Class 8

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 11.4

Question 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.
(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:
(a) Volume of cylinder
= Area of base × height.
(b) Total surface area of cylinder = 2πr (r + h).
(c) Volume of cylinder = πr²h.

Question 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 ?

Solution:
(A) Radius = \(\frac{d}{2}\) = \(\frac{7}{2}\) cm
Height = 14 cm
Volume of cylinder = πr²h

= \(\frac{22}{7}\) × 7 × 7 × 7
= 1078 cm³.

(A) Curved surface area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 14
= 308 cm²

(B) Curved surface area of cylinder
= 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 7
= 308 cm²
Hence the volume of cylinder (B) is greater than cylinder (A)
and surface area of cylinder (A) and (B) are equal.
If radius is greater, then volume of cylinder is greater.

Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm³?
Solution:
Volume of cuboid = l × b × h
Base area = l × b = 180 cm²
Volume of cuboid = 180 × h
900 cm³ = 180 × h
\(\frac{900}{180}\) = h
5 cm = h
∴ Height = 5 cm.

Question 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 ?
Solution:
Volume of cuboid
= 60 × 54 × 30
= 97200 cm²
Volume of cube (6 cm)³ = 216 cm³
Number of small cubes
= \(\frac{97200}{216}\) = 450 cubes.

Question 5.
Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm ?
Solution:
Volume of cylinder = 1.54 m³
Diameter of cylinder = 140 m

1 = h
Hence, height of cylinder = 1 m or 100 cm.

Question 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 ?
Solution:
Radius = 1.5 m
Height = 7 m
Volume of cylinder = πr²h
= \(\frac{22}{7}\) × 1.5 × 1.5 × 7
= 49.5 cm³
We know that
1 m³ = 1000 litres
Hence 49.5 m³ = 1000 × 49.5 m³
= 49500 litres
Hence, the required quantity of milk in litres in tank = 49500 litres.

Question 7.
If each edge of a cube is doubled,
(a) how many times will its surface area increase ?
(b) how many times will its volume increase ?
Solution:

(a) Let the edge of a cube = x
If each edge of a cube is doubled then edge
= 2x
If l = x
Now, surface area of cube
= 6 × (x)²
= 6x²
If l = 2x, then surface area of cube
= 6 × (2x)²
= 6 × 4x²
= 24x².
6x² × 3 = 24x²
Hence, three times will its surface area increase.

(b) If l = x,
Volume of cube = (l)³ = x³
If l = 2x,
Volume of cube = (2x)³ = 8x³
x³ × 8 = 8x³
Hence, eight times will its volume increase.

Question 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.

Solution:
Volume of cuboidal reservoir = 108 m³
Volume of cuboidal reservoir in litres
= 108 m³ = 108 × 1000 litres
(∵ 1 cm³ = 1000 l)
= 108000 litres
Since water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.
Hence no. of the times take to fill the water
= \(\frac{10800}{60}\) = 1800 minutes
= \(\frac{10800}{60}\) = 30 hours.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 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:
(a) Length = 60 cm
Breadth = 40 cm
Height = 50 cm
Total surface area of cuboidal boxes
= 2 (lb + bh + hl)
= 2 × (60 × 40 + 40 × 50 + 50 × 60)
= 2 × (240 + 200 + 300
= 2 × (740) = 1480 cm².

(b) l = b = h = 50 cm
Total surface area of cuboidal boxes = 6l²
= 6 × (50)² = 6 × 2500
= 15000 cm²
Hence (a) box requires the lesser amount of material to make.

Question 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 ?
Solution:
Total surface area of cuboidal suitcase
= 2 (lb + bh + hl)
= 2(80 × 48 + 48 × 24 + 24 × 80)
= 2 × (3840 + 1152 + 1920)
= 2 × 6912
= 13824 cm².

Question 3.
Find the side of a cube whose surface area is 600 cm².
Solution:
The total surface area of cube = 6l²
600 cm² = 6l²
\(\frac{600}{6}\) = l²
100 = l²
\(\sqrt {100}\) = l
10 = l
Hence the required side of a cube = 10 cm.

Question 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.

Solution:
l = 2 m, b = 1 m, h = 1.5 m
Lateral surface area of cuboidal cabinet
= 2 (l + b) × h
= 2(2 + 1) × 1.5
= 2 × 3 × 1.5 = 9 m²
Hence, required surface area covered = 9 m².

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 m2 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
Lateral surface area of cuboidal classrom
= 2 (l + b) × h
= 2(15 + 10) × 7
= 2 × 25 × 7 = 350 m²
Area of ceiling = l × b
= 15 × 10 = 150 m²
Hence, the total area of painted wall and ceiling
= 350+ 150 = 500 m²
Since each can of paint covers 100 m² of area
Hence 500 m² of area needed can = \(\frac{500}{100}\) = 5
Hence, the required 5 cans of paint cover 500 m².

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

Solution:
Lateral surface area of cylinder
= 2 × \(\frac{22}{7}\) × \(\frac{7}{22}\) × 7
[where = 7 cm
∴ r = \(\frac{7}{2}\) cm
h = 7 cm]
= 154 cm²
Lateral surface area of cube
= 4l² [where l = side = 7 cm]
= 4 × (7)²
= 4 × 49 = 196 cm²
Hence cubical box has larger lateral surface area.

Question 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 ?

Solution:
Total surface area of cylindrical tank
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 7 (7 + 3)
= 2 × 22 × 10 = 440 m²
Hence, the required metal sheet = 440 m².

Question 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 ?
Solution:
Width of rectangular sheet = 33 cm
Hence, height of cylinder = 33 cm
The lateral surface area = 2πrh
4224 = 2 × \(\frac{22}{7}\) × r × h
4224 = 2 × \(\frac{22}{7}\) × r × 33
\(\frac{4224 \times 7}{2 \times 22 \times 33}\) = r
20.36 = r
∴ Radius = 20.36
d = 2 × 20.36 = 40.72 cm
Now l = 40.72 and b = 33 cm
Hence, perimeter of rectangular sheet
= 2(l + b)
= 2 × (40.72 + 33)
= 2 × 73.22 = 147.44 cm.

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 road roller = 84 cm
∴ r = \(\frac{84}{2}\) = 2 cm
l = h = 1 m = 100 cm
Total surface area of road roller = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 42 × (42 + 100)
= 2 × 22 × 6 × 142
= 37488 cm²
Hence, the area of road
= 37488 × 750
= 28116000 cm²
\(\frac{28116000}{100×100}\) = 2811.6 m²
Hence, the required area of road = 2811.6 m².

Question 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 ?

Solution:
r = \(\frac{d}{2}\) = \(\frac{14}{2}\) = 7 cm
h = 20 cm
Top and bottom height = 2 + 2 = 4m
Hence required height of cylindrical powdered milk
= 20 – 4 = 16 cm
Lateral surface area of cylindrical powdered milk
= 2πrh = 2 × \(\frac{22}{7}\) × 7 × 16
= 2 × 22 × 16 = 704 cm²
Hence the required area of label = 704 cm².

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 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 m and perpendicular distance between them is 0.8 m.

Solution:
Area of trapezium shape
= \(\frac{1}{2}\)(b₁ + b₂) × h
= \(\frac{1}{2}\)(1.2 + 1) × 0.8 2
= \(\frac{1}{2}\) × 2.2 × 0.8 = 0.88 m².

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 other parallel side.
Solution:
Area of trapezium
= \(\frac{1}{2}\)(b₁ + b₂) × h
34 = \(\frac{1}{2}\)(10.4 + b₂) × 4
34 = (10.4 + b₂) × 2
34 = 20.8 + 2b₂
34 – 20.8 = 2b₂
13.2 = 2b₂
\(\frac{13.2}{2}\) = b₂
6.1 = b₂
Hence another parallel side = 6.6 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:
Perimeter of ABCD
= AB + BC + CD + 40
120 = AB + 48+ 17 + 40
= AB + 105
or 120 – 105 = AB
or 15 = AB
∴ AB = 15 cm

Area of trapezium ABCD
= \(\frac{1}{2}\)(b₁ + b₂) × k
= \(\frac{1}{2}\) × (48 + 40) × 15
= \(\frac{1}{2}\) × 88 × 15 = 660 m².

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 quadrilatral ABCD
= Area of ABC + Area of ADC
= \(\frac{1}{2}\) × AC × OD + \(\frac{1}{2}\) × AC × BP
= \(\frac{1}{2}\) × 24 × 13 + \(\frac{1}{2}\) × 24 × 8 2 2
= 12 × 13 + 12 × 8 = 12(13 + 8)
= 12 × 21 = 252 m².

Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Area of rhombus = \(\frac{1}{2}\) × d₁ × d₂
= \(\frac{1}{2}\) × 7.5 × 12
= 7.5 × 6 = 45.0
= 45 cm².

Question 6.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Area of DADO = \(\frac{1}{2}\) × AO × OD
= \(\frac{1}{2}\) × 4 × 4 = 8 cm²
Hence, Area of 4 triangle = 4 × 8 = 32 cm²

Now, Area of trapezium = \(\frac{1}{2}\) × d₁ × d₂
32 cm² = \(\frac{1}{2}\) × 8 × d₂
32 cm² = 4d₂
\(\frac{32 \mathrm{~cm}^{2}}{4 \mathrm{~cm}}\)= d₂
8 cm = d₂
Hence, the length of other diagonal = 8 cm.

Question 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.
Solution:
Area of rhombus of each tiles
= \(\frac{1}{2}\) × d₁ × d₂
= \(\frac{1}{2}\) × 45 × 30
= 675 cm²

Area of 3000 tiles = 3000 × 675
= \(\frac{2025000 \mathrm{~cm}^{2}}{100 \mathrm{~cm} \times 100 \mathrm{~cm}}\)
= 202.5 m²
Thecost of 3000 tiles = 202.5 × 4
= Rs. 810.0
= Rs.810.

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 Held 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:
Let the one side = x m = b₁
and other side = 2x m = b₂

Area of trapezium = \(\frac{1}{2}\)(b₁ + b₂) × h
= \(\frac{1}{2}\)(x + 2x) × 100
= \(\frac{1}{2}\) × 3x + 100
or 10500m² = 150x
or \(\frac{10500}{150}\) = x
70 = x
∴ x = 70 cm
Hence One side = 70 cm
and Another parallel side = 2 × 70 = 140 m.

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 another trapezium
= \(\frac{1}{2}\)(d₁ + d₂) × h
= \(\frac{1}{2}\)(11 + 5) × 4
= \(\frac{1}{2}\) × 16 × 4 = 32 m²
Area of rectangle
= l × b = 11 m × 5 m = 5m²

= \(\frac{1}{2}\)(d₁ + d₂) × h
= \(\frac{1}{2}\)(11 + 5) × 4 = 32 m²
∴ Area of the octagonal surface = Area of one trapezium + Rectangle + Area of another trapezium
= 32 m² + 55 m² + 32 m² =119 m².

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:
Jyoti’ diagram :
= \(\frac{1}{2}\)(b₁ + b₂) + h
= \(\frac{1}{2}\)(15 + 30) × 7.5
= \(\frac{1}{2}\) × 45 × 7.5
= 168.75 cm²

Hence, Area of two trapezium
= 2 × 168.75 = 237.50 cm²
Kavita’s diagram:
= Area of Kavita’s diagram
= \(\frac{1}{2}\) × 15 × 15 + (15)²
[∵ 30 – 15 = 15]
= \(\frac{225}{2}\) + \(\frac{225}{1}\) = \(\frac{225+50}{2}\)
= \(\frac{675}{2}\)
= 337.5 cm²

Question 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.
Solution:
Area of outer frame
= l × b
= 28 × 24
= 672 cm²

Area of inner frame
= l × b = 16 × 20 = 320 cm²
Hence, area of remaining frame
= 672 – 320 = 352 cm²
Now, area of each section
= \(\frac{352}{4}\) = 88 cm²

Haryana State Board HBSE 8th Class Maths Solutions Chapter 11 Mensuration Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 11 Mensuration Exercise 11.1

Question 1.
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area ?

Solution:
(a) Perimeter of square
= 4 × side
= 4 × 60 = 240 m

(b) Perimeter of rectangle = 2 (l + b)
240 m = 2(80 + b)
\(\frac{240}{2}\) = 80 + b
120 = 80 + b
120 – 80 = b
40 = b
∴ b = 40 m
∴ Area of square = (side)² = (60)²
60 × 60 = 3600 m²
and area of rectangle = l × b = 80 m × 40 m
= 3200 m²
Hence square field is larger than rectangle.

Question 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:
Area of garden
= Area of square plot – Area of middle plot (Rectangular plot)
= (side)² – (l × b)
= (25)² – (20 × 15)
= 625 – 300 = 325 m²
Hence the total cost of developing garden
= 325 m² × Rs. 55 = Rs. 17875.

Question 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) metres].

Solution:
Length of rectangle
= 20 – (3.5 + 3.5)
= 20 – 7 = 13 m
Breadth of rectangle = 7 m
Radius of semi-circle = 3.5 cm
Two semicircle = one circle
Hence perimeter (or circumference) of circle
= 2pr = 2 × \(\frac{22}{7}\) × 3.5 = 22 cm
Perimeter of rectangle
= 2 (l + b)
= 2 × (13 + 7)
= 2 × 20 = 40 cm
Hence, perimeter of this garden
=(22 cm + 40 cm) = 62 cm
Now, area of rectangle
= l × b = 13 × 7 = 91 cm²
Area of circle (two semicircle)
= πr² = \(\frac{22}{7}\) × 3.5 × 3.5
= 38.5 cm²
Hence, area of garden
= 38.5 cm² + 91 cm²
= 129.5 cm².

Question 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:
Area of parallelogram tiles
= base × height
= b × h
= 24 × 10 = 240 cm²
Area of parallelogram
= 1080 m²
= 1080 × 100 × 100 × cm²
= 10800000cm²
∴ Number of tiles
= \(\frac{Area of floor}{7}\)
= \(\frac{10800000}{240}\) = 45000.

Question 5.
An ant is moving around a few food

Solution:
(a) Circumference (perimeter) of semicircle
= \(\frac{2 \pi r}{2}\) = \(\frac{2 \times 22 \times 1.4}{2 \times 7}\)
(d = 2.8 cm, r = 1.4 cm)
= 4.4 cm
Perimeter = 4.4 cm + 2.8 cm = 7.2 cm.

(b) Circumference of semicircle + Perimeter of square
= 4.4 cm + 1.5 cm + 2.8 cm + 1.5 cm
= 10.2 cm.

(c) Circumference of semicircle + Perimeter of triangle
= 4.4 cm + 2 cm + 2 cm
= 8.4 cm
Hence (b) food piece would the ant have to take a longer round.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes InText Questions

Do This (Page 153-154)

Question 1.
Match the following :


Solution:

Do This (Page 154-155)

Question 1.
Match the following pictures (objects) with their shapes :

Solution:

Do This (Page 156)

Question 1.
Observe different things around you from different positions. Discuss with your friends their various views.
Solution:

Do This (Page 162-163)

Question 1.
Look at the following map of a city :

(a) Colour the map as follows: Blue-water, Red-fire station, Orange-Library, Yellow-schools, Green-Parks, Pink-Community Centre, Purple-Hospital, Brown-Cemetry.
(b) Mark a Green X’ at the intersection of 2nd street and Danim street. A black ‘Y’ where the river meets the third street. A red ‘Z’at the intersection of main street and 1st street.
(c) In magenta colour, draw a short street route from the college to the lake.
Solution:
Try yourself.

Question 2.
Draw a map of the route from your house to your school showing important landmarks.
Solution:
Try yourself.

Do This (Page 165)

Question 1.
Tabulate the number of faces, edges and vertices for the following polyhedrons : (Here ‘V’ stands for number of vertices. ‘F’ stands for number of faces and ‘E’ stands for number of edges).

Solution:

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Ex 10.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.3

Question 1.
Can a polyhedron have for its faces
(i) 3 triangles ?
(ii) 4 triangles ?
(iii) a square and four triangles ?
Answer:
Since, each of solids (3-D) is made up of polygonal regions which are called its faces; these faces meet at edges which are line segments; and the edges meet at vertices which are points. Such solids are called polyhedron. Hence prisms and pyramids are polyhedron, but triangle and square are 2-D figures. So (i), (ii) and (iii) cannot polyhedron.

Question 2.
Is it possible to have a polyhedron with any given number of faces ?
[Hint: Think of a pyramid]
Answer:
A pyramid is a solid whose base is a plane rectilinear figure whose side faces are triangles having a common vertex, called the vertex of the pyramid.

A pyramid is a polyhedron whose base is a polygon (of any number of sides) and whose lateral faces are triangles with a common vertex. If we join all the corners of a polygon to a point not in its plane, we get a model for pyramid.

Question 3.
Which are prisms among the following :

Answer:
A solid whose two faces are parallel plane polygons and the side faces are rectangles is called a prism.
Hence (iii) A table weight is prism.

Question 4.
(i) How are prisms and cylinders alike ?
(ii) How are pyramids and cones alike ?
Answer:
(i) A triangular prism is made up of two parallel end faces each one of which is a triangle and three lateral faces each one of which is a rectangle. A right circular cylinder has two plane ends. Each plane end is circular in shape. These two circular regions are congruent and parallel to each other. Hence, prisms and cylinders are alike.

(ii) A pyramid is a polyhedron whose base is a polygon and whose lateral faces are triangles with a common vertex. A triangular pyramid has a triangle as,it base. A right circular cone has a plane end which is circular in shape. This end is called the base of the cone. Hence pyramids and cone are alike.

Question 5.
Is a square prism same as a cube ? Explain.
Answer:
A square prism is made up of two parallel end-faces each one of which is a triangle and three lateral faces each one of which is a square. In a cube, its faces are congruent, regular polygons. Vertices are formed by the same number of faces. Thus a square prism and cube has a square of it base. Clearly their bases are square. Hence square prism and cube are same.

Question 6.
Verify Euler’s formula for these solids.

Solution:
Since, F + V = E or F + V- E = 2
This relationship is called Euler’s formula.
Where, F = Number of faces
V = Number of vertices
E = Number of edges.
(i) F = 7, V = 10, E = 15
From Euler’s formula,
F + V – E = 2 .
L.H.S. = 7 + 10 – 15
= 17 – 15 = 2, R.H.S.
Hence, verified the Euler’s formula.

(ii) F = 5, V = 5, E = 8
∴ F + V – E = 2
L.H.S. = 5 + 5 – 8
= 10 – 8 = 2 R.H.S.
Hence verified the Euler’s formula.

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

Solution:
∵ F + V – E = 2
(i) F = 2 + E – V
= 2 + 12 – 6 = 14 – 6 = 8

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

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

Question 8.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
For any polyhedron,
F + V – E = 2
Hence, L.H.S. = 10 + 15 – 20
= 25 – 20 = 5
Hence L.H.S. ≠ R.H.S.
This is not verified from Euler’s formula.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Ex 10.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.2

Question 1.
Look at the given map of a city.

Answer the following:
(a) Colour the maps as follows : Blue-water, red-fire station, orange-library, yellow-schools, Green-park, Pink-college, Purple-Hospital, Brown-Cemetery.
(b) Mark a green ‘X’at the intersection cf 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 lo 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:
Student fill this map with colours.

Question 2.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:
Try yourself.

Question 3.
Draw a map of your school compound using proper scale and symbols for various features like play ground main building, garden etc.
Solution:
Try yourself.

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

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Ex 10.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.1

Question 1.
For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.


Solution:

Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.


Solution:

Question 3.
For each given solid, identify the top view, front view and side view:


Solution:

Question 4.
Draw the front view, side view and top view of the given objects.


Solution:

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions

Try These (Page 138)

Question 1.
Give five examples of expressions containing one variable and five examples of expression containing two variables.
Solution:
One variable :
(i) x + 3,
(ii) x + 4,
(iii) y + 3,
(iv) 2 + 5,
(v) p + 2.

Two variables :
(i) 2xy + 3,
(ii) 3xy + 4,
(iii) 5xy + 3,
(iv) zx + 9,
(v) 2pq + 3.

Question 2.
Show on the number line x, x – 4, 2x + 1, 3x – 2.
Solution:

Identify the coefficient of each term in the expression
x²y – 10x²y + 5xy – 20
Coefficient of x²y = 1
Coefficient of -10x²y = -10
Coefficient of 5xy = 5.

Question 3.
Classify the following polynomials as monomials, binomials, trinomials
-z + 5, x + y + z, y + z + 100, ab – ac, 17.
Solution:
Monomials : 17
Binomials : -z + 5, ab – ac
Trinomials : x + y + z, y + z + 100.

Question 4.
Construct:
(a) 3 binomials with only x as a variable;
(b) 3 binomials with x and y as variables;
(c) 3 monomials withx and y as variables;
(d) 2 polynomials with 4 or more terms.
Solution:
(a) x + 2, 2x + 5, 3x + 1
(b) 2xy + 3, 3xy + 7, 4xy + 2
(c) xy, x²y², 4x³y³ .
(d) a + b + c + d, 3x + 2y + z + 5

Try These (Page 139)

Question 1.
Write two terms which are like
(i) 7xy,
(ii) 4mn²,
(iii) 2l.
Solution:
(i) (7xy + 2); (14xy + 3)
(ii) (4mn² – 1); (6mn² + 2)
(iii) (2l + 5); (4l + 3).

Try These (Page 142)

Question 1.
Can you think of two more such situations, where we may need to multiply algebraic expressions ?
[Hint: (i) Think of Speed and Time, (ii) Think of interest to be paid, the principal and the rate of simple interest etc.]
Solution:
(i) Speed = \(\frac{Distance}{Time}\)
s = \(\frac{d}{t}\)
⇒ d = s × t

(ii) Simple Interest
= \(\frac{Principal × Rate × Time}{100}\)
⇒ S.I. = \(\frac{P × R × T}{100}\)
⇒ P × R × T = S.I. × 100

Try These (Page 143)

Question 1.
Find 4x × 5y × 7z.
First find 4x × 5y and multiply it by 7r; or first find 5y × 7z and multiply it by 4x.
Is the result the same ? What do you observe?
Does the order in which you carry out the multiplication matter ?
Solution:
4x × 5y × 7z = (4x + by) × 8z
= (20xy) × 7z
= 140 xyz.

Try These (Page 144)

Question 1.
Find the product:
(i) 2x(3x + 5xy)
(ii) a²(2ab – 5c).
Solution:
(i) 2x(3x + 5xy)
= (2 × 3)x² + (2 × 5)x²y
= 6x² + 10x²y.

(ii) a²(2ab – 5c) = 2a³b – 5a²c.

Try These (Page 145)

Question 1.
Find the product (4p² + 5p + 7) × 3p.
Solution:
(4p² + 5p + 7) × 3p
= 12p³ + 15p² + 21p.

Try These (i) (Page 149)

Question 1.
Put -6 in place of 6 in Identity (I). Do you get Identity (II) ?
Solution:
Identity I
⇒ (a + b)² = a² = a² + 2ab + b²
If b = -b.
⇒ [a + (-b)]² = a² + 2a(-b) + (-b)²
⇒ (a – b)² = a² – 2ab + b²
Hence Identity-II verified.

Try These (ii) (Page 149)

Question 1.
Verify identity (IV), for a = 2, b = 3, x = 5.
Solution:
∵ (x + a) (x + b) = x² + (a + b)x + ab
If, a = 2, b = 3, x = 5
Then (x + a) (x + b)
= (5)² + (2 + 3) × 5 + 2 × 3
= 25 + 25 + 6 = 56.

Question 2.
Consider, the special case of identity (IV) with a = b, what do you get ? Is it related to identity (I) ?
Solution:
If a = b,
Then, (x + a) (x + b) = x² + (b + b) x + b × b
= x² + 2bx + b².

Question 3.
Consider, the special case of identity (IV) with a = -c and b = -c. What do you get ? Is it related to identity (II) ?
Solution:
If a = -c and b = -c
Then, (x + a) (x + b)
= x² + [(-c) + (-c)] x + (-c × -c)
= x² – 2cx + c².

Question 4.
Consider the special case of identity (IV) with b = -a. What/ do you get ? Is it related to identity (III) ?
Solution:
If b = -a,
Then, (x + a) (x + b)
= x² + [a + (-a)]x + a × (-a)
= x² + (a – a) x – a²
= x² + 0 – a²
= x² – a²
= (x + a) (x – a).

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Question 1.
Use a suitable identity to get each of the following products :
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\))
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a² + b²) (-a² + b²)
(vii) (6x- 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix)
(x) (7a – 9b) (7a – 9b).
Solution:
(i) ∵ (x + 9) (x + b)
= x² + (a + b) x + ab
∴ (x + 3) (x + 3) = x² + (3 + 3)x +3 × 3
= x² + 6x + 9.

(ii) ∵ (a + b)² = a² + 2ab + b²
∴ (2y + 5) (2y + 5) = (2y + 5)²
= (2y)² + 2 × 2y × 5 + (5)²
= 4y² + 20y + 25.

(iii) (2a – 7) (2a – 7)
= (2a-7)²
[∵ (a – b)² = a² – 2ab + b²]
= (2a)² – 2 × 2a × 7 + (7)²
= 4a² – 28a+ 49.

(iv) (3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\)) = (3a – \(\frac{1}{2}\))²
= (3a)² – 2 × 3a × \(\frac{1}{2}\) + (\(\frac{1}{2}\))²
= 9a² – 3a + \(\frac{1}{4}\)

(v) (1.1m -0.4) (1.1m + 0.4)
∵ (a + b) (a – b) = a² – b²
∴ (1.1m + 0.4) (1.1m – 0.4)
= (1.1m)² – (0.4)²
= 1.21m² – 0.16.

(vi) (a² + b²) (-a² + b²) = (b² + a²) (b² – a²)
= (b²)² – (a²)² – b⁴ – a⁴.

(vii) (6x – 7) (6x + 7) = (6x)² – (7)²
= 36x² – 49.

(viii) (c -a) (c- a) = (c – a)²
= c² – 2ca, + a²

(x) (7a – 9b) (7a – 9b)
= (7a – 9b)²
= (7a)² – 2 x 7a x 9b + (9b)²
= 49a² – 126ab + 81b².

Question 2.
Use the identity (x + a) (x + b) = x² + (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) (2a² + 9) (2a² + 5)
(vii) (xyz – 4) (xyz – 2).
Solution:
(i) (x + 3) (x + 7)
= x² + (3 + 7)x + 3 × 7
= x² + 10x + 21.

(iii) (4x + 5) (4x + 1)
= (4x)² + (5 + 1)x + 5 × 1
= 16x² + 6x + 5.

(iii) (4x – 5) (4x – 1)
= (4x)² + [-5 + (-1)]x + (-5x – 1)
= 16x² – 6x + 5.

(iv) (4x + 5) (4x – 1)
= (4x)² + [5 + (-1)]x + [5 × (-1)]
= 16x² + 4x – 5.

(v) (2x + 5y) (2x + 3y)
= (2x)² + (5y + 3y)x + 5y × 3y
= 4x² + 8xy + 15y².

(vi) (2a² + 9) (2a² + 5)
= (2a²)² + (9 + 5)2a² + 9 × 5
= 4a⁴ + 28a² + 45.

(vii) (xyz – 4) (xyz – 2)
= (xyz)² + [(-4) + (-2)] xyz + (-4) x (-2)
= x²y²z² – 6xyz + 8.

Question 3.
Find the following squares by using the identities :
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y)²
(iv)
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution:
(i) b – 7)² = b² – 2 x b x 7 + 7²
[∵ (a – b)² = a² – 2ab + b²]
= b² – 14 b + 49.

(ii) (xy + 3z)² = (xy)² + 2 × xy × 3z + (3z)²
[∵ (a + b)² = a² + 2ab + b²]
= x²y² + 6xyz² + 9z².

(iii) (6x² – 5y)² = (6x²)² – 2 × 6x² × 5y + (5y)²
= 36x⁴ – 60x²y + 25y².

(iv)

(v) (0.4p – 0.5q)² = (0.4p)² – 2 × 0.4p × 0.5q + (0.5q)²
= 0.16p² – 0.40pq + 0.25q².

(vi) (2xy + 5y)² = (2xy)² + 2 × 2xy × 5y + (5y)²
= 4x²y² + 20xy² + 25y².

Question 4.
Simplify :
(i) (a² – b²)²
(ii) (2x + 5)² – (2x – 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (4m + 5n)² + (5m + 4n)²
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
(vi) (ab + bc)² – 2ab²c
(vii) (m² – n²m)² + 2m³n².
Solution:
(i) (a² – b²)² = (a²)² – 2 × a² × b² + (b²)²
= a⁴ – 2a²b² + b⁴.

(ii) (2x + 5)² – (2x – 5)²
= [(2x + 5) + (2x – 5)] [(2x + 5) – (2x – 5)]
= (2x + 5 + 2x – 5) (2x + 5 – 2x + 5)
= 4x × 10 = 40x.

(iii) (7m – 8n)² + (7m + 8n)²
= (7m)² – 2 × 7m × 8n + (8n)² + (7m)² + 2 × 7m × 8n + (8n)²
= 2 × (7m)² + 2 × (8n)².
= 2 × 49m² + 2 × 64n²
= 98m² + 128n².

(iv) (4m + 5n)² + (5m + 4n)²
= (4m)² + 2 × 4m × 5n + (5n)² + (5m)² + 2 × 5m × 4n + (4n)²
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 41m² + 41n² + 80mn

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= [(2.5p)² – 2 × 2.5p × 1.5q + (1.5q)²] – [(1.5p)² – 2 × 1.5p × 2.5q + (2.5q)²]
= 6.25p² – 0.75pq + 2.25q² – 2.25p² + 0.75pq + 6.25q²
= 12.50p².

(vi) (ab + bc)² – 2ab²c
= (ab)² + 2 × ab × bc + (bc)² – 2ab²c
= a²b² + 2ab²c + b²c² – 2ab²c
= a²b² + b²c².

(vii) (m² – n²m)² + 2m³n²
= (m²)² – 2 × m² × n²m + (n²m)² + 2m³n²
= m⁴ – 2m³n² + n⁴m² + 2m³n² = m⁴ + n⁴m².

Question 5.
Show that:
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0.
Solution:
(i) (3x + 7)² – 84= (3x – 7)²
L.H.S. = (3²)² + 2 × 3x × 7 + (7)² – 84x
= 9x² + 42x + 49 – 84x
= 9x² – 42x + 49 = (3x – 7)²
= R.H.S. Hence proved.

(ii) (9p – 5q)² + 180pg = (9q + 5q)²
L.H.S. = (9p)² + 2 × 9p × 5q + (5q)² + 180pg
= (9p)² + 2 × 9pq × 5q + (5q)²
= (9p + 5q)²
= R.H.S. Hence proved.

(iii) (4pq + 3q)² – (4pq – 3q)² = 48pq²
L.H.S. = [(4pg + 3q) + (4pq – 3q)]
[(4pq + 3q) – (4pq – 3q)]
= (4pq + 3q + 4pq – 3g)
(4pq + 3q – 4pq + 3q)
= 8pq × 6q = 48pq²
= R.H.S. Hence proved.

(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = a² – b² + b² – c² + c² – a² = 0
= R.H.S Hence proved.

Question 6.
Using identities, evaluate.
(i) (71)²
(ii) (99)²
(iii) (102)²
(iv) (998)²
(v) (5.2)²
(vi) 297 × 303
(vii) 78 × 82
(viii) (8.9)²
(ix) 10.5 × 9.5
Solution:
(i) (70 + 1)² = (70)² + 2 × 70 × 1 + (1)²
= 4900 + 140 + 1
= 5041.

(ii) (99)² = (100 – 1)²
= (100)² – 2 × 100 × 1 + (1)²
= 10000 – 200 + 1 = 9801

(iii) (102)² = (100 + 2)²
= (100)² + 2 × 100 × 2 + (2)²
= 10000 + 400 + 4 = 10402.

(iv) (998)² = (1000 – 2)²
= (1000)² – 2 × 1000 × 2 + (2)²
= 1000000 – 4000 + 4
= 996004.

(v) (5.2)² = (5 + 0.2)²
= (5)² + 2 × 5 × 0.2 + (0.2)²
= 25 + 2.0 + 0.04 = 27.04.

(vi) 297 × 303 = (300 – 3) (300 + 3)
= (300)² – (3)²
= 90000 – 9
= 89991

(vii) 78 × 82 = (80 – 2) (80 + 2)
= (80)² – (2)²
= 6400 – 4
= 6396.

(viii) (8.9)² = (9 – 0.1)²
= (9)² – 2 × 9 × 0.1 + (0.1)²
= 81 – 1.8 + 0.01
= 79.21.

(ix) (10.5) × (9.5) = (10 + 0.5) (10 – 0.5)
= (10)² – (0.5)²
= 100 – 0.25
= 99.75.

Question 7.
Using a² – b² = (a + b) (a – b), find
(i) (51)² – (49)²
(ii) (1.02)² – (0.98)²
(iii) (153)² – (147)²
(iv) (12.1)² – (7.9)²
Solution:
(i) (51 + 49) (51 – 49) = 100 × 2 = 200.
(ii) (1.02)² – (0.98)²
= (1.02 + 0.98) (1.02 – 0.98)
= 2.00 × 0.04 = 0.08.

(iii) (153)² – (147)²
= (153 + 147) (153 – 147)
= 200 × 6 = 1200.

(iv) (12.1)² – (7.9)² = (12.1 + 7.9) (12.1 – 7.9)
= 20.0 × 4.2 = 84.0 = 84.

Question 8.
Using
(x + a) (x + b) = x² + (a + b) x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8.
Solution:
(i) (100 + 3) (100 + 4)
= (100)² + (3 + 4) × 100 + 3 × 4
= 10000 + 700 + 12 = 10712.

(ii) (5 + 0.1) (5 + 0.2)
= 5² + (0.1 + 0.2) × 5 + (0.1 × 0.2)
= 25 + 0.02 × 5 + 0.02
= 25 + 0.10 + 0.02
= 25.12.

(iii) (100 + 3) (100 – 2)
= (100)² + [3 + (-2)] × 100 + 3 × (-2)
= 10000 + 100 – 6
= 10094.

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)
= (10)² + (-0.3 + (-o.2)] × 10 + (-0.3) × (-0.2)
= 100 – 5 – 0.06
= 94.94.