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Exercise 12.1 | Class 8th Mathematics

1. Evaluate:
   (i) 3^-2
   (ii) (-4)^-2
   (iii) (12)^-5
   Solution:
   

2. Simplify and express the result in power notation with a positive exponent.

Solution:

3. Find the value of

Solution:

4.

Solution:

5. Find the value of m for which 5^m ÷ 5^-3 = 5⁵.
Solution:
5^m ÷ 5^-3 = 5⁵
⇒ 5^m-(-3) = 5⁵ [∵ a^m ÷ aⁿ = a^m-n]
⇒ 5^m+3 = 5⁵
Comparing the powers of equal bases, we have
m + 3 = 5
m = 5 – 3 = 2, i.e., m = 2

6. Evaluate:

Solution:

7. Simplify:

Solution:

Exercise 12.2 | Class 8th Mathematics

1. Express the following numbers in standard form:
   (i) 0.0000000000085
   (ii) 0.00000000000942
   (iii) 6020000000000000
   (iv) 0.00000000837
   (v) 31860000000
   Solution:
   

(v) 31860000000
= 3.186 × 10000000000
= 3.186 × 10¹⁰
Hence, the required standard from = 3.186 × 10¹⁰

2. Express the following numbers in usual form.
(i) 3.02 × 10^-6
(ii) 4.5 × 10⁴
(iii) 3 × 10^-8
(iv) 1.0001 × 10⁹
(v) 5.8 × 10¹²
(vi) 3.61492 × 10⁶
Solution:

3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 11000000 m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm.
Solution:

4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?
Solution:
Thickness of books = 5 × 20 = 100 mm
Thickness of 5 paper sheets = 5 × 0.016 mm = 0.080 mm.
Total thickness of the stack = 100 mm + 0.080 mm = 100.080 mm = 1.0008 × 10² mm

Extra Questions | Class 8th Mathametics

1. Find the multiplicative inverse of:
   (i) 3^-3
   (ii) 10^-10
   Solution:
   

2. Expand the following using exponents.
(i) 0.0523
(ii) 32.005
Solution:

3. Simplify and write in exponential form.

Solution:

4. Simplify the following and write in exponential form.

Solution:

5. Express 8^-4 as a power with the base 2.
Solution:
We have 8 = 2 × 2 × 2 = 2³
8^-4 = (2³)^-4 = 2^3×(-4) = 2^-12

6. Simplify the following and write in exponential form.
(i) (3⁶ ÷ 3⁸)⁴ × 3^-4
(ii) 127 × 3^-3
Solution:

7. Find the value of k if (-2)^k+1 × (-2)³ = (-2)⁷
Solution:
(-2)^k+1 × (-2)³ = (-2)⁷
⇒ (-2)^k+1+3 = (-2)⁷
⇒ (-2)^k+4 = (-2)⁷
⇒ k + 4 = 7
⇒ k = 3
Hence, k = 3.

8. Simplify the following:

Solution:

9. Find the value of [(−34)−2]2
Solution:

10. Write the following in standard form
(i) 0 0035
(ii) 365.05
Solution:

Exponents and Powers Class 8 Extra Questions Short Answer Type

11. Find the value of P if

Solution:

12.

Solution:

13. Find the value of x if

Solution:

⇒ 3 + x = 18 [Equating the powers of same base]
x = 18 – 3 = 15

14. Solve the following: (81)^-4 ÷ (729)^2-x = 9^4x
Solution:

15.

Solution:

16.

Solution:

17. Find x so that (-5)^x+1 × (-5)⁵ = (-5)⁷ (NCERT Exemplar)
Solution:
(-5)^x+1 × (-5)⁵ = (-5)⁷
(-5)^x+1+5 = (-5)⁷ {a^m × aⁿ = a^m+n}
(-5)^x+6 = (-5)⁷
On both sides, powers have the same base, so their exponents must be equal.
Therefore, x + 6 = 7
x = 7 – 6 = 1
x = 1.

Exercise 11.1 | Class 8th Mathematics

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:
   Perimeter of figure (a) = 4 × side = 4 × 60 = 240 m
   
   Perimeter of figure (b) = 2 [l + b]
   Perimeter of figure (b) = Perimeter of figure (a)
   2[l + b] = 240
   ⇒ 2 [80 + b] = 240
   ⇒ 80 + b = 120
   ⇒ b = 120 – 80 = 40 m
   Area of figure (a) = (side)² = 60 × 60 = 3600 m²
   Area of figure (b) = l × b = 80 × 40 = 3200 m²
   So, area of figure (a) is longer than the area of figure (b)

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 ₹ 55 per m².

Solution:
Area of the plot = side × side = 25 m × 25 m = 625 m²
Area of the house = l × b = 20 m × 15 m = 300 m²
Area of the garden to be developed = Area of the plot – Area of the house = 625 m² – 300 m² = 325 m²
Cost of developing the garden = ₹ 325 × 55 = ₹ 17875

3. The shape of a garden is rectangular in the middle and semicircular 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 the rectangle = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Area of the rectangle = l × b = 13 × 7 = 91 m2
Area of two circular ends = 2(12 πr²)
= πr²
= 227 × 72 × 72
= 772 m²
= 38.5 m²
Total area = Area of the rectangle + Area of two ends = 91 m² + 38.5 m² = 129.5 m²
Total perimeter = Perimeter of the rectangle + Perimeter of two ends
= 2 (l + b) + 2 × (πr) – 2(2r)
= 2 (13 + 7) + 2(227 × 72) – 4 × 72
= 2 × 20 + 22 – 14
= 40 + 22 – 14
= 48 m

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 the floor = 1080 m² = 1080 × 10000 cm² = 10800000 cm² [∵ 1 m² = 10000 cm²]
Area of 1 tile = 1 × base × height = 1 × 24 × 10 = 240 cm²
Number of tiles required

= 45000 tiles

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:

Exercise 11.2 | Class 8th Mathematics

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 the trapezium = 12 × (a + b) × h
   = 12 × (1.2 + 1) × 0.8
   = 12 × 2.2 × 0.8
   = 0.88 m²
   Hence, the required area = 0.88 m²

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 sides.
Solution:
Given: Area of trapezium = 34 cm²
Length of one of the parallel sides a = 10 cm
height h = 4 cm
Area of the trapezium = 12 × (a + b) × h
34 = 12 × (10 + b) × 4
⇒ 34 = (10 + b) × 2
⇒ 17 = 10 + b
⇒ b = 17 – 10 = 7 cm
Hence, the required length = 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:
AB + BC + CD + DA = 120 m .
BC = 48 m, CD = 17 m, AD = 40 m
AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m
Area of the trapezium ABCD = 12 × (BC + AD) × AB
= 12 × (48 + 40) × 15
= 12 × 88 × 15
= 44 × 15 = 660 m².
Hence, the required area = 660 m²

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 = area of ∆ABD + area of ∆BCD

= 12 × b × h + 12 × b × h
= 12 × 24 × 13 + 12 × 24 × 8
= 12 × 13 + 12 × 8
= 12 × (13 + 8)
= 12 × 21
= 252 m²
Hence, the required area of the field = 252 m².

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Here, d1 = 7.5 cm, d2 = 12 cm
Area of the rhombus = 12 × d1 × d2
= 12 × 7.5 × 12
= 7.5 × 6
= 45 cm²
Hence, area of the rhombus = 45 cm².

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:
Given: Side = 5 cm
Altitude = 4.8 cm
Length of one diagonal = 8 cm
Area of the rhombus = Side × Altitude = 5 × 4.8 = 24 cm²
Area of the rhombus = 12 × d₁ × d₂
24 = 12 × d₁ × d₂
24 = 4d₂
d₂ = 6 cm
Hence, the length of other diagonal = 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.
Solution:
Given: Number of tiles = 3000
Length of the two diagonals of a tile = 45 cm and 30 cm
Area of one tile = 12 × d₁ × d₂
= 12 × 45 × 30
= 45 × 15
= 675 cm²
Area covered by 3000 tiles = 3000 × 675 cm² = 2025000 cm² = 202.5 m²
Cost of polishing the floor = 202.5 × 4 = ₹ 810
Hence, the required cost = ₹ 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:
Let the side of the trapezium (roadside) be x cm.
The opposite parallel side = 2x m
h = 100 m
Area = 10500 m²
Area of trapezium = 12 (a + b) × h
10500 = 12 (2x + x) × 100
2 × 10500 = 3x × 100
21000 = 300x
x = 70 m
So, AB = 2x = 2 × 70 = 140 m
Hence, the required length = 140 m.

9. The 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 = area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF
Area of trapezium ABCH = Area of trapezium GDEF
= 12 (a + b) × h
= 12 (11 + 5) × 4
= 12 × 16 × 4
= 32 m²
Area of rectangle HCDG = l × b = 11 m × 5 m = 55 m²
Area of the octagonal surface = 32 m² + 55 m² + 32 m² = 119 m²
Hence, the required area = 119 m².

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:
(i) From Jyoti’s diagram:

Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF
= 2 × Area of trapezium ABCD
= 2 × 12 (a + b) × h
= (15 + 30) × 7.5
= 45 × 7.5
= 337.5 m²
(ii) From Kavita’s diagram:

Area of the pentagonal shape = Area of ∆ABE + Area of square BCDE
= 12 × b × h + 15 × 15
= 12 × 15 × 15 + 225
= 112.5 + 225
= 337.5 m²

Yes, we can also find the other way to calculate the area of the given pentagonal shape.
Join CE to divide the figure into two parts, i.e., trapezium ABCE and right triangle EDC.
Area of ABCDE = Area of ∆EDC + Area of square ABCE

11. Diagram of the 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:

Hence, the areas of the four parts A, B, C, and D are 80 cm², 96 cm², 80 cm^2, and 96 cm² respectively.

Exercise 11.3 | Class 8th Mathematics

1. There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?
   
   Solution:
   (a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 120000 cm³
   (b) Volume of cube = (Side)³ = (50)³ = 50 × 50 × 50 = 125000 cm³
   Cuboidal box (a) requires lesser amount of material.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Measurement of the suitcase = 80 cm × 48 cm × 24 cm
l = 80 cm, b = 48 cm and h = 24 cm
Total surface area of the suitcase = 2[lb + bh + hl]
= 2 [80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]
= 2 × 6912
= 13824 cm²
Area of tarpaulin = length × breadth = l × 96 = 96l cm²
Area of tarpaulin = Area of 100 suitcase
96l = 100 × 13824
l = 100 × 144 = 14400 cm = 144 m
Hence, the required length of the cloth = 144 m.

3. Find the side of a cube whose surface area is 600 cm²?
Solution:
Total surface area of a cube = 6l²
6l² = 600
l² = 100
l = √100 = 10 cm
Hence, the required length of side = 10 cm.

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

Solution:
l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box
= 2 [lb + bh + hl] – lb
= 2[2 × 1.5 + 1.5 × 1 + 1 × 2] – 2 × 1
= 2[3 + 1.5 + 2] – 2
= 2[6.5] – 2
= 13 – 2
= 11 m²
Hence, the required area = 11 m².

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 the area is painted. How many cans of paint will she need to paint the room?
Solution:
Surface area of a cuboidal hall without bottom = Total surface area – Area of base
= 2 [lb + bh + hl] – lb
= 2 [15 × 10 + 10 × 7 + 7 × 15] – 15 × 10
= 2[150 + 70 + 105] – 150
= 2 [325] – 150
= 650 – 150
= 500 m²
Area of the paint in one can = 100 m²
Number of cans required = 500100 = 5 cans.

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:
The two figures given are cylinder and cube.
Both figures are alike in respect of their same height.
Cylinder: d = 1 cm, h = 7 cm
Cube: Length of each side a = 7 cm
Both of the figures are different in respect of their shapes.
Lateral surface of cylinder = 2πrh
= 2 × 227 × 72 × 7 = 154 cm²
Lateral surface of the cube = 4l² = 4 × (7)² = 4 × 49 = 196
So, cube has the larger lateral surface = 196 cm².

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:
Area of metal sheet required = Total surface area of the cylindrical tank = 2πr(h + r)
= 2 × 227 × 7(3 + 7)
= 2 × 227 × 7 × 10
= 440 m²
Hence, the required area of sheet = 440 m².

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 the rectangular sheet.
Solution:
Width of the rectangular sheet = Circumference of the cylinder

h = 128 cm
l = 128 cm, b = 33 cm
Perimeter of the sheet = 2(l + b) = 2(128 + 33) = 2 × 161 = 322 cm
Hence, the required perimeter = 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 is 1 m.

Solution:
The lateral surface area of the road roller = 2πrh

Hence, the area of road = 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:
Here, r = 142 = 7 cm
Height of the cylindrical label = 20 – (2 + 2) = 16 cm
Surface area of the cylindrical shaped label = 2πrh
= 2 × 227 × 7 × 16
= 704 cm²
Hence, the required area of label = 704 cm².

Exercise 11.4 | Class 8th Mathematics

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:
   (a) In this situation, we can find the volume.
   (b) In this situation, we can find the surface area.
   (c) In this situation, we can find the volume.

2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Solution:
Cylinder B has a greater volume.
Verification:
Volume of cylinder A = πr²h

3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.
Solution:
Given: Area of base = lb = 180 cm²
V = 900 cm³
Volume of the cuboid = l × b × h
900 = 180 × h
h = 5 cm
Hence, the required height = 5 cm.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:
Volume of the cuboid = l × b × h = 60 cm × 54 cm × 30 cm = 97200 cm³
Volume of the cube = (Side)³ = (6)³ = 216 cm³
Number of the cubes from the cuboid

Hence, the required number of cubes = 450.

5. Find the height of the cylinder whose volume is 1.54 m³ and the diameter of the base is 140 cm.
Solution:
V = 1.54 m³, d = 140 cm = 1.40 m
Volume of the cylinder = πr2h

Hence, the height of cylinder = 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:
Here, r = 1.5 m
h = 7 m
.’. Volume of the milk tank = πr²h
= 227 × 1.5 × 1.5 × 7
= 22 × 2.25
= 49.50 m³
Volume of milk in litres = 49.50 × 1000 L (∵ 1 m³ = 1000 litres)
= 49500 L
Hence, the required volume = 49500 L.

7. If each edge of a cube is doubled,
(i) how many times will it be surface area increase?
(ii) how many times will its volume increase?
Solution:
Let the edge of the cube = x cm
If the edge is doubled, then the new edge = 2x cm
(i) Original surface area = 6x² cm²
New surface area = 6(2x)² = 6 × 4x² = 24x²
Ratio = 6x² : 24x² = 1 : 4
Hence, the new surface area will be four times the original surface area.

(ii) Original volume of the cube = x³ cm³
New volume of the cube = (2x)³ = 8x³ cm³
Ratio = x³ : 8x³ = 1 : 8
Hence, the new volume will be eight times the original volume.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the reservoir = 108 m³ = 108000 L [∵1 m³ = 1000 L]
Volume of water flowing into the reservoir in 1 minute = 60 L
Time taken to fill the reservoir

Hence, the required hour to fill the reservoir = 30 hours.

Exercise 10.1 | Class 10 | Mathematics

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:
   (a) A bottle → (iii) → (iv)
   (b) A weight → (i) → (v)
   (c) A flask → (iv) → (ii)
   (d) Cup and saucer → (v) → (iii)
   (e) Container → (ii) → (i)

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

Solution:
(a) An Almirah → (i) Front → (ii) Side → (iii) Top
(b) A Match box → (i) Side → (it) Front → (iii) Top
(c) A Television → (i) Front → (ii) Side → (iii) Top
(d) A Car → (i) Front → (ii) Side → (iii) Top

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

Solution:
(a) (i) Top → (ii) Front → (iii) Side
(b) (i) Side → (ii) Front → (iii) Top
(c) (i) Top → (ii) Side → (iii) Front
(d) (i) Side → (ii) Front → (iii) Top
(e) (i) Front → (ii) Top → (iii) Side

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

Solution:

Exercise 10.2 | Class 8th Mathematics

1. Look at the given map of a city.
   
   Answer the following.
   (a) Colour the map 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 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?

2. Draw a map giving instructions to your friend so that she reaches your house without any difficulty.

Solution:
Question 1 to Question 4 each all activities. You can try yourself.

Exercise 10.3 | Class 8th Mathematics

1. Can a polyhedron have for its faces
   (i) 3 triangles?
   (ii) 4 triangles?
   (iii) a square and four triangles?
   Solution:
   (i) No, because polyhedron must have edges meeting at vertices which are points.
   (ii) Yes, because all the edges are meeting at the vertices.
   
   (iii) Yes, because all the eight edges meet at the vertices.
   

2. Is it possible to have a polyhedron with any given number of faces?
(Hint: Think of a pyramid)
Solution:

Yes, it is possible if the number of faces is greater than or equal to 4.
Example: Pyramid which has 4 faces.

3. Which are prisms among the following?

Solution:
Only (ii) unsharpened pencil and (iv) a box are the prism.

4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution:
(i) If the number of sides in a prism is increased to certain extent, then the prism will take the shape of cylinder.
(ii) If the number of sides of the pyramid is increased to same extent, then the pyramid becomes a cone.

5. Is a square prism same as a cube? Explain.
Solution:
Every square prism cannot be cube. It may be cuboid also.

6. Verify Euler’s formula for these solids.

Solution:
(i) Faces = 7
Sides = 15
Vertices = 10
Euler’s formula: F + V – E = 2
⇒ 7 + 10 – 15 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.

(ii) Faces = 9
Sides = 16
Vertices = 9
Euler’s Formula: F + V – E = 2
⇒ 9 + 9 – 16 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.

7. Using Euler’s formula find the unknown.

Faces	?	5	20
Vertices	6	?	12
Edges	12	9	?

Solution:

Faces	8	5	20
Vertices	6	6	12
Edges	12	9	30

Using Eulers Formula: F + V – E = 2

8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
Here faces = 10, Edges = 20, Vertices = 15
According to Euler’s Formula:
F + V – E = 2
⇒ 10 + 15 – 20 = 25 – 20
⇒ 5 ≠ 2
A polyhedron do not have 10 Faces, 20 Edges and 15 Vertices.

Extra Questions | Class 8th Mathematics

Visualising Solid Shapes Class 8 Extra Questions Very Short Answer Type

1. Draw any four 3-dimensional figures.
   Solution:
   

2. Verify Euler’s formula for a right triangular prism.

Solution:
Number of vertices (V) = 6
Number of faces (F) = 5
and number of edges (E) = 9
Euler’s formula:
V + F – E = 2
⇒ 6 + 5 – 9 = 2
⇒ 2 = 2
Hence, the formula is verified.

3. Find the number of vertices of hexagonal prisms.
Solution:
Number of vertices = 2 × Number of sides = 2 × 6 = 12

4. Verify whether a polyhedron can have 10 faces, 20 edges and 15 vertices.
Solution:
We have
Number of faces F = 10
Number of edges E = 20
and number of vertices V = 15
Euler’s formula:
V + F – E = 2
⇒ 15 + 10 – 20 = 2
⇒ 5 ≠ 2
Hence, it is not possible to have a polyhedron satisfying the above data.

5. If F = 18 and V = 10, then find the value of E in Euler’s formula.
Solution:
We know that
V + F – E = 2
⇒ 10 + 18 – E = 2
⇒ 28 – E = 2
⇒ E = 28 – 2 = 26
Hence, the required value of E = 26

6. Draw the front, side and top views of the following 3-D figures.

Solution:

7. Draw the nets of the following polyhedrons.
(i) Cuboid
(ii) Triangular prism with a base equilateral triangle.
(iii) Square pyramid.
Solution:
(i) The net pattern of cuboid

(ii) The net pattern of a triangular prism

(iii) Net pattern of square pyramid

8. The given net is made up of two equilateral triangles and three rectangles.

(i) Name the solid it represents.
(ii) Find the number of faces, edges and vertices.
Solution:
(i) The given figure represents the net prims of the triangular prism
(ii) Number of faces = 5
Number of edges = 9
Number of vertices = 6

Visualising Solid Shapes Class 8 Extra Questions Short Answer Type

9. Match the following:

Solution:
(a) → (vi)
(b) → (v)
(c) → (ii)
(d) → (i)
(e) → (iii)
(f) → (v)

10. Using Euler’s formula, fill in the blanks:

	Faces	Vertices	Edges
(a)	6	8	—
(b)	—	10	15
(c)	4	—	6
(d)	5	6	—
(e)	8	12	—
(f)	7	7	—

Solution:
(a) F + V – E = 2
⇒ 6 + 8 – E = 2
⇒ 14 – E = 2
⇒ E = 14 – 2 = 12

(b) F + V – E = 2
⇒ F + 10 – 15 = 2
⇒ F – 5 = 2
⇒ F = 2 + 5 = 7

(c) F + V – E = 2
⇒ 4 + V – 6 = 2
⇒ V – 2 = 2
⇒ V = 2 + 2 = 4

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

(e) F + V – E = 2
⇒ 8 + 12 – E = 2
⇒ 20 – E = 2
⇒ E = 20 – 2 = 18

(f) F + V – E = 2
⇒ 7 + 7 – E = 2
⇒ 14 – E = 2
⇒ E = 14 – 2 = 12
Hence (a) → 12, (b) → 7, (c) → 4, (d) → 9, (e) → 18, (f) → 12

11. Name the solids that have:
(i) 4 faces
(ii) 8 triangular faces
(iii) 6 faces
(iv) 1 curved surface
(v) 5 faces and 5 vertices
(vi) 6 rectangular faces and 2 hexagonal faces
Solution:
(i) Tetrahedron
(ii) Regular octahedron
(iii) Cube and cuboid
(iv) Cylinder
(v) Square and a rectangular pyramid
(vi) Hexagonal prism

12. Complete the table:

Solid	F	V	E	F + V	E + 2
Cuboid	—	—	—	—	—
Triangular pyramid	—	—	—	—	—
Triangular prism	—	—	—	—	—
Pyramid with square base	—	—	—	—	—
Prism with square base	—	—	—	—	—

Solution:

Solid	F	V	E	F + V	E + 2
Cuboid	6	8	12	14	14
Triangular pyramid	4	4	6	8	8
Triangular prism	5	6	9	11	11
Pyramid with square base	5	5	8	10	10
Prism with square base	6	8	12	14	14

13. Use isometric dot paper to sketch a rectangular prism with length 4 units, height 2 units and width 3 units. (NCERT Exemplar)
Solution:
Steps:
1. Draw a parallelogram with sides 4 units and 3 units. This is the top of the prism (Fig. 1).

2. Start at one vertex. Draw a line passing through two dots. Repeat for the other three vertices. Draw the hidden edges as a dashed line (Fig. 2).

3. Connect the ends of the lines to complete the prism (Fig. 3).

Exercise 9.1 | Class 8th Mathematics

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 – 12) (3a – 12)
   (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) (x2 + 3y4) (x2 + 3y4)
   (x) (7a – 9b) (7a – 9b)
   Solution:
   

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:

3. Find the following squares by using the identities.
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y)²
(iv) (23 m + 32 n)²
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution:

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:

5. Show that:
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) (43 m – 34 n)² + 2mn = 169 m² + 916 n²
(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:

6. Using identities, evaluate:
(i) 71²
(ii) 99²
(iii) 102²
(iv) 998²
(v) 5.2²
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.9²
(ix) 1.05 × 9.5
Solution:

7. Using a² – b² = (a + b) (a – b), find
(i) 51² – 49²
(ii) (1.02)² – (0.98)²
(iii) 153² – 147²
(iv) 12.1² – 7.9²
Solution:
(i) 51² – 49² = (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) 153² – 147² = (153 + 147) (153 – 147) = 300 × 6 = 1800
(iv) 12.1² – 7.9² = (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

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) 103 × 104 = (100 + 3)(100 + 4) = (100)² + (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712
(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)² + (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)² + (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)² – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

Exercise 9.2 | Class 8th Mathematics

1. Find the product of the following pairs of monomials.
   (i) 4, 7p
   (ii) -4p, 7p
   (iii) -4p, 7pq
   (iv) 4p³, -3p
   (v) 4p, 0
   Solution:
   (i) 4 × 7p = (4 × 7) × p = 28p
   (ii) -4p × 7p = (-4 × 7) × p × p = -28p²
   (iii) -4p × 7pq = (-4 × 7) × p × pq = -28p²q
   (iv) 4p³ × -3p = (4 × -3) × p³ × p = -12p⁴
   (v) 4p x 0 = (4 × 0) × p = 0 × p = 0

2. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solution:
(i) Length = p units and breadth = q units
Area of the rectangle = length × breadth = p × q = pq sq units
(ii) Length = 10 m units, breadth = 5n units
Area of the rectangle = length × breadth = 10 m × 5 n = (10 × 5) × m × n = 50 mn sq units
(iii) Length = 20x² units, breadth = 5y² units
Area of the rectangle = length × breadth = 20x² × 5y² = (20 × 5) × x² × y² = 100x²y² sq units
(iv) Length = 4x units, breadth = 3x² units
Area of the rectangle = length × breadth = 4x × 3x² = (4 × 3) × x × x² = 12x³ sq units
(v) Length = 3mn units, breadth = 4np units
Area of the rectangle = length × breadth = 3mn × 4np = (3 × 4) × mn × np = 12mn²p sq units

3. Complete the table of Products.

Solution:
Completed Table

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a², 7a⁴
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c
Solution:
(i) Here, length = 5a, breadth = 3a², height = 7a⁴
Volume of the box = l × b × h = 5a × 3a² × 7a⁴ = 105 a⁷ cu. units
(ii) Here, length = 2p, breadth = 4q, height = 8r
Volume of the box = l × b × h = 2p × 4q × 8r = 64pqr cu. units
(iii) Here, length = xy, breadth = 2x²y, height = 2xy²
Volume of the box = l × b × h = xy × 2x²y × 2xy² = (1 × 2 × 2) × xy × x²y × xy² = 4x⁴y⁴ cu. units
(iv) Here, length = a, breadth = 2b, height = 3c
Volume of the box = length × breadth × height = a × 2b × 3c = (1 × 2 × 3)abc = 6 abc cu. units

5. Obtain the product of
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp
Solution:
(i) xy × yz × zx = x²y²z²
(ii) a × (-a²) × a³ = -a⁶
(iii) 2 × 4y × 8y² × 16y³ = (2 × 4 × 8 × 16) × y × y² × y³ = 1024y⁶
(iv) a × 2b × 3c × 6abc = (1 × 2 × 3 × 6) × a × b × c × abc = 36 a²b²c²
(v) m × (-mn) × mnp = [1 × (-1) × 1 ]m × mn × mnp = -m³n²p

Exercise 9.3 | Class 8th Mathematics

1. Carry out the multiplication of the expressions in each of the following pairs:
   (i) 4p, q + r
   (ii) ab, a – b
   (iii) a + b, 7a²b²
   (iv) a² – 9, 4a
   (v) pq + qr + rp, 0
   Solution:
   (i) 4p × (q + r) = (4p × q) + (4p × r) = 4pq + 4pr
   (ii) ab, a – b = ab × (a – b) = (ab × a) – (ab × b) = a²b – ab²
   (iii) (a + b) × 7a²b² = (a × 7a²b²) + (b × 7a²b²) = 7a³b² + 7a²b³
   (iv) (a² – 9) × 4a = (a² × 4a) – (9 × 4a) = 4a³ – 36a
   (v) (pq + qr + rp) × 0 = 0
   [∵ Any number multiplied by 0 is = 0]

2. Complete the table.

S.No.	First Expression	Second Expression	Product
(i)	a	b + c + d	–
(ii)	x + y – 5	5xy	–
(iii)	p	6p2 – 7p + 5	–
(iv)	4p2q2	p2 – q2	–
(v)	a + b + c	abc	–

Solution:
(i) a × (b + c + d) = (a × b) + (a × c) + (a × d) = ab + ac + ad
(ii) (x + y – 5) (5xy) = (x × 5xy) + (y × 5xy) – (5 × 5xy) = 5x²y + 5xy² – 25xy
(iii) p × (6p² – 7p + 5) = (p × 6p²) – (p × 7p) + (p × 5) = 6p³ – 7p² + 5p
(iv) 4p²q² × (p² – q²) = 4p²q² × p² – 4p²q² × q² = 4p⁴q² – 4p²q⁴
(v) (a + b + c) × (abc) = (a × abc) + (b × abc) + (c × abc) = a²bc + ab²c + abc²

Completed Table:

S.No.	First Expression	Second Expression	Product
(i)	a	b + c + d	ab + ac + ad
(ii)	x + y – 5	5xy	5x2y + 5xy2 – 25xy
(iii)	p	6p2 – 7p + 5	6p3 – 7p2 + 5p
(iv)	4p2q2	p2 – q2	4p4q2 – 4p2q4
(v)	a + b + c	abc	a2bc + ab2c + abc2

3. Find the products.

Solution:

4.
(a) Simplify: 3x(4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 12.
(b) Simplify: a(a² + a + 1) + 5 and find its value for (i) a = 0 (ii) a = 1 (iii) a = -1
Solution:
(a) We have 3x(4x – 5) + 3 = 4x × 3x – 5 × 3x + 3 = 12x² – 15x + 3
(i) For x = 3, we have
12 × (3)² – 15 × 3 + 3 = 12 × 9 – 45 + 3 = 108 – 42 = 66

(b) We have a(a² + a + 1) + 5
= (a² × a) + (a × a) + (1 × a) + 5
= a³ + a² + a + 5
(i) For a = 0, we have
= (0)³ + (0)² + (0) + 5 = 5
(ii) For a = 1, we have
= (1)³ + (1)² + (1) + 5 = 1 + 1 + 1 + 5 = 8
(iii) For a = -1, we have
= (-1)³ + (-1)² + (-1) + 5 = -1 + 1 – 1 + 5 = 4

5.
(a) Add: p(p – q), q(q – r) and r(r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= (p × p) – (p × q) + (q × q) – (q × r) + (r × r) – (r × p)
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – pq – qr – rp
(b) 2x(z – x – y) + 2y(z – y – x)
= (2x × z) – (2x × x) – (2x × y) + (2y × z) – (2y × y) – (2y × x)
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= -2x² – 2y² + 2xz + 2yz – 4xy
= -2x² – 2y² – 4xy + 2yz + 2xz
(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) – (4l × 3m) + (4l × 2l) – (3l × l) – (3l × -4m) – (3l × 5n)
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= (40ln – 15ln) + (-12lm + 12lm) + (8l² – 3l²)
= 25ln + 0 + 5l²
= 25ln + 5l²
= 5l² + 25ln
(d) [4c(-a + b + c)] – [3a(a + b + c) – 2b(a – b + c)]
= (-4ac + 4bc + 4c²) – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc
= -3a² – 2b² + 4c² – ab + 6bc – 7ac

Exercise 9.4 | Class 8th Mathematics

1. Multiply the binomials:
   (i) (2x + 5) and (4x – 3)
   (ii) (y – 8) and (3y – 4)
   (iii) (2.5l – 0.5m) and (2.5l + 0.5m)
   (iv) (a + 3b) and (x + 5)
   (v) (2pq + 3q²) and (3pq – 2q²)
   (vi) (34a² + 3b²) and 4(a² – 23 b²)
   Solution:
   (i) (2x + 5) × (4x – 3)
   = 2x × (4x – 3) + 5 × (4x – 3)
   = (2x × 4x) – (3 × 2x) + (5 × 4x) – (5 × 3)
   = 8x² – 6x + 20x – 15
   = 8x² + 14x – 15

(ii) (y – 8) × (3y – 4)
= y × (3y – 4) – 8 × (3y – 4)
= (y × 3y) – (y × 4) – (8 × 3y) + (-8 × -4)
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= (2.5l × 2.5l) + (2.5l × 0.5m) – (0.5m × 2.5l) – (0.5m × 0.5m)
= 6.25l² + 1.25ml – 1.25ml – 0.25m²
= 6.25l² + 0 – 0.25m²
= 6.25l² – 0.25m²

(iv) (a + 3b) × (x + 5)
= a × (x + 5) + 36 × (x + 5)
= (a × x) + (a × 5) + (36 × x) + (36 × 5)
= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) × (3pq – 2q²)
= 2pq × (3pq – 2q²) + 3q² (3pq – 2q²)
= (2pq × 3pq) – (2pq × 2q²) + (3q² × 3pq) – (3q² × 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴

2. Find the product:
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²)(2p + q)
Solution:
(i) (5 – 2x) (3 + x)
= 5(3 + x) – 2x(3 + x)
= (5 × 3) + (5 × x) – (2x × 3) – (2x × x)
= 15 + 5x – 6x – 2x²

(ii) (x + 7y) (7x – y)
= x(7x – y) + 7y(7x – y)
= (x × 7x) – (x × y) + (7y × 7x) – (7y × y)
= 7x² – xy + 49xy – 7y²
= 7x² + 48xy – 7y²

(iii) (a² + b) (a + b²)
= a² (a + b²) + b(a + b²)
= (a² × a) + (a² × b²) + (b × a) + (b × b²)
= a³ + a²b² + ab + b³

(iv) (p² – q²)(2p + q)
= p²(2p + q) – q²(2p + q)
= (p² × 2p) + (p² × q) – (q² × 2p) – (q² × q)
= 2p³ + p²q – 2pq² – q³

3. Simplify:
(i) (x² – 5) (x + 5) + 25
(ii) (a² + 5)(b³ + 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y)(x² – xy + y²)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c) (a + b – c)
Solution:
(i) (x² – 5) (x + 5) + 25
= x²(x + 5) + 5(x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x + 0
= x³ + 5x² – 5x

(ii) (a² + 5)(b³ + 3) + 5
= a²(b³ + 3) + 5(b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20

(iii) (t + s²) (t² – s)
= t(t² – s) + s²(t² – s)
= t³ – st + s²t² – s³
= t³ + s²t² – st – s³

(iv) (a + b)(c – d) + (a – b) (c + d) + 2(ac + bd)
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac + bc – bc – ad + ad – bd – bd + 2bd
= 4ac + 0 + 0 + 0
= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
= 2x² + x² + xy + 2xy – xy + 2xy + y² – 2y²
= 3x² + 4xy – y²

(vi) (x + y)(x² – xy + y²)
= x(x² – xy + y²) + y(x² – xy + y²)
= x³ – x²y + x²y + xy² – xy² + y³
= x³ – 0 + 0 + y³
= x³ + y³

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x.+ 12y
= 1.5x (1.5x + 4y + 3) – 4y(1.5x + 4y + 3) – 4.5x + 12y
= 2.25x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x + 12y – 12y – 16y²
= 2.25x² + 0 + 0 + 0 – 16y²
= 2.25x² – 16y²

(viii) (a + b + c) (a + b – c)
= a(a + b – c) + b(a + b – c) + c(a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + ab + ab – bc + bc – ac + ac + b² – c²
= a² + 2ab + b² – c² + 0 + 0
= a² + 2ab + b² – c²

Exercise 9.5 | Class 8th Mathematics

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 – 12) (3a – 12)
   (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) (x2 + 3y4) (x2 + 3y4)
   (x) (7a – 9b) (7a – 9b)
   Solution:
   

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:

3. Find the following squares by using the identities.
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y)²
(iv) (23 m + 32 n)²
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution:

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:

5. Show that:
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) (43 m – 34 n)² + 2mn = 169 m² + 916 n²
(iv) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(v) (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:

6. Using identities, evaluate:
(i) 71²
(ii) 99²
(iii) 102²
(iv) 998²
(v) 5.2²
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.9²
(ix) 1.05 × 9.5
Solution:

7. Using a² – b² = (a + b) (a – b), find
(i) 51² – 49²
(ii) (1.02)² – (0.98)²
(iii) 153² – 147²
(iv) 12.1² – 7.9²
Solution:
(i) 51² – 49² = (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) 153² – 147² = (153 + 147) (153 – 147) = 300 × 6 = 1800
(iv) 12.1² – 7.9² = (12.1 + 7.9) (12.1 – 7.9) = 20.0 × 4.2 = 84

Ex 9.5 Class 8 Maths 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) 103 × 104 = (100 + 3)(100 + 4) = (100)² + (3 + 4) (100) + 3 × 4 = 10000 + 700 + 12 = 10712
(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5)² + (0.1 + 0.2) (5) + 0.1 × 0.2 = 25 + 1.5 + 0.02 = 26.5 + 0.02 = 26.52
(iii) 103 × 98 = (100 + 3) (100 – 2) = (100)² + (3 – 2) (100) + 3 × (-2) = 10000 + 100 – 6 = 10100 – 6 = 10094
(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2) = (10)² – (0.3 + 0.2) (10) + (-0.3) (-0.2) = 100 – 5 + 0.06 = 95 + 0.06 = 95.06

Extra Questions | Class 8th Mathematics

Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

1. Write two examples of each of
(i) Monomials
(ii) Binomials
(iii) Trinomials
Solution:
(i) Monomials:
(a) 3x
(b) 5xy²
(ii) Binomials:
(a) p + q
(b) -5a + 2b
(iii) Trinomials:
(a) a + b + c
(b) x² + x + 2

2. Identify the like expressions.
5x, -14x, 3x² + 1, x², -9x², xy, -3xy
Solution:
Like terms: 5x and -14x, x² and -9x², xy and -3xy

3. Identify the terms and their coefficients for each of the following expressions:
(i) 3x²y – 5x
(ii) xyz – 2y
(iii) -x – x²
Solution:

4.
Add: -3a²b², –52 a²b², 4a²b², 23 a²b²
Solution:

5. Add: 8x² + 7xy – 6y², 4x² – 3xy + 2y² and -4x² + xy – y²
Solution:

6. Subtract: (4x + 5) from (-3x + 7)
Solution:
(-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

7. Subtract: 3x² – 5x + 7 from 5x² – 7x + 9
Solution:
(5x² – 7x + 9) – (3x² – 5x + 7)
= 5x² – 7x + 9 – 3x² + 5x – 7
= 5x² – 3x² + 5x – 7x + 9 – 7
= 2x² – 2x + 2

8. Multiply the following expressions:
(a) 3xy² × (-5x²y)
(b) 12 x²yz × 23 xy²z × 15 x²yz
Solution:

9. Find the area of the rectangle whose length and breadths are 3x²y m and 5xy² m respectively.
Solution:
Length = 3x²y m, breadth = 5xy² m
Area of rectangle = Length × Breadth = (3x²y × 5xy²) sq m = (3 × 5) × x²y × xy² sq m = 15x³y³ sq m

10. Multiply x² + 7x – 8 by -2y.
Solution:

Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

11. Simplify the following:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
Solution:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
= a²b² – a²c²) + b²c² – b²a²) + c²a² – c²b²)
= 0
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³
= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³
= 6x³ – x²y² – 2xy³

12. Multiply (3x² + 5y²) by (5x² – 3y²)
Solution:
(3x² + 5y²) × (5x² – 3y²)
= 3x²(5x² – 3y²) + 5y²(5x² – 3y²)
= 15x⁴ – 9x²y² + 25x²y² – 15y⁴
= 15x⁴ + 16x²y² – 15y⁴

13. Multiply (6x² – 5x + 3) by (3x² + 7x – 3)
Solution:
(6x² – 5x + 3) × (3x² + 7x – 3)
= 6x²(3x² + 7x – 3) – 5x(3x² + 7x – 3) + 3(3x² + 7x – 3)
= 18x⁴ + 42x³ – 18x² – 15x³ – 35x² + 15x + 9x² + 21x – 9
= 18x⁴ + 42x³ – 15x³ – 18x² – 35x² + 9x² + 15x + 21x – 9
= 18x⁴ + 27x³ – 44x² + 36x – 9

14. Simplify:
2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)
Solution:
2x²(x + 2) – 3x(x² – 3) – 5x(x + 5)
= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x
= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x
= -x³ – x² – 16x

15. Multiply x² + 2y by x³ – 2xy + y³ and find the value of the product for x = 1 and y = -1.
Solution:
(x² + 2y) × (x³ – 2xy + y³)
= x²(x³ – 2xy + y³) + 2y(x³ – 2xy + y³)
= x⁵ – 2x³y + x²y³ + 2x³y – 4xy² + 2y⁴
= x⁵ + x²y³ – 4xy² + 2y⁴
Put x = 1 and y = -1
= (1)⁵ + (1)² (-1)³ – 4(1)(-1)² + 2(-1)⁴
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

16. Using suitable identity find:
(i) 48² (NCERT Exemplar)
(ii) 96²
(iii) 231² – 131²
(iv) 97 × 103
(v) 181² – 19² = 162 × 200 (NCERT Exemplar)
Solution:

17.

Solution:

18. Verify that (11pq + 4q)² – (11pq – 4q)² = 176pq² (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)² – (11pq – 4q)² = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a² -b² = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq²
= RHS.
Hence Verified.

19. Find the value of 382−22216, using a suitable identity. (NCERT Exemplar)
Solution:

20. Find the value of x, if 10000x = (9982)² – (18)² (NCERT Exemplar)
Solution:
RHS = (9982)² – (18)² = (9982 + 18)(9982 – 18)
[Since a² -b² = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964

Exercise 8.1 | Class 8th Mathematics

1. Find the ratio of the following:
   (a) speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
   (b) 5 m to 10 km
   (c) 50 paise to ₹ 5
   Solution:
   (a) Speed of cycle : Speed of Scooter = 15 km per hour : 30 km per hour
   = 1530 = 12
   Hence, the ratio = 1 : 2
   (b) 5 m to 10 km
   = 5 m : 10 × 1000 m [∵ 1 km = 1000 m]
   = 5 m : 10000 m
   = 1 : 2000
   Hence, the ratio = 1 : 2000
   (c) 50 paise to ₹ 5
   = 50 paise : 5 × 100 paise
   = 50 paise : 500 paise
   ratio = 1 : 10

2. Convert the following ratios to percentages:
(a) 3 : 4
(b) 2 : 3
Solution:

3. 72% of 25 students are good in mathematics. How many are not good in mathematics?3.
Solution:
Number of students who are good in mathematics = 72% of 25

Number of students who are not good in mathematics = 25 – 18 = 7

4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution:
40 matches the team won out of 100 matches
1 match was won out of 10040 matches
10 matches the team will won out of 10040 × 10 = 25 matches
Hence, the total number of matches played by the team = 25

5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Let the money with Chameli be ₹ 100
Money spent by her = 75% of 100
= 75100 × 100 = ₹ 75
The money left with her = ₹ 100 – ₹ 75 = ₹ 25
₹ 25 are left with her out of ₹ 100
₹ 1 is left with her out of ₹ 10025
₹ 600 will be left out of ₹ 10025 × 600 = ₹ 2400
Hence, she had ₹ 2400 in beginning.

6. If 60% of people in a city like a cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.
Solution:
Total number of people = 50,00,000
Number of people who like cricket = 60% of 50,00,000
= 60100 × 50,00,000
= 30,00,000
Number of people who like football = 30% of 50,00,000
= 30100 × 50,00,000
= 15,00,000
Number of people who like other games = 50,00,000 – (30,00,000 + 15,00,000)
= 50,00,000 – 45,00,000
= 5,00,000
Percentage of the people who like other games = 5000005000000 × 100 = 10%
Hence, 10% of people like other game.

Exercise 8.2 | Class 8th Mathematics

1. A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
   Solution:
   The increased salary = ₹ 1,54,000
   Increase in salary = 10%
   Increase salary = Original salary × (1 + Increase100)
   
   Hence, the original salary = ₹ 1,40,000

2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?
Solution:
Number of people visiting the Zoo on Sunday = 845
Number of people visiting the Zoo on Monday = 169
Decrease in number of people visiting the Zoo = 845 – 169 = 676
Decrease per cent

Hence, the decrease per cent = 80%

3. A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
Cost price of 80 articles = ₹ 2,400
Cost of 1 article = ₹ 240080 = ₹ 30
Profit = 16%

Hence, the selling price of one article = ₹ 34.80

4. The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
CP of the article = ₹ 15,500
Money spent on repairs = ₹ 450
Net CP = ₹ 15,500 + ₹ 450 = ₹ 15,950
Profit = 15%

Hence, the selling price of article = ₹ 18342.50

5. A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss per cent on the whole transaction.

Solution:
Cost price of a VCR = ₹ 8,000
Loss = 4%

Hence, the shopkeeper gained 2% profit on the whole transaction.

6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of Jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?
Solution:
Marked Price (MP) of Jeans = ₹ 1,450
MP of two shirts = ₹ 850 × 2 = ₹ 1,700
Total MP = ₹ 1,450 + ₹ 1,700 = ₹ 3,150
Discount = 10%

Thus, the customer will have to pay ₹ 2,835.

7. A milkman sold two of his buffaloes for ₹ 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. [Hint: Find CP of each]

Solution:
SP of a buffalo = ₹ 20,000
Gain = 5%

8. The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Marked price of the TV = ₹ 13,000
ST = 12%

The required amount that Vinod has to pay = ₹ 14,560

9. Arun bought a pair of skates at a sale where the discount is given was 20%. If the amount he pays is ₹ 1,600, find the marked price.
Solution:
Let the MP of the skates be ₹ 100
Discount = ₹ 20% of 100 = ₹ 20
Sale price = ₹ 100 – ₹ 20 = ₹ 80
If SP is ₹ 80 then MP = ₹ 100
If SP is ₹ 1 then MP = ₹ 10080
If SP is ₹ 1,600 then MP = ₹ 10080 × 1600 = ₹ 2,000
Thus the MP = ₹ 2000.

10. I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.
Solution:
Let the original price be ₹ 100
VAT = 8% of 100 = ₹ 8
Sale price = ₹ 100 + ₹ 8 = ₹ 108
If SP is ₹ 108 then original price = ₹ 100
If SP is ₹ 1 then the original price = ₹ 100108
If SP is ₹ 5,400 then the original price = ₹ 100108 × 5,400 = ₹ 5,000
Thus, the price of hair-dryer before the addition of VAT = ₹ 5000.

Exercise 8.3 | Class 8th Mathematics

1. Calculate the amount and compound interest on
   (a) ₹ 10,800 for 3 years at 1212 % per annum compounded annually.
   (b) ₹ 18,000 for 212 years at 10% per annum compounded annually.
   (c) ₹ 62,500 for 112 years at 8% per annum compounded half yearly.
   (d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).
   (e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
   Solution:
   (a) Given:
   P = ₹ 10,800, n = 3 years,
   
   CI = A – P = ₹ 15,377.35 – ₹ 10,800 = ₹ 4,577.35
   Hence amount = ₹ 15,377.34 and CI = ₹ 4,577.34
   (b) Given: P = ₹ 18,000, n = 212 years = 52 years
   R = 10% p.a.
   The amount for 212 years, i.e., 2 years and 6 months can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.
   The amount for 2 years has to be calculated
   
   Total CI = ₹ 3780 + ₹ 1089 = ₹ 4,869
   Amount = P + I = ₹ 21,780 + ₹ 1,089 = ₹ 22,869
   Hence, the amount = ₹ 22,869
   and CI = ₹ 4,869
   (c) Given: P = ₹ 62,500, n = 112 years = 32 years per annum compounded half yearly
   = 32 × 2 years = 3 half years
   R = 8% = 82 % = 4% half yearly
   

CI = A – P = ₹ 70,304 – ₹ 62,500 = ₹ 7,804
Hence, amount = ₹ 70304 and CI = ₹ 7804
(d) Given: P = ₹ 8,000, n = 1 years R = 9% per annum compounded half yearly
Since, the interest is compounded half yearly n = 1 × 2 = 2 half years

CI = A – P = ₹ 8,736.20 – ₹ 8,000 = ₹ 736.20
Hence, the amount = ₹ 8736.20 and CI = ₹ 736.20
(e) Given: P = ₹ 10,000, n = 1 year and R = 8% pa compounded half yearly
Since the interest is compounded half yearly n = 1 × 2 = 2 half years

CI = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816
Hence the amount = ₹ 10,816 and Cl = ₹ 816

2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd
year amount for 412 years).
Solution:
Given:
P = ₹ 26,400
R = 15% p.a. compounded yearly
n = 2 years and 4 months

Amount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70
Hence, the amount to be paid by Kamla = ₹ 36,659.70

3. Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina: P = ₹ 12,500, R = 12% p.a. and n = 3 years

Difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50
Hence, Fabina pays more interest by ₹ 362.50.

4. I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
Given: P = ₹ 12,000, R = 6% p.a., n = 2 years

Difference between two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20
Hence, the extra amount to be paid = ₹ 43.20

5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
(i) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly

Hence, the required amount = ₹ 67416

6. Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after
112 years if the interest is
(i) compounded annually.
(ii) compounded half yearly.
Solution:
(i) Given: P = ₹ 80,000
R = 10% p.a.
n = 112 years
Since the interest is compounded annually

Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the third year.
Solution:
(i) Given: P = ₹ 8,000, R = 5% p.a.
and n = 2 years

Hence, interest for the third year = ₹ 441

8. Find the amount and the compound interest on ₹ 10,000 for 112 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution:
Given: P = ₹ 10,000, n = 112 years
R = 10% per annum
Since the interest is compounded half yearly

Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550
Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25
Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.

9. Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 1212 per annum, interest being compounded half yearly.
Solution:
Given: P = ₹ 4,096, R = 1212 % pa, n = 18 months

Hence, the required amount = ₹ 4913

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population in 2005?
Solution:
(i) Given: Population in 2003 = 54,000
Rate = 5% pa
Time = 2003 – 2001 = 2 years
Population in 2003 = Population in 2001

11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Given: Initial count of bacteria = 5,06,000
Rate = 2.5% per hour
n = 2 hours
Number of bacteria at the end of 2 hours = Number of count of bacteria initially

Thus, the number of bacteria after two hours = 5,31,616 (approx).

12. A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:
Given: Cost price of the scooter = ₹ 42,000
Rate of depreciation = 8% p.a.
Time = 1 year
Final value of the scooter

Hence, the value of scooter after 1 year = ₹ 38,640.

Extra Questions | Class 8th Mathematics

1.Express the following in decimal form:
(a) 12%
(b) 25%
Solution:
(a) 12% = 12100 = 0.12
(b) 25% = 25100 = 0.25

2.Evaluate the following:
(a) 20% of 400
(b) 1212% of 625
Solution:

3. If 20% of x is 25, then find x.
Solution:
20% of x = 25

Hence x = 125

4. Express the following as a fraction
(a) 35%
(b) 64%
Solution:

5. Express the following into per cent
(а) 135
(b) 2 : 5
Solution:

6. There are 24% of boys in a school. If the number of girls is 456, find the total number of students in the school.
Solution:
Let the total number of students be 100.
Number of boys = 24% of 100 = 24100 × 100 = 24
Number of girls = 100 – 24 = 76
⇒ If number of girls is 76, then total number of students = 100
⇒ If Number of girls is 1, then total number of students = 10076
If Number of girls is 456, then total number of students = 100×45676 = 600
Hence, the total number of students in the school = 600

7. The cost of 15 articles is equal to the selling price of 12 articles. Find the profit per cent.
Solution:
Let CP of 15 articles be ₹ 100
CP of 1 article = ₹ 10015
SP of 12 articles = ₹ 100
SP fo 1 article = ₹ 10012
SP > CP


Hence, profit = 25%

8. An article is marked at ₹ 940. If it is sold for ₹ 799, then find the discount per cent.
Solution:
MP = ₹ 940
SP = ₹ 799
Discount = MP – SP = 940 – 799 = ₹ 141

Hence, discount = 15%

9. A watch was bought for ₹ 2,700 including 8% VAT. Find its price before the VAT was added.
Solution:
Cost of watch including VAT = ₹ 2,700
Let the initial cost of the watch be ₹ 100
VAT = 8% of ₹ 100 = ₹ 8
Cost of watch including VAT = ₹ 100 + ₹ 8 = ₹ 108
If cost including VAT is ₹ 108, then its initial cost = ₹ 100
If cost including VAT is ₹ 1, then its initial cost = ₹ 100108
If cost including VAT is ₹ 2,700, then its initial cost = ₹ 100108 × 2700 = ₹ 2500
Hence, the required cost = ₹ 2,500

10. Find the amount if ₹ 2,000 is invested for 2 years at 4% p.a. compounded annually.
Solution:

Comparing Quantities Class 8 Extra Questions Short Answer Tpye

11. A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent. (NCERT Exemplar)
Solution:
Let the number be 100
20% increase = 20100 × 100 = 20
Increased value = 100 + 20 = 120
Now it is decreased by 20%
Decreased value = 120 – 120100 × 20 = 120 – 24 = 96
Net decrease = 100 – 96 = 4
Decrease per cent = 4100 × 100 = 4%
Hence, the net decrease per cent = 4%

12. Two candidates Raman and Rajan contested an election. Raman gets 46% of the valid votes and is#defeated by 1600 votes. Find the total number of valid votes cast in the election.
Solution:
Let the total number of valid votes be 100
Number of votes got by Raman = 46% of 100 = 46100 × 100 = 46
Number of votes got by Rajan = 100 – 46 = 54
Difference between the votes = 54 – 46 = 8
8% of Valid votes = 1,600
⇒ 8100 × Valid votes = 1,600
⇒ Valid votes = 1600×1008 = 20,000
Hence, the total number of valid votes = 20,000

13. A man whose income is ₹ 57,600 a year spends ₹ 43,200 a year. What percentage of his income does he save?
Solution:
Annual income of a man = ₹ 57,600
Amount spent by him in the year = ₹ 43,200
Net amount saved by him = ₹ 57,600 – ₹ 43,200 = ₹ 14,400
Percentage of his annual saving Saving = SavingIncome × 100
= 1440057600 × 100
= 25%
Hence, the saving percentage = 25%

14. A CD player was purchased for ₹ 3,200 and ₹ 560 were spent on its repairs. It was then sold at a gain of 1212 %. How much did the seller receive?
Solution:
Cost price of the CD player = ₹ 3,200
Amount spent on its repairing = ₹ 560
Net cost price = ₹ 3,200 + ₹ 560 = ₹ 3,760

Hence, the required amount = ₹ 4,230

15. A car is marked at ₹ 3,00,000. The dealer allows successive discounts of 6%, 4% and 212 % on it. What is the net selling price of it?
Solution:
Marked price of the car = ₹ 3,00,000
Net selling price after the successive discounts

Hence, the net selling price = ₹ 2,63,952

16. Ramesh bought a shirt for ₹ 336, including 12% ST and a tie for ₹ 110 including 10% ST. Find the list price (without sales tax) of the shirt and the tie together.
Solution:
List price of the shirt = 110112 × 336 = ₹ 300
List price of the tie = 100110 × 110 = ₹ 100
List price of both together = ₹ 300 + ₹ 100 = ₹ 400

17. Find the amount of ₹ 6,250 at 8% pa compounded annually for 2 years. Also, find the compound interest.
Solution:

18. Find the compound interest on ₹ 31,250 at 12% pa for 1212 years.
Solution:

19. Vishakha offers a discount of 20% on all the items at her shop and still makes a profit of 12%. What is the cost price of an article marked at ₹ 280? (NCERT Exemplar)
Solution:
Marked Price = ₹ 280
Discount = 20% of ₹ 280
= 12 × 280 = ₹ 56
So selling price = ₹ (280 – 56) = ₹ 224
Let the cost price be ₹ 100
Profit = 12% of ₹ 100 = ₹ 12
So selling price = ₹ (100 + 12) = ₹ 112
If the selling price is ₹ 112, cost price = ₹ 100
If the selling price is ₹ 224, cost price = ₹ (100112 × 224) = ₹ 200

20.Find the compound interest on ₹ 48,000 for one year at 8% per annum when compounded half yearly. (NCERT Exemplar)
Solution:
Principal (P) = ₹ 48,000
Rate (R) = 8% p.a.
Time (n) = 1 year
Interest is compounded half yearly

Therefore Compound Interest = A – P = ₹ (519,16.80 – 48,000) = ₹ 3,916.80.

Exercise 7.1 | Class 8th Mathematics

1. Which of the following numbers are not perfect cubes?
   (i) 216
   (ii) 128
   (iii) 1000
   (iv) 100
   (v) 46656
   Solution:
   (i) Prime factorisation of 216 is:
   216 = 2 × 2 × 2 × 3 × 3 × 3
   In the above factorisation, 2 and 3 have formed a group of three.
   Thus, 216 is a perfect cube.
   
   (ii) Prime factorisation of 128 is:
   128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
   Here, 2 is left without making a group of three.
   Thus 128 is not a perfect cube.
   
   (iii) Prime factorisation of 1000, is:
   1000 = 2 × 2 × 2 × 5 × 5 × 5
   Here, no number is left for making a group of three.
   Thus, 1000 is a perfect cube.
   
   (iv) Prime factorisation of 100, is:
   100 = 2 × 2 × 5 × 5
   Here 2 and 5 have not formed a group of three.
   Thus, 100 is not a perfect cube.
   
   (v) Prime factorisation of 46656 is:
   46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
   Here 2 and 3 have formed the groups of three.
   Thus, 46656 is a perfect cube.
   

2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Solution:
(i) Prime factorisation of 243, is:
243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3 × 3
Here, number 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3
Thus, the required smallest number to be multiplied is 3.

(ii) Prime factorisation of 256, is:
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2 × 2
Here, a number 2 is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2
Thus, the required smallest number to be multiplied is 2.

(iii) Prime factorisation of 72, is:
72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3 × 3
Here, a number 3 is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3
Thus, the required smallest number to be multiplied is 3.

(iv) Prime factorisation of 675, is:
675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5 × 5
Here, a number 5 is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5
Thus, the required smallest number is 5.

(v) Prime factorisation of 100, is:
100 = 2 × 2 × 5 × 5
Here, number 2 and 5 are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube, i.e., 2 × 2 × 2 × 5 × 5 × 5
Thus, the required smallest number to be multiplied is 2 × 5 = 10.

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 92
(v) 704
Solution:
(i) Prime factorisation of 81, is:
81 = 3 × 3 × 3 × 3 = 3³ × 3
Here, a number 3 is the number by which 81 is divided to make it a perfect cube,
i.e., 81 ÷ 3 = 27 which is a perfect cube.
Thus, the required smallest number to be divided is 3.

(ii) Prime factorisation of 128, is:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2
Here, a number 2 is the smallest number by which 128 is divided to make it a perfect cube,
i.e., 128 ÷ 2 = 64 which is a perfect cube.
Thus, 2 is the required smallest number.

(iii) Prime factorisation of 135 is:
135 = 3 × 3 × 3 × 5 = 3³ × 5
Here, 5 is the smallest number by which 135 is divided to make a perfect cube,
i.e., 135 ÷ 5 = 27 which is a perfect cube.
Thus, 5 is the required smallest number.

(iv) Prime factorisation of 192 is:
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2³ × 2³ × 3
Here, 3 is the smallest number by which 192 is divided to make it a perfect cube,
i.e., 192 ÷ 3 = 64 which is a perfect cube.
Thus, 3 is the required smallest number.

(v) Prime factorisation of 704 is:
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2³ × 2³ × 11
Here, 11 is the smallest number by which 704 is divided to make it a perfect cube,
i.e., 704 ÷ 11 = 64 which is a perfect cube.
Thus, 11 is the required smallest number.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?
Solution:
The sides of the cuboid are given as 5 cm, 2 cm and 5 cm.
Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm³
For the prime factorisation of 50, we have
50 = 2 × 5 × 5
To make it a perfect cube, we must have
2 × 2 × 2 × 5 × 5 × 5
= 20 × (2 × 5 × 5)
= 20 × volume of the given cuboid
Thus, the required number of cuboids = 20.

Exercise 7.2 | Class 8th Mathematics

1. Find the cube root of each of the following numbers by prime factorisation method.
   (i) 64
   (ii) 512
   (iii) 10648
   (iv) 27000
   (v) 15625
   (vi) 13824
   (vii) 110592
   (viii) 46656
   (ix) 175616
   (x) 91125
   Solution:
   

2. State True or False.
(i) Cube of an odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution:
(i) False – Cube of any odd number is always odd, e.g., (7)³ = 343
(ii) True – A perfect cube does not end with two zeros.
(iii) True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5)² = 25 and (5)³ = 625
(iv) False – (12)³ = 1728 (ends with 8)
(v) False – (10)³ = 1000 (4-digit number)
(vi) False – (99)³ = 970299 (6-digit number)
(vii) True – (2)³ = 8 (1-digit number)

3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
The given perfect cube = 1331
Forming groups of three from the rightmost digits of 1331

IInd group = 1
1st group = 331
One’s digit in first group = 1
One’s digit in the required cube root may be 1.
The second group has only 1.
Estimated cube root of 1331 = 11
Thus 1331−−−−√3 = 11
(i) Given perfect cube = 4913
Forming groups of three from the right most digit of 4913

IInd group = 4
1st group = 913
One’s place digit in 913 is 3.
One’s place digit in the cube root of the given number may be 7.
Now in IInd group digit is 4
1³ < 4 < 2³
Ten’s place must be the smallest number 1.
Thus, the estimated cube root of 4913 = 17.
(ii) Given perfect cube = 12167
Forming group of three from the rightmost digits of 12167

We have IInd group = 12
1st group = 167
The ones place digit in 167 is 7.
One’s place digit in the cube root of the given number may be 3.
Now in Ilnd group, we have 12
2³ < 12 < 3³
Ten’s place of the required cube root of the given number = 2.
Thus, the estimated cube root of 12167 = 23.
(iii) Given perfect cube = 32768
Forming groups of three from the rightmost digits of 32768, we have

IInd group = 32
1st group = 768
One’s place digit in 768 is 8.
One’s place digit in the cube root of the given number may be 2.
Now in IInd group, we have 32
3³ < 32 < 4³
Ten’s place of the cube root of the given number = 3.
Thus, the estimated cube root of 32768 = 32.

Extra Questions | Class 8th Mathematics

1. Find the cubes of the following:
   (a) 12
   (b) -6
   (c) 23
   (d) −56
   Solution:
   

2. Find the cubes of the following:
(a) 0.3
(b) 0.8
(c) .001
(d) 2 – 0.3
Sol.
(a) (0.3)³ = 0.3 × 0.3 × 0.3 = 0.027
(b) (0.8)³ = 0.8 × 0.8 × 0.8 = 0.512
(c) (0.001)³ = (0.001) × (0.001) × (0.001) = 0.000000001
(d) (2 – 0.3)³ = (1.7)³ = 1.7 × 1.7 × 1.7 = 4.913

3. Is 135 a perfect cube?
Solution:
Prime factorisation of 135, is:
135 = 3 × 3 × 3 × 5
We find that on making triplet, the number 5 does not make a group of the triplet.
Hence, 135 is not a perfect cube.

4. Find the cube roots of the following:
(a) 1728
(b) 3375
Solution:

5. Examine if (i) 200 (ii) 864 are perfect cubes.
Solution:
(i) 200 = 2 × 2 × 2 × 5 × 5
If we form triplet of equal factors, the number 2 forms a group of three whereas 5 does not do it.
Therefore, 200 is not a perfect cube.

(ii) We have 864 = 2 × 2 × 2 × 2 × 2
If we form triplet of equal factors, the number 2 and 3 form a group of three whereas another group of 2’s does not do so.
Therefore, 864 is not a perfect cube.

6. Find the smallest number by which 1323 may be multiplied so that the product is a perfect cube.
Solution:
1323 = 3 × 3 × 3 × 7 × 7
Since we required one more 7 to make a triplet of 7.
Therefore 7 is the smallest number by which 1323 may be multiplied to make it a perfect cube.

7. What is the smallest number by which 2916 should be divided so that the quotient is a perfect cube?
Solution:
Prime factorisation of
2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Since we required one more 2 to make a triplet
Therefore, the required smallest number by which 2916 should be divided to make it a perfect cube is 2 × 2 = 4, i.e., 2916 ÷ 4 = 729 which is a perfect cube.

8. Check whether 1728 is a perfect cube by using prime factorisation. (NCERT Exemplar)
Solution:
Prime factorisation of 1728 is
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
Since all prime factors can be grouped in triplets.
Therefore, 1728 is a perfect cube.

9. Using prime factorisation, find the cube root of 5832. (NCERT Exemplar)
Solution:

10.

Solution:

Cubes and Cube Roots Class 8 Extra Questions Short Answers Type

11.

Solution:

12. Find the cube roots of
(i) 412125
(ii) -0.729
Solution:

13. Express the following numbers as the sum of odd numbers using the given pattern

Solution:

14. Observe the following pattern and complete the blank spaces.
1³ = 1

Solution:

Exercise 6.1 | Class 8th Mathematics

1. What will be the unit digit of the squares of the following numbers?
   (i) 81
   (ii) 272
   (iii) 799
   (iv) 3853
   (v) 1234
   (vi) 20387
   (vii) 52698
   (viii) 99880
   (ix) 12796
   (x) 55555
   Solution:
   (i) Unit digit of 81² = 1
   (ii) Unit digit of 272² = 4
   (iii) Unit digit of 799² = 1
   (iv) Unit digit of 3853² = 9
   (v) Unit digit of 1234² = 6
   (vi) Unit digit of 26387² = 9
   (vii) Unit digit of 52698² = 4
   (viii) Unit digit of 99880² = 0
   (ix) Unit digit of 12796² = 6
   (x) Unit digit of 55555² = 5

2.The following numbers are not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
Solution:
(i) 1057 ends with 7 at unit place. So it is not a perfect square number.
(ii) 23453 ends with 3 at unit place. So it is not a perfect square number.
(iii) 7928 ends with 8 at unit place. So it is not a perfect square number.
(iv) 222222 ends with 2 at unit place. So it is not a perfect square number.
(v) 64000 ends with 3 zeros. So it cannot a perfect square number.
(vi) 89722 ends with 2 at unit place. So it is not a perfect square number.
(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.
(viii) 505050 ends with 1 zero. So it is not a perfect square number.

3. The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
(i) 431² is an odd number.
(ii) 2826² is an even number.
(iii) 7779² is an odd number.
(iv) 82004² is an even number.

4. Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1…2…1
10000001² = ………
Solution:
According to the above pattern, we have
100001² = 10000200001
10000001² = 100000020000001

5. Observe the following pattern and supply the missing numbers.
11² = 121
101² = 10201
10101² = 102030201
1010101² = ……….
……….² = 10203040504030201
Solution:
According to the above pattern, we have
1010101² = 1020304030201
101010101² = 10203040504030201

6. Using the given pattern, find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + ….² = 21²
5² + ….² + 30² = 31²
6² + 7² + …..² = ……²
Solution:
According to the given pattern, we have
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

7. Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
We know that the sum of n odd numbers = n²
(i) 1 + 3 + 5 + 7 + 9 = (5)² = 25 [∵ n = 5]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)² = 100 [∵ n = 10]
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)² = 144 [∵ n = 12]

8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
(i) 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)
(ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

9. How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution:
(i) We know that numbers between n² and (n + 1)² = 2n
Numbers between 12² and 13² = (2n) = 2 × 12 = 24
(ii) Numbers between 25² and 26² = 2 × 25 = 50 (∵ n = 25)
(iii) Numbers between 99² and 100² = 2 × 99 = 198 (∵ n = 99)

Exercise 6.2 | Class 8th Mathematics

1. Find the square of the following numbers.
   (i) 32
   (ii) 35
   (iii) 86
   (iv) 93
   (v) 71
   (vi) 46
   Solution:
   (i) 32 = 30 + 2
   (32)² = (30 + 2)²
   = 30(30 + 2) + 2(30 + 2)
   = 30² + 30 × 2 + 2 × 30 + 2²
   = 900 + 60 + 60 + 4
   = 1024
   Thus (32)² = 1024

(ii) 35 = (30 + 5)
(35)² = (30 + 5)²
= 30(30 + 5) + 5(30 + 5)
= (30)² + 30 × 5 + 5 × 30 + (5)²
= 900 + 150 + 150 + 25
= 1225
Thus (35)² = 1225

(iii) 86 = (80 + 6)
86² = (80 + 6)²
= 80(80 + 6) + 6(80 + 6)
= (80)² + 80 × 6 + 6 × 80 + (6)²
= 6400 + 480 + 480 + 36
= 7396
Thus (86)² = 7396

(iv) 93 = (90+ 3)
93² = (90 + 3)²
= 90 (90 + 3) + 3(90 + 3)
= (90)² + 90 × 3 + 3 × 90 + (3)²
= 8100 + 270 + 270 + 9
= 8649
Thus (93)² = 8649

(v) 71 = (70 + 1)
71² = (70 + 1)²
= 70 (70 + 1) + 1(70 + 1)
= (70)² + 70 × 1 + 1 × 70 + (1)²
= 4900 + 70 + 70 + 1
= 5041
Thus (71)² = 5041

(vi) 46 = (40+ 6)
46² = (40 + 6)²
= 40 (40 + 6) + 6(40 + 6)
= (40)² + 40 × 6 + 6 × 40 + (6)²
= 1600 + 240 + 240 + 36
= 2116
Thus (46)² = 2116

2. Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
(i) Let m² – 1 = 6
[Triplets are in the form 2m, m² – 1, m² + 1]
m² = 6 + 1 = 7
So, the value of m will not be an integer.
Now, let us try for m² + 1 = 6
⇒ m² = 6 – 1 = 5
Also, the value of m will not be an integer.
Now we let 2m = 6 ⇒ m = 3 which is an integer.
Other members are:
m² – 1 = 3² – 1 = 8 and m² + 1 = 3² + 1 = 10
Hence, the required triplets are 6, 8 and 10

(ii) Let m² – 1 = 14 ⇒ m² = 1 + 14 = 15
The value of m will not be an integer.
Now take 2m = 14 ⇒ m = 7 which is an integer.
The member of triplets are 2m = 2 × 7 = 14
m² – 1 = (7)² – 1 = 49 – 1 = 48
and m² + 1 = (7)² + 1 = 49 + 1 = 50
i.e., (14, 48, 50)

(iii) Let 2m = 16 m = 8
The required triplets are 2m = 2 × 8 = 16
m² – 1 = (8)² – 1 = 64 – 1 = 63
m² + 1 = (8)² + 1 = 64 + 1 = 65
i.e., (16, 63, 65)

(iv) Let 2m = 18 ⇒ m = 9
Required triplets are:
2m = 2 × 9 = 18
m² – 1 = (9)² – 1 = 81 – 1 = 80
and m² + 1 = (9)² + 1 = 81 + 1 = 82
i.e., (18, 80, 82)

Exercise 6.3 | Class 8th Mathematics

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
   (i) 9801
   (ii) 99856
   (iii) 998001
   (iv) 657666025
   Solution:
   (i) One’s digit in the square root of 9801 maybe 1 or 9.
   (ii) One’s digit in the square root of 99856 maybe 4 or 6.
   (iii) One’s digit in the square root of 998001 maybe 1 or 9.
   (iv) One’s digit in the square root of 657666025 can be 5.

2. Without doing any calculation, find the numbers which are surely not perfect squares.
(i) 153
(ii) 257
(iii) 408
(iv) 441
Solution:
We know that the numbers ending with 2, 3, 7 or 8 are not perfect squares.
(i) 153 is not a perfect square number. (ending with 3)
(ii) 257 is not a perfect square number. (ending with 7)
(iii) 408 is not a perfect square number. (ending with 8)
(iv) 441 is a perfect square number.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:
Using the method of repeated subtraction of consecutive odd numbers, we have
(i) 100 – 1 = 99, 99 – 3 = 96, 96 – 5 = 91, 91 – 7 = 84, 84 – 9 = 75, 75 – 11 = 64, 64 – 13 = 51, 51 – 15 = 36, 36 – 17 = 19, 19 – 19 = 0
(Ten times repetition)
Thus √100 = 10

(ii) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69, 69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0
(Thirteen times repetition)
Thus √169 = 13

4. Find the square roots of the following numbers by the prime factorisation Method.
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100
Solution:
(i) We have 729
Prime factors of 729
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3² × 3² × 3²
√729 = 3 × 3 × 3 = 27

(ii) We have 400
Prime factors of 400
400 = 2 × 2 × 2 × 2 × 5 × 5 = 2² × 2² × 5²
√400 = 2 × 2 × 5 = 20

(iii) 1764
1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²
√1764 = 2 × 3 × 7 = 42

(iv) 4096
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

(v) Prime factorisation of 7744 is
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
√7744 = 2 × 2 × 2 × 11 = 88

(vi) Prime factorisation of 9604 is
9604 = 2 × 2 × 7 × 7 × 7 × 7 = 2² × 7² × 7²
√9604 = 2 × 7 × 7 = 98

(vii) Prime factorisation of 5929 is
5929 = 7 × 7 × 11 × 11 = 7² × 11²
√5929 = 7 × 11 = 77

(viii) Prime factorisation of 9216 is
9216 = 2 × 2 × 2 × 2 × 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
√9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96

(ix) Prime factorisation of 529 is
529 = 23 × 23 = 23²
√529 = 23

(x) Prime factorisation of 8100 is
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 = 2² × 3² × 3² × 5²
√8100 = 2 × 3 × 3 × 5 = 90

5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained.
(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768
Solution:
(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factorisation is not in pair. 7 has no pair.
Thus, 7 is the smallest whole number by which the given number is multiplied to get a perfect square number.
The new square number is 252 × 7 = 1764
Square root of 1764 is
√1764 = 2 × 3 × 7 = 42

(ii) Primp factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
New square number = 180 × 5 = 900
The square root of 900 is
√900 = 2 × 3 × 5 = 30
Thus, 5 is the smallest whole number by which the given number is multiplied to get a square number.

(iii) Prime factorisation of 1008 is
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, 7 has no pair.
New square number = 1008 × 7 = 7056
Thus, 7 is the required number.
Square root of 7056 is
√7056 = 2 × 2 × 3 × 7 = 84

(iv) Prime factorisation of 2028 is
2028 = 2 × 2 × 3 × 13 × 13
Here, 3 is not in pair.
Thus, 3 is the required smallest whole number.
New square number = 2028 × 3 = 6084
Square root of 6084 is
√6084 = 2 × 13 × 3 = 78

(v) Prime factorisation of 1458 is
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, 2 is not in pair.
Thus, 2 is the required smallest whole number.
New square number = 1458 × 2 = 2916
Square root of 1458 is
√2916 = 3 × 3 × 3 × 2 = 54

(vi) Prime factorisation of 768 is
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
Thus, 3 is the required whole number.
New square number = 768 × 3 = 2304
Square root of 2304 is
√2304 = 2 × 2 × 2 × 2 × 3 = 48

6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.
(i) 252
(ii) 2925
(iii) 396
(iv) 2645
(v) 2800
(vi) 1620
Solution:
(i) Prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
Here 7 has no pair.
7 is the smallest whole number by which 252 is divided to get a square number.
New square number = 252 ÷ 7 = 36
Thus, √36 = 6

(ii) Prime factorisation of 2925 is
2925 = 3 × 3 × 5 × 5 × 13
Here, 13 has no pair.
13 is the smallest whole number by which 2925 is divided to get a square number.
New square number = 2925 ÷ 13 = 225
Thus √225 = 15

(iii) Prime factorisation of 396 is
396 = 2 × 2 × 3 × 3 × 11
Here 11 is not in pair.
11 is the required smallest whole number by which 396 is divided to get a square number.
New square number = 396 ÷ 11 = 36
Thus √36 = 6

(iv) Prime factorisation of 2645 is
2645 = 5 × 23 × 23
Here, 5 is not in pair.
5 is the required smallest whole number.
By which 2645 is multiplied to get a square number
New square number = 2645 ÷ 5 = 529
Thus, √529 = 23

(v) Prime factorisation of 2800 is
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, 7 is not in pair.
7 is the required smallest number.
By which 2800 is multiplied to get a square number.
New square number = 2800 ÷ 7 = 400
Thus √400 = 20

(vi) Prime factorisation of 1620 is
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, 5 is not in pair.
5 is the required smallest prime number.
By which 1620 is multiplied to get a square number = 1620 ÷ 5 = 324
Thus √324 = 18

7. The students of class VIII of a school donated ₹ 2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Total amount of money donated = ₹ 2401
Total number of students in the class = √2401
= 72×72−−−−−−√
= 7×7×7×7−−−−−−−−−−√
= 7 × 7
= 49

8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
Solution:
Total number of rows = Total number of plants in each row = √2025
= 3×3×3×3×5×5−−−−−−−−−−−−−−−−√
= 32×32×52−−−−−−−−−−√
= 3 × 3 × 5
= 45
Thus the number of rows and plants = 45

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution:
LCM of 4, 9, 10 = 180
The least number divisible by 4, 9 and 10 = 180
Now prime factorisation of 180 is
180 = 2 × 2 × 3 × 3 × 5
Here, 5 has no pair.
The required smallest square number = 180 × 5 = 900

10. Find the smallest number that is divisible by each of the numbers 8, 15 and 20.
Solution:
The smallest number divisible by 8, 15 and 20 is equal to their LCM.
LCM = 2 × 2 × 2 × 3 × 5 = 120
Here, 2, 3 and 5 have no pair.
The required smallest square number = 120 × 2 × 3 × 5 = 120 × 30 = 3600

Exercise 6.4 | Class 8th Mathematics

1. Find the square root of each of the following numbers by Long Division method.
   (i) 2304
   (ii) 4489
   (iii) 3481
   (iv) 529
   (v) 3249
   (vi) 1369
   (vii) 5776
   (viii) 7921
   (ix) 576
   (x) 1024
   (xi) 3136
   (xii) 900
   Solution:
   

2. Find the number of digits in the square root of each of the following numbers (without any calculation)
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution:
We know that if n is number of digits in a square number then
Number of digits in the square root = n2 if n is even and n+12 if n is odd.
(i) 64
Here n = 2 (even)
Number of digits in √64 = 22 = 1
(ii) 144
Here n = 3 (odd)
Number of digits in square root = 3+12 = 2
(iii) 4489
Here n = 4 (even)
Number of digits in square root = 42 = 2
(iv) 27225
Here n = 5 (odd)
Number of digits in square root = 5+12 = 3
(iv) 390625
Here n = 6 (even)
Number of digits in square root = 62 = 3

3. Find the square root of the following decimal numbers.
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:

4. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution:
(i)

Here remainder is 2
2 is the least required number to be subtracted from 402 to get a perfect square
New number = 402 – 2 = 400
Thus, √400 = 20

(ii)

Here remainder is 53
53 is the least required number to be subtracted from 1989.
New number = 1989 – 53 = 1936
Thus, √1936 = 44

(iii)

Here remainder is 1
1 is the least required number to be subtracted from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
Thus, √3249 = 57

(iv)

Here, the remainder is 41
41 is the least required number which can be subtracted from 825 to get a perfect square.
New number = 825 – 41 = 784
Thus, √784 = 28

(v)

Here, the remainder is 31
31 is the least required number which should be subtracted from 4000 to get a perfect square.
New number = 4000 – 31 = 3969
Thus, √3969 = 63

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution:
(i)

Here remainder is 41
It represents that square of 22 is less than 525.
Next number is 23 an 23² = 529
Hence, the number to be added = 529 – 525 = 4
New number = 529
Thus, √529 = 23

(ii)

Here the remainder is 69
It represents that square of 41 is less than in 1750.
The next number is 42 and 42² = 1764
Hence, number to be added to 1750 = 1764 – 1750 = 14
Require perfect square = 1764
√1764 = 42

(iii)

Here the remainder is 27.
It represents that a square of 15 is less than 252.
The next number is 16 and 16² = 256
Hence, number to be added to 252 = 256 – 252 = 4
New number = 252 + 4 = 256
Required perfect square = 256
and √256 = 16

(iv)

The remainder is 61.
It represents that square of 42 is less than in 1825.
Next number is 43 and 43² = 1849
Hence, number to be added to 1825 = 1849 – 1825 = 24
The required perfect square is 1848 and √1849 =43

(v)

Here, the remainder is 12.
It represents that a square of 80 is less than in 6412.
The next number is 81 and 81² = 6561
Hence the number to be added = 6561 – 6412 = 149
The require perfect square is 6561 and √6561 = 81

6. Find the length of the side of a square whose area = 441 m²
Solution:
Let the length of the side of the square be x m.
Area of the square = (side)² = x² m²
x² = 441 ⇒ x = √441 = 21

Thus, x = 21 m.
Hence the length of the side of square = 21 m.

7. In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Solution:
(a) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]
⇒ AC² = (6)² + (8)² = 36 + 64 = 100
⇒ AC = √100 = 10
Thus, AC = 10 cm.
(b) In right triangle ABC

AC² = AB² + BC² [By Pythagoras Theorem]
⇒ (13)² = AB² + (5)²
⇒ 169 = AB² + 25
⇒ 169 – 25 = AB²
⇒ 144 = AB²
AB = √144 = 12 cm
Thus, AB = 12 cm.

8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this.
Solution:
Let the number of rows be x.
And the number of columns also be x.
Total number of plants = x × x = x²
x² = 1000 ⇒ x = √1000

Here the remainder is 39
So the square of 31 is less than 1000.
Next number is 32 and 32² = 1024
Hence the number to be added = 1024 – 1000 = 24
Thus the minimum number of plants required by him = 24.
Alternative method:

The minimum number of plants required by him = 24.

9. There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
Let the number of children in a row be x. And also that of in a column be x.
Total number of students = x × x = x²
x² = 500 ⇒ x = √500

Here the remainder is 16
New Number 500 – 16 = 484
and, √484 = 22
Thus, 16 students will be left out in this arrangement.

Extra Questions | Class 8th Mathematics

Squares and Square Roots Class 8 Extra Questions Very Short Answer Type

1. Find the perfect square numbers between 40 and 50.
Solution:
Perfect square numbers between 40 and 50 = 49.

2. Which of the following 24², 49², 77², 131² or 189² end with digit 1?
Solution:
Only 49², 131² and 189² end with digit 1.

3. Find the value of each of the following without calculating squares.
(i) 27² – 26²
(ii) 118² – 117²
Solution:
(i) 27² – 26² = 27 + 26 = 53
(ii) 118² – 117² = 118 + 117 = 235

4. Write each of the following numbers as difference of the square of two consecutive natural numbers.
(i) 49
(ii) 75
(iii) 125
Solution:
(i) 49 = 2 × 24 + 1
49 = 25² – 24²
(ii) 75 = 2 × 37 + 1
75 = 38² – 37²
(iii) 125 = 2 × 62 + 1
125 = 63² – 62²

5. Write down the following as sum of odd numbers.
(i) 7²
(ii) 9²
Solution:
(i) 7² = Sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 9² = Sum of first 9 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

6. Express the following as the sum of two consecutive integers.
(i) 15²
(ii) 19²
Solution:

7. Find the product of the following:
(i) 23 × 25
(ii) 41 × 43
Solution:
(i) 23 × 25 = (24 – 1) (24 + 1) = 24² – 1 = 576 – 1 = 575
(ii) 41 × 43 = (42 – 1) (42 + 1) = 42² – 1 = 1764 – 1 = 1763

8. Find the squares of:
(i) −37
(ii) −917
Solution:

9. Check whether (6, 8, 10) is a Pythagorean triplet.
Solution:
2m, m² – 1 and m² + 1 represent the Pythagorean triplet.
Let 2m = 6 ⇒ m = 3
m² – 1 = (3)² – 1 = 9 – 1 = 8
and m² + 1 = (3)² + 1 = 9 + 1 = 10
Hence (6, 8, 10) is a Pythagorean triplet.
Alternative Method:
(6)² + (8)² = 36 + 64 = 100 = (10)²
⇒ (6, 8, 10) is a Pythagorean triplet.

10. Using property, find the value of the following:
(i) 19² – 18²
(ii) 23² – 22²
Solution:
(i) 19² – 18² = 19 + 18 = 37
(ii) 23² – 22² = 23 + 22 = 45

Squares and Square Roots Class 8 Extra Questions Short Answer Type

11. Using the prime factorisation method, find which of the following numbers are not perfect squares.
(i) 768
(ii) 1296
Solution:

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, 3 is not in pair.
768 is not a perfect square.

1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
Here, there is no number left to make a pair.
1296 is a perfect square.

12. Which of the following triplets are Pythagorean?
(i) (14, 48, 50)
(ii) (18, 79, 82)
Solution:
We know that 2m, m² – 1 and m² + 1 make Pythagorean triplets.
(i) For (14, 48, 50),
Put 2m =14 ⇒ m = 7
m² – 1 = (7)² – 1 = 49 – 1 = 48
m² + 1 = (7)² + 1 = 49 + 1 = 50
Hence (14, 48, 50) is a Pythagorean triplet.
(ii) For (18, 79, 82)
Put 2m = 18 ⇒ m = 9
m² – 1 = (9)² – 1 = 81 – 1 = 80
m² + 1 = (9)² + 1 = 81 + 1 = 82
Hence (18, 79, 82) is not a Pythagorean triplet.

13. Find the square root of the following using successive subtraction of odd numbers starting from 1.
(i) 169
(ii) 81
(iii) 225
Solution:
(i) 169 – 1 = 168, 168 – 3 = 165, 165 – 5 = 160, 160 – 7 = 153, 153 – 9 = 144, 144 – 11 = 133, 133 – 13 = 120, 120 – 15 = 105, 105 – 17 = 88, 88 – 19 = 69,
69 – 21 = 48, 48 – 23 = 25, 25 – 25 = 0
We have subtracted odd numbers 13 times to get 0.
√169 = 13
(ii) 81 – 1 = 80, 80 – 3 = 77, 77 – 5 = 72, 72 – 7 = 65, 65 – 9 = 56, 56 – 11 = 45, 45 – 13 = 32, 32 – 15 = 17, 17 – 17 = 0
We have subtracted 9 times to get 0.
√81 = 9
(iii) 225 – 1 = 224, 224 – 3 = 221, 221 – 5 = 216, 216 – 7 = 209, 209 – 9 = 200, 200 – 11 = 189, 189 – 13 = 176, 176 – 15 = 161, 161 – 17 = 144, 144 – 19 = 125,
125 – 21 = 104, 104 – 23 = 81, 81 – 25 = 56, 56 – 27 = 29, 29 – 29 = 0
We have subtracted 15 times to get 0.
√225 = 15

14. Find the square rootofthe following using prime factorisation
(i) 441
(ii) 2025
(iii) 7056
(iv) 4096
Solution:
(i) 441 = 3 × 3 × 7 × 7
√441 = 3 × 7 = 21

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5
√2025 = 3 × 3 × 5 = 45

(iii) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
√7056 = 2 × 2 × 3 × 7 = 84

(iv) 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64

15. Find the least square number which is divisible by each of the number 4, 8 and 12.
Solution:
LCM of 4, 8, 12 is the least number divisible by each of them.
LCM of 4, 8 and 12 = 24
24 = 2 × 2 × 2 × 3
To make it perfect square multiply 24 by the product of unpaired numbers, i.e., 2 × 3 = 6
Required number = 24 × 6 = 144

16. Find the square roots of the following decimal numbers
(i) 1056.25
(ii) 10020.01
Solution:

17. What is the least number that must be subtracted from 3793 so as to get a perfect square? Also, find the square root of the number so obtained.
Solution:
First, we find the square root of 3793 by division method.

Here, we get a remainder 72
61² < 3793
Required perfect square number = 3793 – 72 = 3721 and √3721 = 61

18. Fill in the blanks:
(а) The perfect square number between 60 and 70 is …………
(b) The square root of 361 ends with digit …………..
(c) The sum of first n odd numbers is …………
(d) The number of digits in the square root of 4096 is ………..
(e) If (-3)² = 9, then the square root of 9 is ……….
(f) Number of digits in the square root of 1002001 is …………
(g) Square root of 36625 is ………..
(h) The value of √(63 × 28) = …………
Solution:
(a) 64
(b) 9
(c) n²
(d) 2
(e) ±3
(f) 4
(g) 625
(h) 42

19. Simplify: √900 + √0.09 + √0.000009
Solution:
We know that √(ab) = √a × √b
√900 = √(9 × 100) = √9 × √100 = 3 × 10 = 30
√0.09 = √(0.3 × 0.3) = 0.3
√0.000009 = √(0.003 × 0.003) = 0.003
√900 + √0.09 + √0.000009 = 30 + 0.3 + 0.003 = 30.303

Squares and Square Roots Class 8 Extra Questions Higher Order Thinking Skills (HOTS)

20. Find the value of x if
1369−−−−√+0.0615+x−−−−−−−−−√=37.25
Solution:

21. Simplify:

Solution:

22. A ladder 10 m long rests against a vertical wall. If the foot of the ladder is 6 m away from the wall and the ladder just reaches the top of the wall, how high is the wall? (NCERT Exemplar)

Solution:
Let AC be the ladder.
Therefore, AC = 10 m
Let BC be the distance between the foot of the ladder and the wall.
Therefore, BC = 6 m
∆ABC forms a right-angled triangle, right angled at B.
By Pythagoras theorem,
AC² = AB² + BC²
10² = AB² + 6²
or AB² = 10² – 6² = 100 – 36 = 64
or AB = √64 = 8m
Hence, the wall is 8 m high.

23. Find the length of a diagonal of a rectangle with dimensions 20 m by 15 m. (NCERT Exemplar)
Solution:
Using Pythagoras theorem, we have Length of diagonal of the rectangle = l2+b2−−−−−√ units

Hence, the length of the diagonal is 25 m.

24. The area of a rectangular field whose length is twice its breadth is 2450 m2. Find the perimeter of the field.
Solution:
Let the breadth of the field be x metres. The length of the field 2x metres.
Therefore, area of the rectangular field = length × breadth = (2x)(x) = (2x²) m²
Given that area is 2450 m².
Therefore, 2x² = 2450
⇒ x² = 1225
⇒ x = √1225 or x = 35 m
Hence, breadth = 35 m
and length = 35 × 2 = 70 m
Perimeter of the field = 2 (l + b ) = 2(70 + 35) m = 2 × 105 m = 210 m.

Exercise 5.1 | Class 8th Mathematics

1. For which of these would you use a histogram to show the data?
   (i) The number of letters for different areas in a postman’s bag.
   (ii) The height of competitors in an athletics meet.
   (iii) The number of cassettes produced by 5 companies.
   (iv) The number of passengers boarding trains from 7 a.m to 7 p.m at a station.
   Give a reason for each.
   Solution:
   (i) The number of areas cannot be represented in class-intervals. So, we cannot use the histogram to show the data.
   (ii) Height of competitors can be divided into intervals. So, we can use histogram here.
   For example:
   
   (iii) Companies cannot be divided into intervals. So, we cannot use histogram here.
   (iv) Time for boarding the train can be divided into intervals. So, we can use histogram here.
   For example:
   

2. The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning.
W W W G B W W M G G M M W W W W
G B M W B G G M W W M M W W W
M W B W G M W W W W G W M M W
W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution:

3.The weekly wages (in ₹) of 30 workers in a factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890,
820, 860, 832, 833, 855, 845, 804, 808, 812, 840,
885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on.
Solution:

4. Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions:
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
Solution:
Refer to the frequency table of Question No. 3.

(i) Group 830-840 has the maximum number of workers, i.e., 9.
(ii) 10 workers earn equal and more than ₹ 850.
(iii) 20 workers earn less than ₹ 850.

5. The number of hours for which students of a particular class watched television during holidays is shown through the given graph.
Answer the following questions.
(i) For how many hours did the maximum number of students watch TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours watching TV?

Solution:
(i) 32 is the maximum number of students who watched TV for 4 to 5 hours.
(ii) 4 + 8 + 22 = 34 students watched TV for less than 4 hours.
(iii) 8 + 6 = 14 students watched TV for more than 5 hours.

Exercise 5.2 | Class 8th Mathematics

1. A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey.
   
   From this pie chart answer the following:
   (i) If 20 people liked classical music, how many young people were surveyed?
   (ii) Which type of music is liked by the maximum number of people?
   (iii) If a cassette company were to make 1000 CDs. How many of each type would they make?
   Solution:
   (i) Number of young people who were surveyed = 100×2010 = 200 people.
   (ii) Light music is liked by the maximum people, i.e., 40%
   (iii) Total number of CD = 1000
   Number of viewers who like classical music = 10×1000100 = 100
   Number of viewer who like semi-classical music = 20×1000100 = 200
   Number of viewers who like light music = 40×1000100 = 400
   Number of viewers who like folk music = 30×1000100 = 300

2. A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer.

(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
Solution:
(i) Winter season got the most votes, i.e. 150

(iii) Pie chart

3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Solution:
Table to find the central angle of each sector

4.The following pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

(i) In which subject did the student score 105 marks?
(Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
(Hint: Just study the central angles).
Solution:

(i) For 540 marks, the central angle = 360°
For 105 marks the central angle = 360540×105 = 70°
Corresponding subject = Hindi
(ii) Marks obtained in Mathematics = 90360×540 = 135
Marks obtained in Mathematics more than Hindi = 135 – 105 = 30
(iii) Central angle of Social Science + Mathematics = 65° + 90° = 155°
Central angle of Science + Hindi = 80° + 70° = 150°

5.Marks obtained in Social Science and Mathematics are more than that of the marks obtained in Science and Hindi.
The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Solution:

Exercise 5.3 | Class 8th Mathematics

1. List the outcomes you can see in these experiments.
   (i) Spinning a wheel
   
   (ii) Tossing two coins together
   Solution:
   (i) On spinning the wheel, we can get the following outcomes B, C, D, E and A.
   (ii) When two coins are tossed together, we get the following outcomes
   HH, HT, TH, TT (Where H denotes Head and T denotes Tail)

2.When a die is thrown, list the outcomes of an event of getting
(i) (a) a prime number
(b) not a prime number
(ii) (a) a number greater than 5
(b) a number not greater than 5
Solution:
(i) (a) The prime number are 2, 3 and 5
Required outcomes = 2, 3 and 5
(b) Outcomes for not a prime number are 1, 4 and 6
Required outcomes = 1, 4, 6.
(ii) (a) Outcomes for a number greater than 5 = 6
Required outcome = 6
(b) Outcomes for a number not greater than 5 are 1, 2, 3, 4, 5
Required outcomes = 1, 2, 3, 4, 5.

3.Find the
(i) Probability of the pointer stopping on D in (Question 1-(a))?
(ii) Probability of getting an ace from a well-shuffled deck of 52 playing cards?
(iii) Probability of getting a red apple, (see figure below)

Solution:
(i) Refer to fig. Question 1-(a)
Total number of sectors = 5
Number of sector where the pointer stops = 1, i.e. D
Probability of pointer stopping at D = 15
(ii) Number of aces = 4 (one from each suit i.e. heart, diamond, club and spade)
Total number of playing cards = 52
Probability of getting an ace

(iii) Total number of apples = 7
Number of red apples = 4
Probability of getting red apples

4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is choosen from the box without looking into it. What is the probability of:
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
Solution:
(i) Probability of getting a number 6 = 110
(ii) Probability of getting a number less than 6 = 510 = 12 [∵ Numbers less than 6 are 1, 2, 3, 4, 5]
(iii) Probability of getting a number greater than 6 = 410 = 25 [∵ Number greater than 6 are 7, 8, 9, 10]
(iv) Probability of getting a 1-digit number = 910
[∵ 1-digit numbers are 9, i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9]

5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:
Total number of sectors are = 3 green + 1 blue + 1 red = 5 sectors
Probability of getting a green sector

Number of non-blue sectors are = 3 green + 1 red = 4 sectors
Probability of getting non-blue sector

6. Find the probabilities of the events given in Question 2.
Solution:
Refer to Question 2, we have
(i) (a) Probability of getting a prime number

(b) Probability of getting a non-prime number

(ii) (a) Probability of getting a number greater than 5 = 16
(b) Probability of a number not greater than 5 = 56 or, 1 – 16 = 56

Extra Questions | Class 8th Mathematics

1.In the class interval 5-10, find the
(i) lower limit
(ii) upper limit
(iii) class mark
(iv) class size
Solution:
(i) lower limit = 5
(ii) upper limit = 10
(iii) Class mark = 5+102 = 152 = 7.5
(iv) Class size = 10 – 5 = 5

2.A group of 20 students recorded their heights (in cm). The data received were as given below. What is the range?
150, 120, 112, 160, 155, 151, 158, 142, 148, 149, 161, 165, 140, 157, 156, 146, 148, 153, 138, 135
Solution:
The minimum height =112 cm
Maximum height = 165 cm
Range = Maximum height – Minimum height = 165 cm – 112 cm = 47 cm

3. In the given pie chart, which colour is most popular? Which colour is the least popular?

Solution:
Red colour is the most popular and the blue colour is the least popular.

4. A die is thrown once. Find the probability of getting a number greater than 4.
Solution:
Number greater than 4 = 5, 6
n(E) = 2
Sample space n(S) = 6
Probability of getting a number greater than 4
= n(E)n(S = 26 = 13
Where re(E): Number of favourable outcomes
n(S): Total number of outcomes

5. A class consists of 21 boys and 9 girls. A student is to be selected for social work. Find the probability that
(i) a girl is selected
(ii) a boy is selected
Solution:
Sample space n(S) = 21 + 9 = 30
Number of girls n(E) = 9
(i) Probability of selecting a girl
= n(E)n(S = 930 = 310
(ii) Probability of selecting a boy
= n(E)n(S = 2130 = 710

6. The following pie chart depicts the percentage of students, nationwide. What is the percentage of
(i) Indian students
(ii) African students?

Solution:
(i) Percentage of Indian students = 180×100360 = 50%
(ii) Percentage of African students = 45×100360 = 1212%

Short Answer (SA) Questions

7. Fill in the blanks:

Solution:
Class-marks are
Class-mark

8. Construct a frequency table for the following marks obtained by 50 students using equal Intervals taking 16-24 (24 not included) as one of the class-intervals.
52, 16, 18, 20, 42, 48, 39, 38, 54, 58, 47, 37, 25, 16, 42, 49, 36, 35, 53, 21, 30, 43, 56, 34, 33, 17, 22, 24, 37, 41, 40, 50, 54, 56, 54, 36, 38, 42, 44, 56, 17, 18, 22, 24, 17, 48, 58, 23, 29, 58
Solution:

9. The double bar graph shows the average monthly temperatures of two cities over 4 months period. Read the graph carefully and answer the questions given below:
(i) What does each 1 cm block on the vertical axis represent?

(ii) What was the average monthly temperature in Dehradun in
(a) March
(b) April
(c) May
(d) June?
(iii) What was the average monthly temperature in Delhi for the whole 4 months?
(iv) In which month was the difference between the temperature of Delhi and Dehradun maximum and how much?
Solution:
(i) 1 cm block on vertical axis = 10°C
(ii) The average monthly temperature in Dehradun in the month of
(a) March was 25°C
(b) April was 34°C
(c) May was 40°C
(d) June was 36°C
(iii) The average monthly temperature in Delhi in the 4 months

(iv) Difference between the average monthly temperature of Delhi and Dehradun was maximum in the month of June, i.e. (50° – 36°) = 14°C.

10. The following table represents the number of students in a school playing six different games.

Present the above information on a bar graph.
Solution:

11. Prepare a grouped frequency table for the given histogram.

Solution:

12. A bag contains 144 coloured balls represented by the following table. Draw a pie chart to show this information.

Solution:

13. Mrs Verma spends her allowance in the following way.

Represent the above information by a pie chart.
Solution:

14. What is the probability of getting a marble which is not red from a bag containing 3 black, 8 yellow, 2 red and 5 white marbles?
Solution:
Total number of balls = 3 black + 8 yellow + 2 red + 5 white = 18
n( S) = 18
Number of the balls which are not red = 3 + 8 + 5 = 16
n(E) = 16
Probability = n(E)n(S) = 1618 = 89

15. From a well shuffled deck of 52 playing cards, a card is selected at random. Find the probability of getting
(i) a black card
(ii) a black king
(iii) an ace
(iv) a card of diamond
Solution:
Here, n(S) = 52
(i) Total number of black card = 26
n(E) = 26
Probability of getting a black card = n(E)n(S) = 2652 = 12
(ii) Number of black king = 2
n(E) = 2
Probability of getting a black king = n(E)n(S) = 252 = 126
(iii) Number of aces = 4
n(E) = 4
Probability of getting an ace = n(E)n(S) = 452 = 113
(iv) Number of diamond cards = 13
n(E) = 13
Probability of getting a card of diamond = n(E)n(S) = 1352 = 14

Exercise 4.1 | Class 8th Mathematics

1. Construct the following quadrilaterals.
   (i) Quadrilateral ABCD
   AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm, AC = 7 cm
   (ii) Quadrilateral JUMP
   JU = 3.5 cm, UM = 4 cm, MP = 5 cm, PJ = 4.5 cm, PU = 6.5 cm
   (iii) Parallelogram MORE
   OR = 6 cm, RE = 4.5 cm, EO = 7.5 cm
   (iv) Rhombus BEST
   BE = 4.5 cm, ET = 6 cm
   Solution:
   (i) We have to draw first rough sketch.
   
   Construction:
   Step I: Draw AB = 4.5 cm
   Step II: Draw an arc with centre B and radius 5.5 cm.
   Step III: Draw another arc with centre A and radius 7 cm to meet the previous arc at C.
   Step IV: Draw an arc with centre C and radius 4 cm.
   Step V: Draw another arc with centre A and radius 6 cm to cut the former arc at D.
   Step VI: Join BC, AC, CD and AD.
   (ii) We have to draw the first rough sketch.
   
   Thus ABCD is the required quadrilateral.
   Construction:
   Step I: Draw JU = 3.5 cm.
   Step II: Draw an arc with centre J and radius 4.5 cm.
   Step III: Draw another arc with centre U and radius 6.5 cm to meet the previous arc at P.
   Step IV: Join JP and UP.
   Step V: Draw an arc with centre U and radius 4 cm.
   Step VI: Draw another arc with centre P and radius 5 cm to meet the previous arc at M.
   Step VII: Join UM and PM.
   Thus, JUMP is the required quadrilateral.
   (iii) We have to draw the first rough sketch.
   
   Construction: (Opposite sides of a parallelogram are equal)
   Step I: Draw OR = 6 cm.
   Step II: Draw an arc with centre R and radius 4.5 cm.
   Step III: Draw another arc with centre O and radius 7.5 cm to meet the previous arc at E.
   Step IV: Join RE and OE.
   Step V: Draw an arc with centre E and radius 6 cm.
   Step VI: Draw another arc with centre O and radius 4.5 cm to meet the former arc at M.
   Step VII: Join EM and OM.
   Thus, MORE is the required parallelogram.
   (iv) We have to draw first rough sketch.
   
   Construction: (All sides of a rhombus are equal)
   Step I: Draw BE = 4.5 cm
   Step II: Draw an arc with centre B and radius 4.5 cm.
   Step III: Draw another arc with centre E and radius 6 cm to meet the previous arc at T.
   Step IV: Join BT and ET.
   Step V: Draw two arcs with centres E and T with equal radii 4-5 cm to meet each other at S. .
   Step VI: Join ES and TS.
   Thus, BEST is the required rhombus.

Exercise 4.2 | Class 8th Mathematics

1. Construct the following quadrilaterals.
   (i) Quadrilateral LIFT
   LI = 4 cm
   IF = 3 cm
   TL = 2.5 cm
   LF = 4.5 cm
   IT = 4 cm
   (ii) Quadrilateral GOLD
   OL = 7.5 cm
   GL = 6 cm
   GD = 6 cm
   LD = 5 cm
   OD = 10 cm
   (iii) Rhombus BEND
   BN = 5.6 cm
   DE = 6.5 cm
   Solution:
   (i) Construction:
   
   Step I: Draw LI = 4 cm.
   Step II: Draw an arc with centre I and radius 3 cm.
   Step III: Draw another arc with centre L and radius 4.5 cm to meet the former arc at F.
   Step IV: Join LF and IF.
   Step V: Draw an arc with centre L and radius 2.5 cm.
   Step VI: Draw another arc with centre I and radius 4 cm to meet the previous arc at T.
   Step VII: Join LT and IT.
   Thus LIFT is the required quadrilateral.
   (ii) Construction:
   
   Step I: Draw OL = 7.5 cm
   Step II: Draw an arc with centre O and radius 10 cm.
   Step III: Draw another arc with centre L and radius 5 cm to meet the previous arc at D.
   Step IV: Join OD and LD.
   Step V: Draw an arc with centre L and D with equal radii of 6 cm to meet each other at G.
   Step VI: Join LG and DG.
   Thus GOLD is the required quadrilateral.
   (iii) Construction: (The diagonals of a rhombus bisect each other at the right angle)
   
   Step I: Draw BN = 5.6 cm.
   Step II: Draw the right bisector of BN at O.
   Step III: Draw two arcs with centre O and radius 12 × DE, i.e., 12 × 6.5 = 3.25 cm to meet the right bisector at D and E.
   Step IV: Join BE, EN, ND and BD.
   Thus, BEND is the required rhombus.

Exercise 4.3 | Class 8th Mathematics

1. Construct the following quadrilaterals:
   (i) Quadrilateral MORE
   MO = 6 cm, ∠R = 105°, OR = 4.5 cm, ∠M = 60°, ∠O = 105°
   (ii) Quadrilateral PLAN
   PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, ∠N = 85°
   (iii) Parallelogram HEAR
   HE = 5 cm, EA = 6 cm, ∠R = 85°
   (iv) Rectangle OKAY
   OK = 7 cm, KA = 5 cm
   Solution:
   (i) Construction:
   Step I: Draw OR = 4.5 cm
   Step II: Draw two angles of 105° each at O and R with the help of protactor.
   Step III: Cut OM = 6 cm.
   
   Step IV: Draw an angle of 60° at M to meet the angle line through R at E.
   Thus, MORE is the required quadrilateral.

(ii) Construction:

Step I: Draw LA = 6.5 cm
Step II: Draw an angle of 75° at L and 110° at A with the help of a protractor.
[∵ 360° – (110° + 90° + 85°) = 75°]
Step III: Cut LP = 4 cm.
Step IV: Draw an angle of 90° at P which meets the angle line through A at N.
Thus PLAN is the required quadrilateral.

(iii) Construction: (Opposite sides of a parallelogram are equal)

Step I: Draw HE = 5 cm.
Step II: Draw an angle of 85° at E and cut EA = 6 cm.
Step III: Draw an arc with centre A and radius 5 cm.
Step IV: Draw another arc with centre H and radius 6 cm to meet the previous arc at R.
Step V: Join HR and AR
Thus, HEAR is the required parallelogram.

(iv) Construction:
(Each angle of a rectangle is 90° and opposite sides are equal.)

Step I: Draw OK = 7 cm.
Step II: Draw the angle of 90° at K and cut KA = 5 cm.
Step III: Draw an arc with centre O and radius 5 cm.
Step IV: Draw another arc with centre A and radius 7 cm to meet the previous arc at Y.
Step V: Join OY and AY.
Thus OKAY is the required rectangle.

Exercise 4.3 | Class 8th Mathematics

1. Construct the following quadrilaterals:
   (i) Quadrilateral DEAR
   DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E = 60°, ∠A = 90°
   (ii) Quadrilateral TRUE
   TR = 3.5 cm, RU = 3 cm, UE = 4.5 cm, ∠R = 75°, ∠U = 120°
   Solution:
   (i) Construction:
   
   Step I: Draw DE = 4 cm.
   Step II: Draw an angle of 60° at E.
   Step III: Draw an arc with centre E and radius 5 cm to meet the angle line at A.
   Step IV: Draw an angle of 90° at A and cut AR = 4.5 cm.
   Step V: Join DR.
   Thus, DEAR is the required quadrilateral.

(ii) Construction:

Step I: Draw TR = 3.5 cm
Step II: Draw an angle of 75° at R and cut RU = 3 cm.
Step III: Draw an angle of 120° at U and cut UE = 4.5 cm.
Step IV: Join TE.
Thus, TRUE is the required quadrilateral.

Exercise 4.5 | Class 8th Mathematics

1. The square READ with RE = 5.1 cm.
   Solution:
   Construction:
   
   Step I: Draw RE = 5.1 cm.
   Step II: Draw an angle of 90° at E and cut EA = 5.1 cm.
   Step III: Draw two arcs from A and R with radius 5.1 cm to cut each other at D.
   Step IV: Join RD and AD.
   Thus, READ is the required square.

2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Solution:
Construction:

Step I: Draw AC = 6.4 cm.
Step II: Draw the right bisector of AC at E.
Step III: Draw two arcs with centre E and radius = 5.22 = 2.6 cm to cut the previous diagonal at B and D.
Step IV: Join AD, AB, BC and DC.
Thus ABCD is the required rhombus.

3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Solution:
Construction: Let the two adjacent sides of a rectangle PQRS be PQ = 5 cm and QR = 4 cm.

Step I: Draw PQ = 5 cm.
Step II: Draw an angle of 90° at Q and cut QR = 4 cm.
Step III: Draw an arc with centre R and radius 5 cm.
Step IV: Draw another arc with centre P and radius 4 cm to meet the previous arc at S.
Step V: Join RS and PS.
Thus, PQRS is the required rectangle.

4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Solution:
Construction:
Step I: Draw OK = 5.5 cm.
Step II: Draw an angle of any measure (say 60°) at K and cut KA = 4.2 cm.

Step III: Draw an arc with centre A and radius of 5.5 cm.
Step IV: Draw another arc with centre O and radius 4.2 cm to cut the previous arc at Y.
Step V: Join AY and OY.
Thus, OKAY is the required parallelogram.
No, it is not a unique parallelogram. The angle at K can be of measure other than 60°.

Extra Questions | Class 8th Mathematics

Question 1.
Construct a quadrilateral PQRS, given that QR = 4.5 cm, PS = 5.5 cm, RS = 5 cm and the diagonal PR = 5.5 cm and diagonal SQ = 7 cm.

Solution:
Construction:
Step I: Draw QR = 4.5 cm.
Step II: Draw an arc with centre R and radius 5 cm.
Step III: Draw another arc with centre Q and radius 7 cm to meet the previous arc at S.
Step IV: Join RS and QS.
Step V: Draw two arcs with centre S and R and radius 5.5 cm each to meet each other at P.
Step VI: Join RP, SP and PQ.
Thus PQRS is the required quadrilateral.

2. Construct a quadrilateral ABCD in which AB = 4 cm, BC = 3.5 cm, CD = 5 cm, AD = 5.5 cm and ∠B = 75°.

Solution:
Construction:
Step I: Draw AB = 4 cm.
Step II: Draw an angle of 75° at B and cut BC = 3.5 cm.
Step III: Draw an arc with centre C and radius 5 cm.
Step IV: Draw another arc with centre A and radius 5.5 cm to meet the previous arc at D.
Step V: Join CD and AD.
Thus ABCD is the required quadrilateral.

3. Construct a square whose side is 5 cm.

Solution:
Construction:
Step I: Draw AB = 5 cm.
Step II: Draw an angle of 90° at B and cut BC = 5 cm.
Step III: Draw two arcs with centre A and C and same radii of 5 cm which meet each other at D.
Step IV: Join AD and CD.
Thus, ABCD is the required square.

4. Construct a rhombus ABCD in which AB = 5.8 cm and AC = 7.5 cm.

Solution:
Construction:
Step I: Draw AB = 5.8 cm.
Step II: Draw an arc with centre B and radius 5.8 cm.
Step III: Draw another arc with centre A and radius 7.5 cm to meet the previous arc at C.
Step IV: Draw two arcs with centres A and C and of the same radius 5.8 cm to meet each other at D.
Step V: Join BC, AC, CD and AD.
Thus ABCD is the required rhombus.

5. Construct a rhombus whose diagonals are 6 cm and 8 cm.

Solution:
Construction:
Step I: Draw SQ = 8 cm.
Step II: Draw a right bisector of SQ at O.
Step III: Draw two arcs with centre O and radius 3 cm each to cut the right bisector at P and R.
Step TV: Join PQ, QR, RS and SP.
Thus PQRS is the required rhombus.

6. Construct a rectangle whose diagonal is 5 cm and the angle between the diagonal is 50°.


Solution:
Construction:
Step I: Draw AC = 5 cm.
Step II: Draw the right bisector of AC at O.
Step III: Draw an angle of 50° at O and product both sides.
Step IV: Draw two arcs with centre O and of the same radius 2.5 cm to cut at B and D.
Step V: Join AB, BC, CD and DA.
Thus, ABCD is the required rectangle.

7. Construct a quadrilateral ABCD in which BC = 4 cm, ∠B = 60°, ∠C = 135°, AB = 5 cm and ∠A = 90°.

Solution:
Construction:
Step I: Draw AB = 5 cm.
Step II: Draw the angle of 60° at B and cut BC = 4 cm.
Step III: Draw an angle of 135° at C and angle of 90° at A which meet each other at D.
Thus, ABCD is the required quadrilateral.

8. Construct a parallelogram ABCD in which AB = 5.5 cm, AC = 7 cm and BD = 8 cm.

Solution:
Construction:
Step I: Draw AB = 5.5 cm.
Step II: Draw an arc with centre B and radius 82 cm = 4 cm.
Step III: Draw another arc with centre A and radius 72 cm = 3.5 cm which cuts the previous arc at O.
Step IV: Join AO and produce to C such that AO = OC.
Step V: Join BO and produce to D such that BO = OD.
Step VI: Join BC, CD and AD.
Thus ABCD is the required parallelogram.

9. Construct a rhombus PAIR, given that PA = 6 cm and angle ∠A = 110°.
Solution:
Since in a rhombus, all sides are equal, so PA = AI = IR = RP = 6 cm
Also, rhombus is a parallelogram
so, adjacent angle, ∠I = 180° – 110° = 70°

Steps of construction
Step I. Draw AI = 6 cm
Step II. Draw ray AX¯ such that ∠IAX = 110° and draw IY¯ such that ∠AIY = 70°.
Step III. With A and I as centres and radius 6 cm draw arcs intersecting AX and IY at P and R respectively.
Step IV. Join PR.
Thus, PAIR is the required rhombus.

Exercise 3.2 | Class 8th Mathematics

Exercise 3.2 | Class 8th Mathematics

Q1. Find x in the following figures.

Solution:
(a) We know that the sum of all the exterior angles of a polygon = 360°
125° + 125° + x = 360°
⇒ 250° + x = 360°

x = 360° – 250° = 110°
Hence x = 110°
(b) Here ∠y = 180° – 90° = 90°

and ∠z = 90° (given)
x + y + 60° + z + 70° = 360° [∵ Sum of all the exterior angles of a polygon = 360°]
⇒ x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310° = 50°
Hence x = 50°

Q2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) We know the sum of all the exterior angles of polygon = 360°
Measure of each angle of 9 sided regular polygon = 3609 = 40°
(ii) Sum of all the exterior angles of a polygon = 360°
Measure of each angle of 15 sided regular polygon = 36015 = 24°

Q3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
Sum of all exterior angles of a regular polygon = 360°
Number of sides

Hence, the number of sides = 15

Q4. How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Let re be the number of sides of a regular polygon.
Sum of all interior angles = (n – 2) × 180°
and, measure of its each angle

Hence, the number of sides = 24

Q5.  What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
The measure of each interior angle

The minimum measure the angle of an equilateral triangle (n = 3) = 60°.
(b) From part (a) we can conclude that the maximum exterior angle of a regular polygon = 180° – 60° = 120°.

Exercise 3.3 |Class 8th Mathematics

Q1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …………
(ii) ∠DCB = ………
(iii) OC = ………
(iv) m∠DAB + m∠CDA = ……..

Solution:
(i) AD = BC [Opposite sides of a parallelogram are equal]
(ii) ∠DCB = ∠DAB [Opposite angles of a parallelogram are equal]
(iii) OC = OA [Diagonals of a parallelogram bisect each other]
(iv) m∠DAB + m∠CDA = 180° [Adjacent angles of a parallelogram are supplementary]

Q2. Consider the following parallelograms. Find the values of the unknowns x, y, z.

Solution:
(i) ABCD is a parallelogram.

∠B = ∠D [Opposite angles of a parallelogram are equal]
∠D = 100°
⇒ y = 100°
∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
⇒ z + 100° = 180°
⇒ z = 180° – 100° = 80°
∠A = ∠C [Opposite angles of a ||gm]
x = 80°
Hence x = 80°, y = 100° and z = 80°
(ii) PQRS is a parallelogram.

∠P + ∠S = 180° [Adjacent angles of parallelogram]
⇒ x + 50° = 180°
x = 180° – 50° = 130°
Now, ∠P = ∠R [Opposite angles are equal]
⇒ x = y
⇒ y = 130°
Also, y = z [Alternate angles]
z = 130°
Hence, x = 130°, y = 130° and z = 130°
(iii) ABCD is a rhombus.
[∵ Diagonals intersect at 90°]

x = 90°
Now in ∆OCB,
x + y + 30° = 180° (Angle sum property)
⇒ 90° + y + 30° = 180°
⇒ y + 120° = 180°
⇒ y = 180° – 120° = 60°
y = z (Alternate angles)
⇒ z = 60°
Hence, x = 90°, y = 60° and z = 60°.
(iv) ABCD is a parallelogram

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
Now, ∠D = ∠B [Opposite angles of a |jgm]
⇒ y = 80°
Also, z = ∠B = 80° (Alternate angles)
Hence x = 100°, y = 80° and z = 80°
(v) ABCD is a parallelogram.

∠D = ∠B [Opposite angles of a ||gm]
y = 112°
x + y + 40° = 180° [Angle sum property]
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180° – 152 = 28°
z = x = 28° (Alternate angles)
Hence x = 28°, y = 112°, z = 28°.

Q3. Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) For ∠D + ∠B = 180, quadrilateral ABCD may be a parallelogram if following conditions are also fulfilled.
(a) The sum of measures of adjacent angles should be 180°.

Q4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
ABCD is a rough figure of a quadrilateral in which m∠A = m∠C but it is not a parallelogram. It is a kite.

Q5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD is parallelogram such that
m∠B : m∠C = 3 : 2

Let m∠B = 3x° and m∠C = 2x°
m∠B + m∠C = 180° (Sum of adjacent angles = 180°)
3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Thus, ∠B = 3 × 36 = 108°
∠C = 2 × 36° = 72°
∠B = ∠D = 108°
and ∠A = ∠C = 72°
Hence, the measures of the angles of the parallelogram are 108°, 72°, 108° and 72°.

Q6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which
∠A = ∠B

We know ∠A + ∠B = 180° [Sum of adjacent angles = 180°]
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Thus, ∠A = ∠C = 90° and ∠B = ∠D = 90°
[Opposite angles of a parallelogram are equal]

Q7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:
∠y = 40° (Alternate angles)
∠z + 40° = 70° (Exterior angle property)
⇒ ∠z = 70° – 40° = 30°
z = ∠EPH (Alternate angle)
In ∆EPH
∠x + 40° + ∠z = 180° (Adjacent angles)
⇒ ∠x + 40° + 30° = 180°
⇒ ∠x + 70° = 180°
⇒ ∠x = 180° – 70° = 110°
Hence x = 110°, y = 40° and z = 30°.

Q8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:
(i) GU = SN (Opposite sides of a parallelogram)

⇒ y + 7 = 20
⇒ y = 20 – 7 = 13
Also, ON = OR
⇒ x + y = 16
⇒ x + 13 = 16
x = 16 – 13 = 3
Hence, x = 3 cm and y = 13 cm.

Q9.

In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
Here RISK and CLUE are two parallelograms.

∠1 = ∠L = 70° (Opposite angles of a parallelogram)
∠K + ∠2 = 180°
Sum of adjacent angles is 180°
120° + ∠2 = 180°
∠2 = 180° – 120° = 60°
In ∆OES,
∠x + ∠1 + ∠2 = 180° (Angle sum property)
⇒ ∠x + 70° + 60° = 180°
⇒ ∠x + 130° = 180°
⇒ ∠x = 180° – 130° = 50°
Hence x = 50°

Q10. Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:
∠M + ∠L = 100° + 80° = 180°
∠M and ∠L are the adjacent angles, and sum of adjacent interior angles is 180°
KL is parallel to NM
Hence KLMN is a trapezium.

Ex 3.3 Class 8 Maths Question 11.
Find m∠C in below figure if AB¯ || DC¯
Solution:
Given that AB¯ || DC¯
m∠B + m∠C = 180° (Sum of adjacent angles of a parallelogram is 180°)

120° + m∠C = 180°
m∠C = 180° – 120° = 60°
Hence m∠C = 60°

Exercise 3.4 | Class 8th Mathematics

Q1. State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True

Q2. Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares and rhombuses.
(b) Rectangles and squares.

Q3. Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) Square is a quadrilateral because it is closed with four line segments.
(ii) Square is a parallelogram due to the following properties:
(a) Opposite sides are equal and parallel.
(b) Opposite angles are equal.
(iii) Square is a rhombus because its all sides are equal and opposite sides are parallel.
(iv) Square is a rectangle because its opposite sides are equal and has equal diagonal.

Q4. Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) Parallelogram, rectangle, square and rhombus
(ii) Square and rhombus
(iii) Rectangle and square

Q5. Explain why a rectangle is a convex quadrilateral.
Solution:
In a rectangle, both of its diagonal lie in its interior. Hence, it is a convex quadrilateral.

Q6. ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Solution:
Since the right-angled triangle ABC makes a rectangle ABCD by the dotted lines.
Therefore OA = OB = OC = OD [Diagonals of a rectangle are equal and bisect each other]
Hence, O is equidistant from A, B and C.

Extra Question | Class 8th Mathematics

Q1. In the given figure, ABCD is a parallelogram. Find x.

Solution:
AB = DC [Opposite sides of a parallelogram]
3x + 5 = 5x – 1
⇒ 3x – 5x = -1 – 5
⇒ -2x = -6
⇒ x = 3

Q2. In the given figure find x + y + z.

Solution:
We know that the sum of all the exterior angles of a polygon = 360°
x + y + z = 360°

Q3. In the given figure, find x.

Solution:
∠A + ∠B + ∠C = 180° [Angle sum property]
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
⇒ 6x = 150
⇒ x = 25

Q4. The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution:
Sum of all interior angles of a quadrilateral = 360°
Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
⇒ x = 20°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.

Q5. Find the measure of an interior angle of a regular polygon of 9 sides.
Solution:
Measure of an interior angle of a regular polygon

Q6. Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
= 2[9 + 7] = 2 × 16 = 32 cm.

Now perimeter of the square = Perimeter of rectangle = 32 cm.
Side of the square = 324 = 8 cm.
Hence, the length of the side of square = 8 cm.

Q7. In the given figure ABCD, find the value of x.

Solution:
Sum of all the exterior angles of a polygon = 360°
x + 70° + 80° + 70° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220° = 140°

Q8. In the parallelogram given alongside if m∠Q = 110°, find all the other angles.

Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°

Q9. In the given figure, ABCD is a rhombus. Find the values of x, y and z.

Solution:
AB = BC (Sides of a rhombus)
x = 13 cm.
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.

Q10In the given figure, ABCD is a parallelogram. Find x, y and z.

Solution:
∠A + ∠D = 180° (Adjacent angles)
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
x = 55°
∠A = ∠C [Opposite angles of a parallelogram]
⇒ 125° = y + 56°
⇒ y = 125° – 56°
⇒ y = 69°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°