NCERT Solutions for Class 8 Maths

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3.

• Mensuration Class 8 Ex 11.1
• Mensuration Class 8 Ex 11.2
• Mensuration Class 8 Ex 11.4

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

Question 1.
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Solution.

Question 2.
A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpau¬lin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution.

Question 3.
Find the side of a cube whose surface area is 600 \({ cm }^{ 2 }\).
Solution.
Let the side of the cube be a cm.
Then, Total surface area of the cube = 6\({ a }^{ 2 }\)
According to the question,
6\({ a }^{ 2 }\)= 600
⇒ \({ a }^{ 2 }\) = \(\frac { 600 }{ 6 } \)
⇒ \({ a }^{ 2 }\) = 100
⇒ a = \(\sqrt { 100 } \)
⇒ a = 10 cm
Hence, the side of the cube is 10 cm.

Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution.

Question 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 area is painted.
How many cans of paint will she need to paint the room?
Solution.
l = 15 m
b = 10 m
h = 7 m
Surface area to be painted

Hence, she will need 5 cans of paint to paint the room.

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Solution.
Similarity → Both have the same heights.
Difference → One is a cylinder, the other is a cube;
The cylinder is a solid obtained by revolving a rectangular area about its one side whereas a cube is a solid enclosed by six square faces; a cylinder has two circular faces whereas a cube has six square faces.

Question 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.

Question 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?
Solution.
Lateral surface area of the hollow cylinder = 4224 \({ cm }^{ 2 }\)

Question 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.
Solution.
Diameter of the road roller = 84 cm

Question 10.
A company packages its milk powder in the 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.

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2.

• Mensuration Class 8 Ex 11.1
• Mensuration Class 8 Ex 11.3
• Mensuration Class 8 Ex 11.4

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

Question 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 man the d perpendicular distance between them is 0.8 m.
Solution.
Area of the top surface of the table

= \(\frac { 1 }{ 2 } h(a+b)\)
= \(\frac { 1 }{ 2 } \times 0.8\times (1.2+1)\)
= \(0.88{ m }^{ 2 }\)

Question 2.
The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the another parallel side.
Solution.
Area of trapezium
= \(\frac { 1 }{ 2 } h(a+b)\)

⇒ \(34=\frac { 1 }{ 2 } \times 4(10+b)\)
⇒ \(34=2\times (10+b)\)
⇒ \(10+b=\frac { 34 }{ 2 } \)
⇒ 10 + b=17
⇒ b = 17 – 10
⇒ b = 7 cm
Hence, the length of another parallel side is 7 cm.

Question 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.
Fence of the trapezium shaped field ABCD = 120 m
⇒ AB + BC + CD + DA = 120
⇒ AB + 48 + 17 + 40 = 120
⇒ AB + 105 = 120
⇒ AB = 120 – 105
⇒ AB = 15 m
∴ Area of the field
= \(\frac { (BC+AD)\times AB }{ 2 } \)
= \(\frac { (48+40)\times 16 }{ 2 } \) = 660 \({ m }^{ 2 }\)

Question 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.

Solution.
Area of the field
= \(\frac { 1 }{ 2 } d({ h }_{ 1 }+{ h }_{ 2 })\)
= \(\frac { 24\times (8+13) }{ 2 } \) = \(\frac { 24\times 21 }{ 2 } \)
= 12 x 21 = 252\({ m }^{ 2 }\)

Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution.
Area of the rhombus

= \(\frac { 1 }{ 2 } \times { d }_{ 1 }\times { d }_{ 2 }\)
= \(\frac { 1 }{ 2 } \times 7.5\times 12\)
= 45 \({ m }^{ 2 }\)

Question 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.
Solution.
Area of the rhombus
= base (b) x altitude (h) = 5
= 5 x 4.8 = 24 \({ cm }^{ 2 }\)

Question 7.
The floor of a building consists of 3,000 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.
Solution.
Area of a tile

Question 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 10,500 \({ m }^{ 2 }\) and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution.
Let the length of the side along the road be x m. Then, the length of the side along the river is 2x m.
Area of the field = 10,500 square metres

Question 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.
Area of the octagonal surface

Question 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 ?
Solution.

Question 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 the same.

Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3.

• Visualising Solid Shapes Class 8 Ex 10.1
• Visualising Solid Shapes Class 8 Ex 10.2

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3

Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution.
(i) No
(ii) Yes
(iii) Yes

Question 2.
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Solution.
Possible, only if the number of faces is greater than or equal to 4.

Question 3.
Which are prisms among the following?


Solution.
We know that a prism is a polyhedron whose base and top faces are congruent and parallel and other (lateral) faces are parallelograms in shape. So, only (ii) and (iv) are prisms.

Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution.
(i) The prisms and cylinder, both, have their base and top faces as congruent and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.

(ii) The pyramids and cones are alike in the sense that their lateral faces meet at a point (called vertex). Also, a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Question 5.
Is a square prism same as a cube ? Explain.
Solution.
No; not always as it can be a cuboid also.

Question 6.
Verify Euler’s formula for these solids

Solution.
(i)
F = 7
V= 10
E = 15
F + V = 7 + 10 = 17
E + 2 = 15 + 2 = 17
So, F + V = E + 2
Hence, Euler’s Formula is verified,

(ii)
F = 9
V = 9
E = 16
F + V = 9 + 9 = 18
E + 2 = 16 + 2 = 18
So, F + V = E + 2
Hence, Euler’s Formula is verified

Question 7.
Using Euler’s formula find the unknown.

Solution.
(i)
F + V = E + 2
⇒ F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8

(ii)
F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V= 11-5 = 6

(iii)
F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2
⇒ E = 30

Question 8.
Can a polyhedron have 10 faces, 20 edges, and 15 vertices?
Solution.
Here F = 10
E = 20
V= 15
So, F + V = 10 + 15 = 25
E + 2 = 20 + 2 = 22
∵ F + V ≠ E + 2
∴ Such a polyhedron is not possible.

We hope the NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2.

• Visualising Solid Shapes Class 8 Ex 10.1
• Visualising Solid Shapes Class 8 Ex 10.3

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2

Question 1.
Look at the given map of a city.
Answer the following.
(a) Colour the map as follows: Bluewater, Red-fire station, Orange-Library, Yellow-schools, Green-Parks, Pink-Community Centre, Purple-Hospital, Brown-Cemetery.
(b) Mark a green X’ at the intersection of Road ‘C’ and Nehru Road, Green Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?

Solution.
(a) Please color yourself.
(b) See the above figure.
(c) See the above figure.
(d) City Park.
(e) Senior Secondary school.

Question 2.
Draw a map of your classroom using a proper scale and symbols for different objects.
Solution.
Please draw yourself.

Question 3.
Draw a map of your school compound using a proper scale and symbols for various features like playground main building, garden etc.
Solution.
Please draw yourself.

Question 4.
Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Solution.
Please draw yourself.

We hope the NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5.

• Algebraic Expressions and Identities Class 8 Ex 9.1
• Algebraic Expressions and Identities Class 8 Ex 9.2
• Algebraic Expressions and Identities Class 8 Ex 9.3
• Algebraic Expressions and Identities Class 8 Ex 9.4

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 1.
Use a suitable identity to get each of the following products:


Solution.

Question 2.
Use the identity \((x+a)(x+b)={ x }^{ 2 }+(a+b)x+ab\) to find the following products:
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1)
(iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) \((2{ a }^{ 2 }+9)(2{ a }^{ 2 }+5)\)
(vii) (xyz – 4) (xyz – 2).
Solution.

Question 3.
Find the following squares by using the identities.

Solution.

Question 4.
Simplify:

Solution.

Question 5.
Show that:

Solution.

Question 6.
Using identities, evaluate:

Solution.

Question 7.
Using \({ a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b)\), find

Solution.

Question 8.
Using \((x+a)(x+b)={ x }^{ 2 }+(a+b)x+ab\), find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5, drop a comment below and we will get back to you at the earliest.