 26400 Curved surface area of the road roller is 26400 cm2 Again, Area covered by the road roller in 750 revolutions = 26400×750cm2 = 1,98,00,000cm2 = 1980 m2 [∵ 1 m2= 10,000 cm2] Therefore, the area of the road is 1980 m2.
10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. The 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?
Solution: Diameter of the cylindrical container, d = 14 cm Radius of the cylindrical container, r = d/2 = 14/2 = 7 cm Height of the cylindrical container = 20 cm Height of the label, says h = 20–2–2 (from the figure) = 16 cm The curved surface area of the label = 2πrh = 2×(22/7)×7×16 = 704 Therefore, the area of the label is 704 cm2.
Exercise 11.4 Page No: 191
1. Given a cylindrical tank, in which situation will you find the surface area 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. Solution: We find the area when a region is covered by a boundary, such as the outer and inner surface area of a cylinder, a cone, a sphere or a surface of a wall or floor. When the amount of space is occupied by an object such as water, milk, coffee, tea, etc., then we have to find out the volume of the object. (a) Volume (b) Surface area (c) Volume
2. Diameter of cylinder A is 7 cm and the height is 14 cm. The 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 cylinders. Check whether the cylinder with greater volume also has a greater surface area.
Solution: Yes, we can say that the volume of cylinder B is greater because the radius of cylinder B is greater than that of cylinder A. Find Volume for cylinders A and B Diameter of the cylinder A = 7 cm Radius of the cylinder A = 7/2 cm And Height of the cylinder A = 14 cm Volume of the cylinder A = πr2h = (22/7 )×(7/2)×(7/2)×14 = 539 Volume of the cylinder A is 539 cm3 Now, Diameter of the cylinder B = 14 cm Radius of the cylinder B = 14/2 = 7 cm And Height of the cylinder B = 7 cm Volume of the cylinder B = πr2h = (22/7)×7×7×7 = 1078 Volume of the cylinder B is 1078 cm3 Find surface area for cylinders A and B Surface area of the cylinder A = 2πr(r+h ) = 2 x 22/7 x 7/2 x (7/2 + 14) = 385 The surface area of the cylinder A is385 cm2 Surface area of the cylinder B = 2πr(r+h) = 2×(22/7)×7(7+7) = 616 Surface area of the cylinder B is 616 cm2 Yes, the cylinder with the greater volume also has the greater surface area.
3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3? Solution: Given, the base area of cuboid = 180 cm2 and Volume of cuboid = 900 cm3 We know that, the volume of the cuboid = lbh 900 = 180×h (substituting the values) h= 900/180 = 5 Therefore, the height of the cuboid is 5 cm.
4. A cuboid is of dimensions 60 cm×54 cm×30 cm. How many small cubes with a side of 6 cm can be placed in the given cuboid? Solution: Given, the length of the cuboid, l = 60 cm, Breadth of the cuboid, b = 54 cm and Height of the cuboid, h = 30 cm We know that, the volume of the cuboid = lbh = 60 ×54×30 = 97200 cm3 And Volume of the cube = (Side)3 = 6×6×6 = 216 cm3 Also, Number of small cubes = volume of the cuboid/volume of the cube = 97200/216 = 450 Therefore, the required number of cubes is 450.
5. Find the height of the cylinder whose volume if 1.54 m3 and the diameter of the base is 140 cm. Solution: Given that the volume of the cylinder = 1.54 m3 and the diameter of the cylinder = 140 cm Radius ( r )= d/2 = 140/2 = 70 cm Volume of the cylinder = πr2h 1.54 = (22/7)×0.7×0.7×h After simplifying, we get the value of h which is h = (1.54×7)/(22×0.7×0.7) h = 1 Therefore, the height of the cylinder is 1 m.
6. A milk tank is in the form of a 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.
Solution: Given, Radius of the cylindrical tank, r = 1.5 m and Height of the cylindrical tank, h = 7 m Volume of cylindrical tank, V = πr2h = (22/7)×1.5×1.5 ×7 = 49.5 cm3 = 49.5×1000 litres = 49500 litres [∵ 1 m3 = 1000 litres] Hence, the required quantity of milk is 49500 litres.
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? Solution: (i) Consider the edge of the cube to be “ l”. The formula for the Surface area of the cube, A = 6 l2 When the edge of the cube is doubled, then Surface area of the cube, say A’ = 6(2l)2 = 6×4l2 = 4(6 l2) A’ = 4A Hence, surface area will increase by four times. (ii) Volume of the cube, V = l3 When the edge of the cube is doubled, then Volume of the cube, say V’ = (2l)3 = 8(l3) V’ = 8×V Hence, the volume will increase 8 times.
8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m^3, find the number of hours it will take to fill the reservoir. Solution: Given, the volume of the reservoir = 108 m3 Rate of pouring water into the cuboidal reservoir = 60 litres/minute = 60/1000 m3per minute As 1 liter = (1/1000 )m3 = (60×60)/1000 m3 per hour Therefore, (60×60)/1000 m3 water filled in the reservoir will take = 1 hour Therefore 1 m3 water filled in the reservoir will take = 1000/(60×60) hours Therefore, 108 m3 water filled in the reservoir will take = (108×1000)/(60×60) hours = 30 hours Answer: It will take 30 hours to fill the reservoir.
NCERT Solutions for Class 8 Maths: Mensuration Chapter Description
In NCERT Solutions for Class 8 Maths Ch 11, students will learn about the measurement of geometric figures and their parameters like length, volume, shape, surface area, lateral surface area, etc. In the other half of the chapter, Solid Shapes like the cube, cuboid, and cylinder are discussed along with diagrams. The important mensuration formulas and properties of different geometric shapes and figures are also covered in this chapter. This chapter is based on formulas and thus, requires knowledge of the concepts and application of the formulas.
Along with Chapter 8 Class 11 Maths Solutions, Embibe also provides you with the Mensuration Chapter-wise Mock Test with which you can know how well you are prepared for the chapter and identify the areas you need to improve on. Solve Class 8 Maths Mensuration Practice Questions as well. Embibe gives you a summary based on how your attempt was, for example – Perfect Attempt, Too Fast Correct, Overtime Correct, Wasted Attempt, etc. This will help you improve your speed and accuracy.
CBSE Class 8 Maths Chapter 11 Important Questions
Some of the important questions from this chapter are as under:
1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm. 2. Find the area of a square whose side is 30m. 3. Calculate the area of a trapezium whose height is 4cm and the length of parallel sides are 5cm and 3cm respectively. 4. Find the perimeter of a triangle with two equal sides of 5 cm and one side of 10 cm. 5. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm. 6. Calculate the height of a cuboid which has a base area of 180 cm2 and volume is 900 cm3. 7. Find the area of the special quadrilateral whose diagonal length is 6 cm. And the lengths of two perpendiculars on the given diagonal from the vertices are 3 cm and 4 cm. 8. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted. 9. Find the area of the rhombus in which the length of its two diagonals is 8.5cm and 10 cm. 10. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed. 11. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.
The frequently asked questions regarding Chapter 11 Class 8 Maths Solutions are given below:
Q. What is the use of Embibe’s Chapter 8 Maths Class 11 NCERT Solutions? A. We have provided you with the CBSE Class 8 Maths – Mensuration Solutions. They are easy to follow and understand by any student of Grade 8. Students must make the best use of the solutions provided here as they solve the exercise questions.
Q. From where to download Chapter 11 Maths Class 8 NCERT Solutions? A.Click here to download the Class 8 Maths Chapter 11 NCERT Solutions.
Q. From where can I download Mensuration Formulas? A.Mensuration Formulas can be downloaded from here.
Q. How will NCERT Solutions for Class 8 Maths Chapter 11 help me in board exams? A. NCERT Solutions for Class 8 Maths Chapter 11 offers answers with detailed descriptions and is created by experts at Embibe for self-evaluation. Practising these questions will ensure that students have good preparation for all sorts of questions that can be devised in the finals.
Q. How can I score full marks in Mensuration Formulas? A. By using the NCERT Solutions for Class 8 Maths Chapter 11 provided in the Embibe platform will help you score full marks in class tests as well as in the final exams. These solutions offer easy and quick revision for the class tests and exams. Meanwhile, this serves as the best study material for learners.
We hope this article on NCERT Solutions For Class 8 Maths Chapter 11– Mensuration helps you. If you have any queries, feel free to ask in the comment section below. We will get back to you at the earliest.
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