Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 Cubes and Cube Roots Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2

1. Find the cube root of each of the following numbers by prime factorisation method.

Question (i).
64
Solution:
\(\begin{array}{l|l}
2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
By prime factorisation,
64 = 2 × 2 × 2 × 2 × 2 × 2
∴ \(\sqrt[3]{64}\) = 2 × 2
= 4
Thus, cube root of 64 is 4.

Question (ii).
512
Solution:
\(\begin{array}{l|l}
2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
By prime factorisation,
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ \(\sqrt[3]{512}\) = 2 × 2 × 2
= 8
Thus, cube root of 512 is 8.

Question (iii).
10648
Solution:
\(\begin{array}{r|l}
2 & 10648 \\
\hline 2 & 5324 \\
\hline 2 & 2662 \\
\hline 11 & 1331 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
By prime factorisation,
10648 = 2 × 2 × 2 × 11 × 11 × 11
∴ \(\sqrt[3]{10648}\) = 2 × 11
= 22
Thus, cube root of 10648 is 22.

Question (iv).
27000
Solution:
\(\begin{array}{l|l}
2 & 27000 \\
\hline 2 & 13500 \\
\hline 2 & 6750 \\
\hline 3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5
= 30
Thus, cube root of 27000 is 30.

Question (v).
15625
Solution:
\(\begin{array}{l|l}
5 & 15625 \\
\hline 5 & 3125 \\
\hline 5 & 625 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation,
15625 = 5 × 5 × 5 × 5 × 5 × 5
∴ \(\sqrt[3]{15625}\) = 5 × 5
= 25
Thus, cube root of 15625 is 25.

Question (vi).
13824
Solution:
\(\begin{array}{l|l}
2 & 13824 \\
\hline 2 & 6912 \\
\hline 2 & 3456 \\
\hline 2 & 1728 \\
\hline 2 & 864 \\
\hline 2 & 432 \\
\hline 2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ \(\sqrt[3]{13824}\) = 2 × 2 × 2 × 3
= 24

Question (vii).
110592
Solution:
\(\begin{array}{l|l}
2 & 110592 \\
\hline 2 & 55296 \\
\hline 2 & 27648 \\
\hline 2 & 13824 \\
\hline 2 & 6912 \\
\hline 2 & 3456 \\
\hline 2 & 1728 \\
\hline 2 & 864 \\
\hline 2 & 432 \\
\hline 2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ \(\sqrt[3]{110592}\) = 2 × 2 × 2 × 2 × 3
= 48
Thus, cube root of 110592 is 48.

Question (viii).
46656
Solution:
\(\begin{array}{l|l}
2 & 46656 \\
\hline 2 & 23328 \\
\hline 2 & 11664 \\
\hline 2 & 5832 \\
\hline 2 & 2916 \\
\hline 2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
∴ \(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3
= 36
Thus, cube root of 46656 is 36.

Question (ix).
175616
Solution:
\(\begin{array}{l|l}
2 & 175616 \\
\hline 2 & 87808 \\
\hline 2 & 43904 \\
\hline 2 & 21952 \\
\hline 2 & 10976 \\
\hline 2 & 5488 \\
\hline 2 & 2744 \\
\hline 2 & 1372 \\
\hline 2 & 686 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
By prime factorisation,
175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
∴ \(\sqrt[3]{175616}\) = 2 × 2 × 2 × 7
= 56
Thus, cube root of 175616 is 56.

Question (x).
91125
Solution:
\(\begin{array}{l|l}
3 & 91125 \\
\hline 3 & 30375 \\
\hline 3 & 10125 \\
\hline 3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation,
91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
∴ \(\sqrt[3]{91125}\) =3 × 3 × 5
= 45
Thus, cube root of 91125 is 45.

2. State true or false.

Question (i).
Cube of any odd number is even.
Solution:
False, cube of any odd number is odd.

Question (ii).
A perfect cube does not end with two zeros.
Solution:
True, a perfect cube ending with zero will always have a triplet of zeros.

Question (iii).
If square of a number ends with 5, then its cube ends with 25.
Solution:
False, let us understand with an example.
15² = 15 × 15 = 225
15³ = 15 × 15 × 15 = 3375

Question (iv).
There is no perfect cube which ends with 8.
Solution:
False, let us understand with an example.
8 = 2³, 1728 = 12³

Question (v).
The cube of a two-digit number may be a three-digit number.
Solution:
False, let us take an example.
The smallest two-digit number is 10.
10³ = 1000
which is a four-digit number and not a three-digit number.

Question (vi).
The cube of a two-digit number may have seven or more digits.
Solution:
False, let us take an example.
The greatest two-digit number is 99.
99³ = 970299
which is a six-digit number.

Question (vii).
The cube of a single digit number may be a single digit number.
Solution:
True, let us take examples.
1³ = 1 and 2³ = 8

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:
Yes, we can guess without prime factorisation.
1331 : Separate given number into two groups.
1331 → 1 and 331
331 → Units place digit of 331 is 1.
∴ Unit place digit of cube root of 1331 = 1 (∵ 1³ = 1)
1 → 1³ = 1 and 2³ = 8
∴ Tens digit of cube root of 1331 = 1
Thus, the cube root of 1331 is 11.

(i) 4913 : Separate given number into two groups.
4913 → 4 and 913
913 → Units place digit of 913 is 3.
∴ Unit digit of cube root of 4913 = 7 (∵ 7³ = 343)
4 → 1³ = 1 and 2³ = 8
1 < 4 < 8 (∴ 1³ < 4 < 2³)
∴ The tens digit of cube root of 4913 = 1
Thus, the cube root of 4913 is 17.

(ii) 12167 : Separate given number into two groups.
12167 → 12 and 167
167 → Units place digit of 167 is 7.
∴ Unit digit of cube root of 12167 = 3
(∵ 3³ = 27)
12 → 2³ = 8 and 3³ = 27
8 < 12 < 27 (∵ 2³ < 12 < 3³)
∴ Tens place digit of cube root of 12167 = 2
Thus, the cube root of 12167 is 23.

(iii) 32768 : Separate given number into two groups.
32768 → 32 and 768
768 → Units place digit of 768 is 8.
∴ Unit digit of cube root of 32768 = 2 (∵ 2³ = 8)
32 → 3³ = 27 and 4³ = 64
27 < 32 < 64 (∵ 3³ < 32 < 4³)
∴ The tens digit of cube root of 32768 = 3
Thus, the cube root of 32768 is 32.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 10 Visualising Solid Shapes InText Questions

Do This : [Textbook Page No. 153 – 154]

1. Match the following: (First one is done for you.)

Solution:
(A) → (ii) → (c),
(B) → (iii) → (g),
(C) → (i) → (b),
(D) → (iv) → ( h ),
(E) → (v) → (f),
(F) → (vii) → (d),
(G) → (vi ) → (e),
(H) → (viii) → (a).

Do This : [Textbook Page No. 154 – 155]

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

Solution:
(i) → (c),
(ii) → (d),
(iii) → (e),
(iv) → (b),
(v) → (a).

Do This : [Textbook Page No. 162-163]

1. Look at the following map of a city:

(a) Colour the map as follows: Blue – Water, Red – Fire Station, Orange – Library, Yellow – Schools, Green -Park, Pink – Community Centre, Purple – Hospital, Brown – Cemetery.
(b) Mark a Green ‘X’ at the intersection of 2nd street and Danim street. A Black ‘Y’ where the river meets the third street. A red ‘Z’ at the intersection of main street and 1st street.
(c) In magenta colour, draw a short street route from the college to the lake.

2. Draw a map of the route from your house to your school showing important landmarks.
[Note: Friends, do activity 1 and 2 by yourself. Enjoy drawing and colouring.]

Do This : [Textbook Page No. 165-166]

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

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

What do you infer from the last two columns ? In each case, do you find F + V = E + 2, i.e., F + V – E = 2? This relationship is called Euler’s formula. Infact this formula is true for any polyhedron.
Solution:

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

Yes, in each case Euler’s formula F + V = E + 2 or F + V – E = 2 is true.

Think, Discuss and Write : [Textbook Page No. 166]

1. What happens to F, V and E if some parts are sliced off from a solid ? (To start with, you may take a plasticine cube, cut a corner off and investigate.)
Solution:

Suppose ABCDEFGH is a cube. It has 6 faces, 8 vertices and 12 edges.
V = 8, F = 6 and E = 12
∴ V + F – E
= 8 + 6 – 12 = 2
So the Euler’s formula is verified.

Now, suppose we sliced off ∆ PQR from this cube, then F = 6 + 1 = 7
V = 8 – 1 + 3
= 7 + 3 = 10
E = 12 + 3 = 15
∴ V + F – E = 10 + 7 – 15 = 2
∴ Here also Euler’s formula is verified. Thus, if some parts are sliced off a solid, then the number of vertices, edges and faces will be changed but still the Euler’s formula is verified.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.3 Textbook Exercise Questions and Answers.

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

1. Can a polyhedron have for its faces:

Question (i)
3 triangles?
Solution:
[Note: Polyhedron is a solid, which is made by polygonal regions and these polygonal regions are called its faces.]
No, a polyhedron cannot have 3 triangles for its faces because it have atleast 4 faces.

Question (ii)
4 triangles?
Solution:
Yes, a polyhedron can have 4 triangles for its faces as triangular pyramid.

Question (iii)
a square and four triangles?
Solution:
Yes, a polyhedron can have a square and four triangles for its faces as a pyramid with square base.

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

3. Which are prisms among the following?

Solution:
Note: A prism is a polyhedron whose base and top are congruent polygons and lateral faces are rectangles.
(i) No, a nail is not a prism because its base and top are not congruent polygons.
(ii) Yes, an unsharpened pencil is a prism because its base and top are congruent polygons and faces are rectangles.
(iii) No, a table weight is not a prism because its top and base are not congruent polygons.
(iv) Yes, box is a prism because its base and top are congruent polygons and lateral faces are rectangles.

4.

Question (i)
How are prisms and cylinders alike?
Solution:
Top and base of a prism and of a cylinder are congruent and parallel to each other and a prism becomes a cylinder, if the number of sides of its base becomes larger and larger.

Question (ii)
How are pyramids and cones alike?
Solution:
The pyramids and cones are alike because their lateral faces meet at a vertex. Also, a pyramid becomes a cone if the number of sides of its base becomes larger and larger.

5. Is a square prism same as a cube? Explain.
Solution:
No, a square prism is not same as a cube. It can be a cuboid also.

6. Verify Euler’s formula for these solids:

Solution:
(i) For figure (i)
F = 7, V= 10 and E = 15
∴ F + V = 7 + 10 = 17
Now, F + V – E = 17 – 15 = 2
Thus, F + V – E = 2
Hence, Euler’s formula is verified.

(ii) For figure (ii)
F = 9, V = 9 and E = 16
∴ F + V = 9 + 9 = 18
Now, F + V – E = 18 – 16 = 2
Thus, F + V – E = 2
Hence, Euler’s formula is verified.

7. Using Euler’s formula find the unknown:

	(i)	(ii)	(iii)
Faces	?	5	20
Vertices	6	?	12
Edges	12	9	?

Solution:
(i) Here, F = ?, V = 6 and E = 12
Now, F + V – E = 2 (∵ Euler’s formula)
∴ F + 6 – 12 = 2
∴ F – 6 = 2
∴ F = 2 + 6
∴ F = 8

(ii) Here, F = 5, V = ? and E = 9
Now, F + V – E = 2 (∵ Euler’s formula)
∴ 5 + V – 9 = 2
∴ V – 4 = 2
∴ V = 2 + 4
∴ V = 6

(iii) Here, F = 20, V = 12 and E = ?
Now, F + V – E = 2 (∵ Euler’s formula)
∴ 20 + 12 – E = 2
∴ 32 – E = 2
∴ – E = 2 – 32
∴ – E = – 30
∴ E = 30

8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
Here, F = 10, E = 20 and V = 15
By Euler’s formula, F + V – E = 2
LHS = F + V – E
= 10 + 15 – 20
= 25 – 20 = 5
Thus, F + V – E ≠ 2
Hence, such a polyhedron is not possible.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.2 Textbook Exercise Questions and Answers.

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

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?
Solution:
[Note: Enjoy (a) to (c) by doing oneself. (d) city park (e) Sr. Secondary School]

2. Draw a map of your classroom using proper scale and symbols for different objects.

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

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

[Note: Enjoy while doing 2, 3, 4 by yourself.]

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1

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) → (iii) → (iv)
(b) → (i) → (v), (c) → (iv) → (ii)
(d) → (v) → (iii), (e) → (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)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(d)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

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


Solution:
(a)

Object	(i)	(ii)	(iii)
View	Top view	Front view	Side view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Top view	Side view	Front view

(d)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(e)

Object	(i)	(ii)	(iii)
View	Front view	Top view	Side view

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

Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 11 Mensuration InText Questions

From Textbook : [Textbook Page No. 170]

Can you write an expression for the perimeter of each of the above shapes?
Solution:
1. Rectangle → a × b
2. Square → a × a
3. Triangle → \(\frac {1}{2}\) × b × h
4. Parallelogram → b × h
5. Circle → πb²

Yes, perimeter of above shapes are as follows:
1. Perimeter of rectangle = 2 (length + breadth)
2. Perimeter of square = 4 × (side)
3. Perimeter of triangle = sum of lengths of three sides
4. Perimeter of parallelogram = 2 × (sum of adjacent sides)
5. Perimeter (circumference) of circle = 2πr

Try These : [Textbook Page No. 170]

(a) Match the following figures with their respective areas in the box.

Solution:

Area of square
= l × l = 7 × 7

(b) Write the perimeter of each shape.
Solution:
Perimeter of given shapes:
1. Perimeter of this figure can’t be found as breadth is not given [7 cm is the height, not breadth].

2. Perimeter of semicircle = πr + 2r
= \(\frac {22}{7}\) × 7 + 2 × 7
= 22 + 14
= 36 cm

3. Perimeter of triangle = sum of lengths of three sides
= 14 + 11 + 19 = 34cm

4. Perimeter of rectangle = 2(l + b)
= 2 (14 + 7)
= 2 × 21
= 42cm

5. Perimeter of square = 4l
= 4 × 7 = 28cm

Try These : [Textbook Page No. 172]

1. Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown in figure. Show that the area of trapezium WXYZ = \(\frac {a + b}{2}\).

Solution:
Area of ∆ PWZ = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × c × h
= \(\frac {1}{2}\)ch
Area of rectangle PQYZ = length × breadth
= b × h = bh
Area of ∆ QXY = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × d × h
= \(\frac {1}{2}\)dh
Now, Area of trapezium WXYZ
Area of ∆ PWZ + Area of rectangle PQYZ + Area of ∆ QXY
= \(\frac {1}{2}\) ch + bh + \(\frac {1}{2}\)dh
= \(\frac {1}{2}\) ch + \(\frac {1}{2}\)dh + bh
= \(\frac {1}{2}\) (c + d)h + bh
= \(\frac {1}{2}\)(a – b)h + bh (∵ c + d = a – b)
= [\(\frac {1}{2}\)(a – b) + b]h
= [\(\frac {a – b}{2}\) + b] h
= \(\frac {a-b+2b}{2}\) h
= \(\frac {h (a + b)}{2}\)
Thus, area of trapezium WXYZ = \(\frac{h(a+b)}{2}\)

2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression \(\frac{h(a+b)}{2}\).
Solution:
h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm.
Area of ∆ PWZ = \(\frac {1}{2}\) × c × h
= (\(\frac {1}{2}\) × 6 × 10)cm²
= 30 cm²
Area of ∆ QXY = \(\frac {1}{2}\) × d × h
= (\(\frac {1}{2}\) × 4 × 10)cm²
= 20 cm²
Area of rectangle PQYZ
= length × breadth
= 12 × 10
= 120 cm²
∴ Area of trapezium WXYZ = Area of ∆ PWZ + Area of ∆ QXY + Area of rectangle PQYZ
= (30 + 20 + 120) cm²
= 170 cm²
Now, Area of trapezium WXYZ
= \(\frac{h(a+b)}{2}\)
= \(\frac{10(22+12)}{2}\)
= \(\frac{10(34)}{2}\)
= 170cm²
∵ a = c + b + d
= 6 + 12 + 4
= 22 cm
Thus, area of trapezium verified.

Try These : [Textbook Page No. 173]

1. Find the area of the following trapeziums:

Solution:
(i) Area of given trapezium = \(\frac {1}{2}\) × (9 + 7) × 3
= \(\frac {1}{2}\) × 16 × 3 cm² = 24 cm²
(ii) Area of given trapezium = \(\frac {1}{2}\) × (10 + 5) × 6
= \(\frac {1}{2}\) × 15 × 6 cm² = 45 cm²

Try These : [Textbook Page No. 174]

1. We know that parallelogram is also a quadrilateral.

Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already?
Solution:
Let XYZW be a given quadrilateral, which is a parallelogram.

Join the diagonal WY of the parallelogram XYZW. It divides the parallelogram into two triangles ∆WXY and ∆YZW.
Then, area of parallelogram WXYZ
= Area of ∆ WXY + Area of ∆ YZW
= (\(\frac {1}{2}\) × xy × h) + (\(\frac {1}{2}\) × wz × h)
= (\(\frac {1}{2}\) × b × h) + (\(\frac {1}{2}\) × b × h)
= \(\frac {1}{2}\) bh + \(\frac {1}{2}\) bh
= bh sq units
Area of parallelogram = base × height
= b × h
= bh sq units
As we have studied that, a parallelogram is a special case of a trapezium, where parallel sides are equal.
∴ Area of trapezium XYZW
= \(\frac {1}{2}\) × (b + b) × h
= (\(\frac {1}{2}\) × 2b × h) sq. units
= bh sq units
Thus, the above relation prooves the formula which already we know.

Think, Discuss and Write: [Textbook Page No. 175]

1. A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?
Solution:
No, by dividing trapezium by diagonal two congruent triangles cannot be obtained.

Let us understand from given figures. Here, by drawing diagonals of a quadrilateral ABCD (which is a trapezium) congruent triangles are not obtained.

Try These : [Textbook Page No. 175]

1. Find the area of these quadrilaterals:

Solution:
(i) h₁ = 3 cm, h₂ = 5 cm
d = length of diagonal AC = 6 cm
∴ Area of quadrilateral ABCD
= \(\frac {1}{2}\)d(h₁ + h₂)
= \(\frac {1}{2}\) × 6 × (3 + 5)
= 3 × 8 cm²
= 24 cm²

(ii) d₁ = 7 cm, d₂ = 6 cm
∴ Area of rhombus ABCD
= \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 7 × 6
= 7 × 3 cm²
= 21 cm²

(iii) [Note : Here, given figure is a parallelogram. It’s diagonals divides it into two congruent triangles. From this figure, we can see that base of the triangle is 8 cm and height is 2 cm.]
∴ Area of parallelogram
= 2 (area of ∆ ADC)
= 2 × (\(\frac {1}{2}\) × b × h)
= 2 × (\(\frac {1}{2}\) × 8 × 2) cm²
= 2 × 8
= 16 cm²

Try These : [Textbook Page No. 176]

(i) Divide the following polygons into parts (triangles and trapezium) to find out its area.

Solution:
(a)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{FI}}\).
Here, GA ⊥ FI, EB ⊥ FI and HC ⊥ FI are drawn.
Area of polygon EFGHI = Area of ∆ GFA + Area of the trapezium ACHG + Area of ∆ HCI + Area of ∆ BIE + Area of ∆ FBE
= [\(\frac {1}{2}\) × FA × GA] + [\(\frac {1}{2}\) (AG + CH) × AC] + [\(\frac {1}{2}\) × CI × HC] + [\(\frac {1}{2}\) × BI × BE] + [\(\frac {1}{2}\) × FB × BE]
[Note : We can also use Area of ∆ EFI in place of Area of ∆ BIE + Area of ∆ FBE]

(b)

Let’s draw perpendicular on diagonal \(\overline{\mathrm{NQ}}\).
\(\overline{\mathrm{OE}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{MF}}\) ⊥ \(\overline{\mathrm{NQ}}\), \(\overline{\mathrm{PG}}\) ⊥ \(\overline{\mathrm{NQ}}\) and \(\overline{\mathrm{RH}}\) ⊥ \(\overline{\mathrm{NQ}}\) are drawn.
Area of polygon MNOPQR = Area of ∆ NEO + Area of trapezium EGPO + Area of ∆ GQP + Area of ∆ HQR + Area of trapezium MRHF + Area of ∆ NFM
= [\(\frac {22}{7}\) × NE × OE] + [\(\frac {22}{7}\) × (OE + PG) × EG] + [\(\frac {22}{7}\) × GQ × PG] + [\(\frac {22}{7}\) × HQ × HR] + [\(\frac {22}{7}\) × (FM + HR) × FH] + [\(\frac {22}{7}\) × NF × FM]

(ii) Polygon ABODE is divided into parts as shown in figure.

Find its area if AD = 8 cm, AH = 6 cm,
AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.
Area of polygon ABODE = area of ∆ AFB + ….
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = ………..
Area of trapezium FBCH
= FH × \(\frac{(BF + CH)}{2}\)
= 3 × \(\frac{(2+3)}{2}\) [FH = AH – AF]
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH= …………;
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE = ………….
So the area of polygon ABCDE = …………
Solution:

Here, AD = 8 cm,
AH = 6 cm,
HD = 2 cm,
AG = 4 cm,
GD = 4 cm,
AF = 3 cm and
GF = 1 cm
Area of polygon ABCDE = Area of ∆ AFB + Area of trapezium FBCH + Area of ∆ CHD + Area of ∆ ADE
Now,
Area of ∆ AFB = \(\frac {1}{2}\) × AF × BF
= \(\frac {1}{2}\) × 3 × 2 = 3 cm2 … (i)
Area of trapezium FBCH
= \(\frac {1}{2}\) × (BF × CH) × FH
= \(\frac {1}{2}\) × (2 + 3) × 3 [∵ FH = AH – AF]
= \(\frac {1}{2}\) × 5 × 3 = \(\frac {15}{2}\)
= 7.5 cm² … (ii)
Area of ∆ CHD = \(\frac {1}{2}\) × HD × CH
= \(\frac {1}{2}\) × (AD – AH) × CH
[∵ HD = AD – AH]
= \(\frac {1}{2}\) × (8 – 6) × 3
= \(\frac {1}{2}\) × 2 × 3
= 3 cm² … (iii)
Area of ∆ ADE = \(\frac {1}{2}\) × AD × GE
= \(\frac {1}{2}\) × 8 × 2.5
= 4 × 2.5
= 10 cm² ………. (iv )
∴ Area of polygon ABCDE = Area of [(i) + (ii) + (iii) + (iv)]
= (3 + 7.5 + 3 + 10) cm²
= 23.5 cm²

(iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm NA, OC, QD and RB are perpendiculars to diagonal MP.

Solution:

Here,
MP = 9 cm,
MD = 7 cm
∴ DP = MP – MD
= (9 – 7) cm
= 2 cm
MC = 6 cm ∴ CP = 3 cm
MB = 4 cm ∴ BP = 5 cm
MA = 2 cm ∴ AC = MC – MA
= (6 – 2) cm = 4 cm
MB + BD = MD
∴ BD = (7 – 4) cm
= 3 cm
Area of ∆ MAN = \(\frac {1}{2}\) × MA × AN
= \(\frac {1}{2}\) × 2 × 2.5
= 2.5 cm² ……. (i)
Area of trapezium ACON
= \(\frac {1}{2}\) × (AN + OC) × AC
= \(\frac {1}{2}\) × (2.5 + 3) × 4
= \(\frac {1}{2}\) × 4 × 5.5
= 2 × 5.5
= 11 cm² … (ii)
Area of ∆ CPO = \(\frac {1}{2}\) × CP × CO
= \(\frac {1}{2}\) × 3 × 3
= \(\frac {9}{2}\) = 4.5 cm² … (iii)
Area of ∆ MBR = \(\frac {1}{2}\) × MB × BR
= \(\frac {1}{2}\) × 4 × 2.5
= 2 × 2.5
= 5 cm² … (iv )
Area of trapezium BRQD
= \(\frac {1}{2}\) × (BR + DQ) × BD
= \(\frac {1}{2}\) × (2.5 + 2) × 3
= \(\frac {1}{2}\) × 4.5 × 3
= 6.75 cm² ………. (v)
Area of ∆ DQP = \(\frac {1}{2}\) × DP × DQ
= \(\frac {1}{2}\) × 2 × 2
= 2 cm² … ( vi )
∴ Area of polygon MNOPQR = Area of [(i) + (ii) + (iii) + (iv) + (v) + (vi)]
= (2.5 + 11 + 4.5 + 5 + 6.75 + 2) cm²
= 31.75 cm²

Think, Discuss and Write : [Textbook Page No. 180]

1. Why is it incorrect to call the solid shown here a cylinder?
Solution:

It is incorrect to call given solid as a cylinder because, a cylinder has two identical circular faces, parallel to each other. The radii of both the faces are the same.

Try These :[Textbook Page No. 181]

1. Find the total surface area of the following cuboids:

Solution:
(i) Here, length (i) = 6 cm,
breadth (b) = 4 cm and
height (h) = 2 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2(6 × 4 + 4 × 2 + 2 × 6)
= 2(24 + 8 + 12)
= 2 (44)
= 88 cm²

(ii) Here, length (l) = 4 cm,
breadth (b) = 4 cm and
height (h) = 10 cm.
Total surface area of a cuboid = 2 (lb + bh + hl)
= 2 (4 × 4 + 4 × 10 + 10 × 4)
= 2 (16 + 40 + 40)
= 2(96) = 192 cm²

Think, Discuss and Write :[Textbook Page No. 181]

1. Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base?
Solution:
Yes, the total surface area of cuboid = lateral surface area + 2 × area of base [Note: Area of base and area of top of a cuboid, both are same.]

2. If we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change?

Solution:
(a) Lateral surface area of cuboid (i)
= 2(l + b) × h
(b) Lateral surface area of cuboid (ii)
= 2 (h + b) × l
Thus, it is clear that by interchanging the lengths of the base and the height of a cuboid, its lateral surface area will change.

Try These : [Textbook Page No. 182]

1. Find the surface area of cube A and lateral surface area of cube B.

Solution:
For cube A:
Side = 10 cm
Surface area of cube A = 6 × (side)²
= 6 × (10)²
= 6 × 100
= 600 cm²

For cube B:
Side = 8 cm
Lateral surface area of a cube B
= 4 × (side)²
= 4 × (8)²
= 4 × 64
= 256 cm²

Think, Discuss and Write : [Textbook Page No. 183]

Question (i)
Two cubes each with side b are joined to form a cuboid. What is the surface area of this cuboid? Is it 12b²? Is the surface area of cuboid formed by joining three such cubes, 18b²? Why?

Solution:
(a) By joining two cubes end-to-end with side b, a cuboid shape formed.
For cuboid:
length = b + b = 2b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(2b × b) + (b × b) + (2b × b)]
= 2 (2b² + b² + 2b²)
= 2 (5b²)
= 10b²
So surface area of the cuboid formed by joining two cubes is not equal to 12b²), it is 10b²).

(b) By joining three cubes end-to-end with side b, a cuboid shape forms.
For cuboid:
length = b + b + b = 3b, breadth = b and height = b
Total surface area of this cuboid = 2 (lb + bh + lh)
= 2 [(3b × b) + (b × b) + (3b × b)]
= 2 (3b² + b² + 3b²)
= 2(7b²) = 14b²
So the surface area of cuboid formed by joining three such cubes is not equal to 18b², it is 14b².

(ii) How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?
Solution:
Let us arrange 12 cubes of equal length say b to form a cuboid in different situations.
(a)

Here, length = 12b, breadth = b and height = b
Total surface area of a cuboid = 2 (lb + bh + lh)
= 2 [(12b × b) + (b × b) + (12b × b)]
= 2 (12b² + b² + 12b²)
= 2 (25b²) – 50b²

(b)

Here, length = 3b, breadth = 2b and height = 2b
Total surface area of a cuboid
= 2 (lb+ bh+ lh)
= 2 [(3b × 2b) + (2b × 2b) + (2b × 3b)]
= 2 (6b² + 4b² + 6b²)
= 2 (16b²) = 32b²

(c)

Here, length = 6b, breadth = 2b and height = b
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(6b × 2b) + (2b × b) + (b × 6b)]
= 2 (12b² + 2b² + 6b²)
= 2 (20b²)
= 40b²
Hence, from above results we can conclude that, if we arrange 12 cubes of equal length according to situation (b), we get smallest surface area.

(iii) After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions. How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted?
Solution:
1. 8 cubes, which have no face painted. (∵ Middle 4 × 2)
2. 24 cubes, which have 1 face painted. (∵ on each surface 4 × 6)
3. 24 cubes, which have 2 faces painted. (∵ on each surface 4 × 6)
4. 8 cubes, which have 3 faces painted. (∵ on each surface 2 × 4)

Try These : [Textbook Page No. 184]

1. Find total surface area of the following cylinders:

Solution:
(i) Radius of cylinder r = 14 cm
Height of cylinder h = 8 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 14(14 + 8)
= 2 × 22 × 2 × 22
= 44 × 44
= 1936 cm²

(ii) Radius of cylinder r = \(\frac{\text { diameter }}{2}=\frac{2}{2}\) = 14 cm
Height of cylinder h = 2 cm
Total surface area of a cylinder = 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 1(1 + 2)
= 2 × \(\frac {22}{7}\) × 1 × 3
= \(\frac {132}{7}\)
= 18\(\frac {6}{7}\) m²

Think, Discuss and Write: [Textbook Page No. 184]

1. Note that lateral surface area of a cylinder is the circumference of base × height of cylinder. Can we write lateral surface area of a cuboid as perimeter of base × height of cuboid?
Solution:

Let l be the height, b be the breadth and h be the height of cuboid.
∴ Lateral surface area of a cuboid = Area of 4 walls of the cuboid
= (l × h) + (l × h) + (b × h) + (b × h)
= 2 lh + 2 bh
= 2(1 + b) × h
= Perimeter of base × height
Yes, we can write lateral surface area of a cuboid as perimeter of base × height of cuboid.

Try These : [Textbook Page No. 188]

1. Find the volume of the following cuboids:

Solution:
(i) For given cuboid:
length (l) = 8 cm, breadth (b) = 3 cm s and height (h) = 2 cm
Volume of a cuboid
= Area of base × height
= (l × b) × h
= (8 × 3) × 2
= 24 × 2 = 48 cm³
OR
Volume of cuboid = l × b × h
= (8 × 3 × 2) cm³
= 48 cm³

(ii) Area of base of cuboid = 24 m³
height (h) = 3 cm = \(\frac {3}{100}\) m
Volume of cuboid
= Area of base × height
= 24 × \(\frac {3}{100}\) = 0.72 m³

Try These: [Textbook Page No. 189]

1. Find the volume of the following cubes
(a) With a side 4 cm
Solution:
For given cube : side = 4 cm
∴ Volume of the cube = (side)³
= (4)³ = 4 × 4 × 4
= 64 cm³

(b) With a side 1.5m
Solution:
For given cube : side = 1.5 m
∴ Volume of the cube = (side)³
= (1.5)³ = 1.5 × 1.5 × 1.5
= \(\frac{15}{10} \times \frac{15}{10} \times \frac{15}{10}=\frac{3375}{1000}\)
= 3.375 m³

Think, Discuss and Write: [Textbook Page No. 189]

1. A company sells biscuits. For packing purpose they are using cuboidal boxes:
box A → 3 cm × 8 cm × 20 cm,
box B → 4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Can you suggest any other size (dimensions) which has the same volume but is more economical than these?
Solution:
For box A:
Given dimension = 3 cm × 8 cm × 20 cm
length = 20 cm, breadth = 8 cm and height = 3 cm
∴ Volume of the box A = l × b × h
= 20 × 8 × 3
= 480 cm³
Total surface area of the box A
= 2 (lb + bh + lh)
= 2 [(20 × 8) + (8 × 3) + (20 × 3)]
= 2 (160 + 24 + 60 )
= 2 (244)
= 488 cm³

For box B:
Given dimension = 4 cm × 12 cm × 10 cm
length = 12 cm, breadth = 10 cm and height = 4 cm
∴ Volume of the box B = l × b × h
= 12 × 10 × 4
= 480 cm³
Total surface area of the box B
= 2 (lb + bh + lh)
= 2 [(12 × 10) + (10 × 4) + (12 × 4)]
= 2(120 + 40 + 48 )
= 2(208)
= 416 cm³
From above results, we can conclude that both boxes have same volume, but surface area of box B is less than that of box A.
∴ Box B is more economical than box A.
Now, let another box of size be 8 cm × 6 cm × 10 cm.
∴ length = 8 cm, breadth = 6 cm and height =10 cm
Volume = l × b × h
= 8 × 6 × 10 = 480 cm³
It’s surface area = 2 (lb + bh + Ih)
= 2 [(8 × 6) + (6 × 10) + (8 × 10)]
= 2 (48 + 60 + 80)
= 2 (188) = 376 cm²
Surface area of this box is less than that of box B.
∴ This box is more economical for company.

Try These: [Textbook Page No. 189]

1. Find the volume of the following cylinders:

Solution:
(i) For given cylinder :
radius (r) = 7 cm, height (h) = 10 cm
Volume of the cylinder = πr²h
= \(\frac {22}{7}\) × 7² × 10
= \(\frac {22}{7}\) × 7 × 7 × 10
= 22 × 70
= 1540cm³

(ii) For given cylinder:
base area 250 m², height (h) = 2m
Volume of the cylinder
= base area × height
= 250 × 2
= 500 m³

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

1. Find the square root of each of the following numbers by Division method.

Question (i).
2304
Solution:

Question (ii).
4489
Solution:

Question (iii).
3481
Solution:

Question (iv).
529
Solution:

Question (v).
3249
Solution:

Question (vi).
1369
Solution:

Question (vii).
5776
Solution:

Question (viii).
7921
Solution:

Question (ix).
576
Solution:

Question (x).
1024
Solution:

Question (xi).
3136
Solution:

Question (xii).
900
Solution:

2. Find the number of digits in the square root of each of the following numbers (without any calculation).

Question (i).
64
Solution:
Here, number of digits, n = 2
(Which is an even number.)
∴ Number of digits in the square root of 64 = \(\frac{n}{2}=\frac{2}{2}\) = 1

Question (ii).
144
Solution:
Here, number of digits, n = 3
(Which is an odd number.)
∴ Number of digits in the square root of 144 = \(\frac{n+1}{2}=\frac{3+1}{2}\)
= \(\frac {4}{2}\)
= 2

Question (iii).
4489
Solution:
Here, number of digits, n = 4
(Which is an even number.)
∴ Number of digits in the square root of 4489 = \(\frac{n}{2}=\frac{4}{2}\)
= 2

Question (iv).
27225
Solution:
Here, number of digits, n = 5
(Which is an odd number.)
∴ Number of digits in the square root of 27225 = \(\frac{n+1}{2}=\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Question (v).
390625
Solution:
Here, number of digits, n = 6
(Which is an even number.)
∴ Number of digits in the square root of 390625 = \(\frac{n}{2}=\frac{6}{2}\)
= 3

3. Find the square root of the following decimal numbers.

Question (i).
2.56
Solution:

Here, number of decimal places are two.
∴ The number of decimal places in square root should be one.

Question (ii).
7.29
Solution:

Here, number of decimal places are two.
∴ The number of decimal places in square root should be one.

Question (iii).
51.84
Solution:

Here, number of 51.84 decimal places are two.
∴ The number of decimal places in square root should be one.

Question (iv).
42.25
Solution:

Question (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.

Question (i).
402
Solution:

Here, the remainder is 2. It shows that 20² is less than 402 by 2.
So, to get a perfect square, 2 must be subtracted from given number.
∴ Required perfect square number = 402 – 2 = 400
\(\sqrt{400}\) = 20

Question (ii).
1989
Solution:

Here, the remainder is 53. It shows that 44² is less than 1989 by 53.
So, to get a perfect square, 53 must be subtracted from the given number.
∴ Required perfect square number = 1989 – 53 = 1936
\(\sqrt{1936}\) = 44

Question (iii).
3250
Solution:

Here, the remainder is 1. It shows that 57² is less them 3250 by 1.
So, to get a perfect square, 1 must be subtracted from the given number.
∴ Required perfect square number = 3250 – 1 = 3249
\(\sqrt{3249}\) = 57

Question (iv).
825
Solution:

Here, the remainder is 41. It shows that 28² is less than 825 by 41.
So, to get a perfect square, 41 must be subtracted from the given number.
∴ Required perfect square number = 825 – 41 = 784
\(\sqrt{784}\) = 28

Question (v).
4000
Solution:

Here, the remainder is 31. It shows that 63² is less than 4000 by 31.
So, to get a perfect square, 31 must be subtracted from the given number.
∴ Required perfect square number = 4000 – 31 = 3969
\(\sqrt{3969}\) = 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.

Question (i).
525
Solution:

Here, the remainder is 41.
525 > 22²,
and the number next to 22 is 23.
23² = 529.
∴ The required number to be added
= 23² – 525
= 529 – 525 = 4
Now, 525 + 4 = 529
\(\sqrt{529}\) = 23
Thus, 4 is the least number which must be added to 525 to get a perfect square.

Question (ii).
1750
Solution:

Here, the remainder is 69.
1750 > 41²,
and the number next to 41 is 42.
42² = 1764.
∴ The required number to be added = 42² – 1750
= 1764 – 1750
= 14
Now, 1750 + 14 = 1764
\(\sqrt{1764}\) = 42
Thus, 14 is the least number which must be added to 1750 to get a perfect square.

Question (iii).
252
Solution:

Here, the remainder is 27.
252 > 15²,
and the number next to 15 is 16.
16² = 256.
∴ The required number to be added = 16² – 252
= 256 – 252
= 4
Now, 252 + 4 = 256
\(\sqrt{256}\) = 16
Thus, 4 is the least number which must be added to 252 to get a perfect square.

Question (iv).
1825
Solution:

Here, the remainder is 61.
1825 > 42²,
and the number next to 42 is 43.
43² = 1849.
∴ The required number to be added 43² – 1825
= 1849 – 1825
= 24
Now, 1825 + 24 = 1849
\(\sqrt{1849}\) = 43
Thus, 24 is the least number which must be added to 1825 to get a perfect square.

Question (v).
6412
Solution:

Here, the remainder is 12.
6412 > 80²,
and the number next to 80 is 81.
81² = 6561.
∴ The required number to be added = 81² – 6412
= 6561 – 6412
= 149
Now, 6412 + 149 = 6561
\(\sqrt{6561}\) = 81
Thus, 149 is the least number which must be added to 6412 to get a perfect square.

6. Find the length of the side of a square whose area is 441 m².
Solution:
Let the side of a square be x m.
Area of a square = x × x = x²
Area of a square = 441 (given)

∴x² = 441
∴ x = \(\sqrt{441}\)
∴ x = 21
Thus, the length of the side of the square is 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:
[Note: In a right triangle, the longest side is called the hypotenuse. The square of the hypotenuse is equal to the sum of the squares of the remaining two sides.]
Let us use this theorem here.
(a) Here, ∠B = 90°, AB = 6 cm, BC = 8 cm
In ΔABC, AC is hypotenuse.

AC² = AB² + BC²
= (6)² + (8)²
= 36 + 64
=100
∴ AC = \(\sqrt{100}\)
= 10cm

(b) Here, ∠B= 90°, AC = 13cm, BC = 5cm
In ΔABC, AC Is hypotenuse.


AC² = AB² + BC²
∴ (13)² = AB² + (5)²
∴ 169 = AB² + 25
∴ AB² = 169 – 25
= 144
∴ AB = \(\sqrt{144}\)
= 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 same. Find the minimum number of plants he needs more for this.
Solution:
Total number of plants = 1000
The number of plants in a row = The number of plants in a column
Let the number of plants planted in a row be x.
So, the number of plants planted in a column is x.
∴ Total plants = x × x = x²
∴ x² > 1000
∴ x > \(\sqrt{1000}\)
Here, the remainder is 39.
(31)² < 1000
The next square number would be 32.
32² = 1024
∴ The number of plants required to be added = 1024 – 1000
= 24
Thus, 24 more plants needed.

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 number of columns. How many children would be left out in this arrangement?
Solution:
Total number of children in a school = 500
The number of rows = The number of columns
Let the number of children in a row be x.
So, the number of children in a column is x.
Total number of children = x × x = x².
∴ x² < 500
∴ x < \(\sqrt{500}\)
Here, the remainder is 16.
500 > 22²
∴ 500 > 484
500 – 484 = 16
Thus, 16 children would be left out in this arrangement.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 6 Squares and Square Roots InText Questions

Try These (Textbook Page No. 90)

1. Find the perfect square numbers between (i) 30 and 40 (ii) 50 and 60
Solution:
(i) Since,
1 × 1 = 1,
2 × 2 = 4,
3 × 3 = 9,
4 × 4 = 16,
5 × 5 = 25,
6 × 6 = 36,
7 × 7 = 49.
Thus, 36 is the only perfect square number between 30 and 40.

(ii) Since, 7 × 7 = 49 and 8 × 8 = 64.
∴ There is no perfect square number between 49 and 64.
Thus, there is no perfect square number between 50 and 60.

Try These (Textbook Page No. 90 – 91)

1. Can we say whether the following numbers are perfect squares? How do we know ?
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 1069
(vi) 2061
Write five numbers which you can decide by looking at their units digit that they are not square numbers.
Solution:
(i) 1057
The ending digit is 7. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 1057 cannot be a perfect square.

(ii) 23453
The ending digit is 3. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 23453 cannot be a perfect square.

(iii) 7928
The ending digit is 8. All square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 7928 cannot be a perfect square.

(iv) 222222
The ending digit is 2. All square numbers end with 0, 1, 4, 5, 6 or s 9 at unit’s place.
∴ 222222 cannot be a perfect square.

(v) 1069
The ending digit is 9. All square jj numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 1069 may or may not be a square number.
30 × 30 = 900, 31 × 31 = 961,
32 × 32 = 1024 and 33 × 33 = 1089.
e.g. No natural number between 1024 and 1089 is a perfect square.
∴ 1069 cannot be a perfect square.

(vi) 2061
The ending digit is 1. All square s numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
∴ 2061 may or may not be a square number.
45 × 45 = 2025, 46 × 46 = 2116,
e.g. No natural number between 2025 and 2116 is a square number.
∴ 2061 is not a square number.

2. Write five numbers which you cannot decide just by looking at their unit’s digit (or units place) whether they are square numbers or not.
Solution:
Any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number.
Five such numbers are :
719, 2431, 524, 215, 326
[Note: You can write many more numbers.]

Try These (Textbook Page No. 91)

1. Which of 1232, 772, 822, 1612, 1092 would end with digit 1 ?
Solution:
The square of those numbers end in 1 which end in either 1 or 9. Here, the square of 161 and 109 would end in 1.
OR
The number having 1 or 9 in units place has the digit 1 in units place of its square.

Try These (Textbook Page No. 91)

Which of the following numbers would have digit 6 at unit place ?
(i) 19²
(ii) 24²
(iii) 26²
(iv) 36²
(v) 34²
Solution:
The number having 4 or 6 in the units place has the digit 6 in units place of its square. Except 19² all other numbers have 6 in their units place.
OR
(i) 19²
Units place digit is 9.
∴ 19² would not have units digit as 6. (9 × 9 = 81)

(ii) 24²
Units place digit is 4.
∴ 24² would have units digit as 6. (4 × 4= 16)

(iii) 26²
Units place digit is 6.
∴ 26² would have 6 at units place. (6 × 6 = 36)

(iv) 36²
Units place digit is 6.
∴ 36² would have 6 at units place. (6 × 6 = 36)

(v) 34²
Units place digit is 4.
∴ 34² would have 6 at units place. (4 × 4 = 1²)

Try These (Textbook Page No. 92)

What will be the “one’s digit” in the square of the following numbers ?

Question (i).
1234
Solution:
1234
The ending digit is 4.
4 × 4 = 16
∴ (1234)² will have 6 as the one’s place.

Question (ii).
26387
Solution:
26387
The ending digit is 7.
7 × 7 = 49
∴ (26387)² will have 9 as the one’s place.

Question (iii).
52698
Solution:
52698
The ending digit is 8.
8 × 8 = 64
∴ (52698)² will have 4 as the one’s place.

Question (iv).
99880
Solution:
99880
The ending digit is 0.
0 × 0 = 0
∴ (99880)² will have 0 as the one’s place.

Question (v).
21222
Solution:
21222
The ending digit is 2.
2 × 2 = 4
∴ (21222)² will have 4 as the one’s place.

Question (vi).
9106
Solution:
9106
The ending digit is 6.
6 × 6 = 36
∴ (9106)² will have 6 as the one’s place.

Try These (Textbook Page No. 92)

1. The square of which of the following numbers would be an odd number/an even number? Why?

Question (i).
727
Solution:
727
Here, the ending digit is 7.
It is an odd number.
∴ Its square is also an odd number.

Question (ii).
158
Solution:
158
Here, the ending digit is 8.
It is an even number.
∴ Its square is also an even number.

Question (iii).
269
Solution:
269
Here, the ending digit is 9.
It is an odd number.
∴ Its square is also an odd number.

Question (iv).
1980
Solution:
1980
Here, the ending digit is 0.
It is an even number.
∴ Its square is also an even number.

2. What will be the number of zeros in the square of the following numbers?

Question (i).
60
Solution:
60
In 60, number of zero is 1.
∴ (60)² will have 2 zeros. (∵ 60² = 3600)

Question (ii).
400
Solution:
400
In 400, number of zeros are 2.
∴ (400)² will have 4 zeros.
(∵ 400² = 160000)

Try These (Textbook Page No. 94)

1. How many natural numbers lie between 9² and 10²? Between 11² and 12²?
Solution:
(a) We can find 2n natural numbers between two consecutive natural numbers, n² and (n + 1)².
Here, n = 9, n + 1 = 9 + 1 = 10.
∴ Natural numbers between 9² and 10² are 2n = 2 × 9 = 18.
Thus, 18 natural numbers lie between 9² and 10².

(b) We can find 2n natural numbers between two consecutive natural numbers, n² and (n + 1)².
Here, n = 11, n + 1 = 11 + 1 = 12.
∴ Natural numbers between 11² and 12² are 2n = 2 × 11 = 22.
Thus, 22 natural numbers lie between 112 and 122.

2. How many non square numbers lie between the following pairs of numbers:

Question (i).
100² and 101²
Solution:
100² and 101²
Here, n = 100 and n + 1 = 100 + 1
= 101
∴ non square numbers between 100² and 101² = 2 × n
= 2 × 100
= 200
Thus, 200 non square numbers lie between 100² and 101².

Question (ii).
90² and 91²
Solution:
90² and 91²
Here, n = 90 and n + 1 = 90 + 1 = 91.
∴ Natural numbers between 90² and 91²
= 2 × n
= 2 × 90
= 180.
Thus, 180 non square numbers lie between 90² and 91².

Question (iii).
1000² and 1001²
Solution:
1000² and 1001²
Here, n = 1000 and
n + 1 = 1000 + 1 = 1001.
∴ Natural numbers between 1000² and 1001²
= 2 × n
= 2 × 1000
= 2000.
Thus, 2000 non square numbers lie between 1000² and 1001².

Try These (Textbook Page No. 94)

Find whether each of the following numbers is a perfect square or not:

Question (i).
121
Solution:
121
121 – 1 = 120, 120 – 3 = 117
117-5 = 112, 112 – 7 = 105
105-9 = 96, 96 – 11 = 85
85 – 13 = 72, 72 – 15 = 57
57- 17 = 40, 40 – 19 = 21
21 – 21=0
e.g. 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Thus, 121 is a perfect square.

Question (ii).
55
Solution:
55
55 – 1 = 54, 54 – 3 = 51
51 – 5 = 46, 46 – 7 = 39
39 – 9 = 30, 30 – 11 = 19
19 – 13 = 6, 6 – 15 = – 9
Thus, 55 cannot be expressed as the sum of successive odd numbers starting from 1.
∴ 55 is not a perfect square.

Question (iii).
81
Solution:
81
81 – 1 = 80, 80 – 3 = 77
77 – 5 = 72, 72 – 7 = 65
65 – 9 = 56, 56 – 11 = 45
32 – 15 = 17, 32 – 15 = 17
17 – 17 = 0
∴ 81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Thus, 81 is a perfect square.

Question (iv).
49
Solution:
49
49 – 1 = 48, 48 – 3 = 45
45 – 5 = 40, 40 – 7 = 33
33 – 9 = 24, 24 – 11 = 13
13 – 13 = 0
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
Thus, 49 is a perfect square.

Question (v).
69
Solution:
69
69 – 1 = 68, 68 – 3 = 65
65 – 5 = 60, 60 – 7 = 53
53-9 = 44, 44 – 11 = 33
33- 13 = 20, 20 – 15 = 5
5 – 17 = – 12
Thus, 69 cannot be expressed as the sum of successive odd numbers starting from 1.
∴ 69 is not a perfect square.

Try These (Textbook Page No. 95)

1. Express the following as the sum of two consecutive integers:

Question (i).
21²
Solution:
Remember : n² = \(\frac{n^{2}-1}{2}+\frac{n^{2}+1}{2}\)
21²
n = 21

∴ 21² = 220 + 221 = 441

Question (ii).
13²
Solution:
13²
n = 13

∴ 13² = 84 + 85 = 169

Question (iii).
11²
Solution:
11²
n = 11

∴ 11² = 60 + 61 = 121

Question (iv).
19²
Solution:
19²
n = 19

∴ 19² = 180 + 181 = 361

2. Do you think the reverse is also true, e.g. is the sum of any two consecutive positive integers is perfect square of a number? Give example to support your answer.
Solution:
No, the reverse is not always true.
(i) 3 + 4 = 7, 7 is not a perfect square.
(ii) 10 + 11 = 21, 21 is not a perfect square.
But,
(i) 4 + 5 = 9, 9 is a perfect square.
(ii) 12 + 13 = 25, 25 is a perfect square.

Try These (Textbook Page No. 95)

Write the square, making use of the above pattern:

Question (i).
111111²
Solution:
(111111)²
The given number is a six-digit number,
∴ middle number 6

Question (ii).
1111111²
Solution:
(1111111)²
The given number is a seven-digit number.
∴ middle number 7

Try These (Textbook Page No. 95)

Can you find the square of the following numbers using the above pattern ?
(i) 6666667²
(ii) 66666667²
Solution:
(i) Yes, 6666667² = 44444448888889
(ii) Yes, 66666667² = 4444444488888889

Try These (Textbook Page No. 97)

Find the squares of the following numbers containing 5 in unit’s place:

Question (i).
15
Solution:
[Note : A number with unit digit 5, e.g. a5
(a5)² = a(a + 1) × 100 + 25]
( i ) (15)² = 1 × (1 + 1) × 100 + 25
= 1 × 2 × 100 + 25
= 200 + 25
= 225
Hint:

Question (ii).
95
Solution:
(95)² = 9 (9 + 1) × 100 + 25
= 9 × 10 × 100 + 25
= 9000 + 25
= 9025

Question (iii).
105
Solution:
(105)² = 10 × (10 + 1) × 100 + 25
= 10 × 11 × 100 + 25
= 11000 + 25
= 11025
Hint:

Question (iv).
205
Solution:
(205)² = 20 × (20 + 1) × 100 + 25
= 20 × 21 × 100 + 25
= 42000 + 25
= 42025

Try These (Textbook Page No. 99)

Question (i).
11² = 121. What is the square root of 121 ?
Solution:
Square root of 121 is 11.

Question (ii).
14² = 196. What is the square root of 196 ?
Solution:
Square root of 196 is 14.

Think, Discuss and Write (Textbook Page No. 99)

Question (i).
(-1)² = 1. Is – 1, a square root of 1 ?
Solution:
[Note: There are two (positive as well as negative) integral square roots of a perfect square number.]
(- 1) × (- 1) = 1, (- 1)² = 1
∴ Square root of 1 can also be (-1).

Question (ii).
(-2)² = 4. Is -2, a square root of 4 ?
Solution:
(- 2) × (- 2) = 4, (-2)² = 4
∴ Square root of 4 can also be (-2).

Question (iii).
(-9)² = 81. Is – 9, a square root of 81 ?
Solution:
(-9) × (-9) = 81, (- 9)² = 81
∴ Square root of 81 can also be (-9).

Try These (Textbook Page No. 100)

By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root:

Question (i).
121
Solution:
Subtracting the successive odd numbers from 121 :
121 – 1 = 120, 120 – 3 = 117
117 – 5 = 112, 112 – 7 = 105
105 – 9 = 96, 96 – 11 = 85
85 – 13 = 72, 72 – 15 = 57
57 – 17 = 40, 40 – 19 = 21
21 – 21 = 0
∴ \(\sqrt{121}\) = 11
∴ 121 is a perfect square.

Question (ii).
55
Solution:
∵ 55 – 1 = 54, 54 – 3 = 51
51 – 5 = 46, 46 – 7 = 39
39 – 9 = 30, 30 – 11 = 19
19 – 13 = 6, 6 – 15 = – 9
55 does not reduced to 0 after subtracting odd numbers starting from 1.
∴ 55 is not a perfect square.

Question (iii).
36
Solution:
∵ 36 – 1 = 35, 35 – 3 = 32
32 – 5 = 27, 27 – 7 = 20
20 – 9 = 11, 11 – 11 = 0
∴ \(\sqrt{36}\) = 6
∴ 36 is a perfect square.

Question (iv).
49
Solution:
49 – 1 = 48, 48 – 3 = 45
45 – 5 = 40, 40 – 7 = 33
33 – 9 = 24, 24 – 11 = 13
13- 13 = 0
∴ \(\sqrt{49}\) = 7
∴ 49 is a perfect square.

Question (v).
90
Solution:
90 – 1 = 89, 89 – 3 = 86
86 – 5 = 81, 81 – 7 = 74
74 – 9 = 65, 65 – 11 = 54
54 – 13 = 41, 41 – 15 = 26
26 – 17 = 9, 9 – 19 = – 10
90 does not reduced to 0 after subtracting odd numbers starting from 1.
∴ 90 is not a perfect square.

Think, Discuss and Write (Textbook Page No. 103)

Can we say that if a perfect square is of n-digits, then its square root will have \(\frac{n}{2}\) digits if n is even or \(\frac{(n+1)}{2}\) if n is odd ?
Solution:
Yes, we can say that if a perfect square is of n-digits, then its square root will have
(a) \(\frac{n}{2}\) digits, if n is even.
(b) \(\frac{(n+1)}{2}\) digits, if n is odd.

Try These (Textbook Page No. 105)

Without calculating square roots, find the number of digits in the square root of the following numbers:

Question (i).
25600
Solution:
Here, the number of digits, n = 5 (an odd number.)
∴ Number of digits in the square root of 25600 = \(\frac{(n+1)}{2}\)
= \(\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Question (ii).
100000000
Solution:
Here, the number of digits, n =9 (an odd number.)
∴ Number of digits in the square root of 100000000 = \(\frac{(n+1)}{2}\)
= \(\frac{9+1}{2}\)
= \(\frac {10}{2}\)
= 5

Question (iii).
36864
Solution:
Here, the number of digits, n = 5 (an odd number.)
∴ Number of digits in the square root of 36864 = \(\frac{(n+1)}{2}\)
= \(\frac{5+1}{2}\)
= \(\frac {6}{2}\)
= 3

Try these (Textbook Page No. 107)

Estimate the value of the following to the nearest whole number:

Question (i).
\(\sqrt{80}\)
Solution:
\(\sqrt{80}\)
10² = 100, 9² = 81, 8² = 64
∴ 80 is between 64 and 81.
∴ 64 < 80 < 81
∴ 8² < 80 < 9²
∴ 8 < \(\sqrt{80}\) < 9
Thus, \(\sqrt{80}\) lies between 8 and 9.
\(\sqrt{80}\) is much closer to 81 than 64.
∴ \(\sqrt{80}\) is approximately 9.

Question (ii).
\(\sqrt{1000}\)
Solution:
\(\sqrt{1000}\)
30² = 900, 31² = 961, 32² = 1024
∴ 1000 is between 961 and 1024.
∴ 961 < 1000 < 1024
∴ 31² < 1000 < 32²
∴ 31 < \(\sqrt{1000}\) < 32
Thus, \(\sqrt{1000}\) lies between 31 and 32.
\(\sqrt{1000}\) is much closer to 32 than 31.
∴ \(\sqrt{1000}\) is approximately 32.

Question (iii).
\(\sqrt{350}\)
Solution:
\(\sqrt{350}\)
18² = 324 and 19² = 361
∴ 350 is between 324 and 361.
∴ 324 < 350 < 361
∴ 18² < 350 < 19²
∴ 18 < \(\sqrt{350}\) < 19
Thus, \(\sqrt{350}\) lies between 18 and 19.
\(\sqrt{350}\) is much closer to 19 than 18.
∴ \(\sqrt{350}\) is approximately 19.

Question (iv).
\(\sqrt{500}\)
Solution:
\(\sqrt{500}\)
22² = 484 and 23² = 529
∴ 500 is between 484 and 529.
∴ 484 < 500 < 529
∴ 22² < 500 < 23²
∴ 22 < \(\sqrt{500}\) < 23
Thus, \(\sqrt{500}\) lies between 22 and 23.
\(\sqrt{500}\) is much closer to 22 than 23.
∴ \(\sqrt{500}\) is approximately 22.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.3

1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers ?

Question (i).
9801
Solution:
The possible digit at one’s place of the square root of 9801 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (ii).
99856
Solution:
The possible digit at one’s place of the square root of 99856 can be either 4 or 6.
(∵ 4 × 4= 16 and 6 × 6 = 36)

Question (iii).
998001
Solution:
The possible digit at one’s place of the square root of 998001 can be either 1 or 9.
(∵ 1 × 1 = 1 and 9 × 9 = 81)

Question (iv).
657666025
Solution:
The possible digit at one’s place of the square root of 657666025 can be 5.
(∵ 5 × 5 = 25)

2. Without doing any calculation, find the numbers which are surely not perfect squares:
[Note : The ending digit of perfect square is 0, 1, 4, 5, 6 or 9.
∴ A number having end digit 2, 3, 7 or 8 can never be a perfect square.

Question (i).
153
Solution:
153
Here, the end digit is 3.
∴ 153 cannot be a perfect square.

Question (ii).
257
Solution:
257
Here, the end digit is 7.
∴ 257 cannot be a perfect square.

Question (iii).
408
Solution:
408
Here, the end digit is 8.
∴ 408 cannot be a perfect square.

Question (iv).
441
Solution:
441
Here, the end digit is 1.
∴ 441 can be a perfect square.

3. Find the square roots of 100 and 169 by the method of repeated subtraction.

Question (i).
100
Solution:
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
∴ 100 is a perfect square.
∴ \(\sqrt{100}\) = 10

Question (ii).
169
Solution:
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
∴ 169 is a perfect square.
∴ \(\sqrt{169}\) = 13

4. Find the square roots of the following numbers by the Prime Factorisation Method:

Question (i).
729
Solution:
729
\(\begin{array}{l|r}
3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
729 = 3 × 3 × 3 × 3 × 3 × 3
= 3² × 3² × 3²
∴ \(\sqrt{729}\) = 3 × 3 × 3
= 27

Question (ii).
400
Solution:
400
\(\begin{array}{l|r}
2 & 400 \\
\hline 2 & 200 \\
\hline 2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5
= 20

Question (iii).
1764
Solution:
1764
\(\begin{array}{l|r}
2 & 1764 \\
\hline 2 & 882 \\
\hline 3 & 441 \\
\hline 3 & 147 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7
= 42

Question (iv).
4096
Solution:
4096
\(\begin{array}{l|r}
2 & 4096 \\
\hline 2 & 2048 \\
\hline 2 & 1024 \\
\hline 2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2² × 2² × 2² × 2² × 2² × 2²
∴ \(\sqrt{4096}\) = 2 × 2 × 2 × 2 × 2 × 2
= 64

Question (v).
7744
Solution:
7744
\(\begin{array}{r|r}
2 & 7744 \\
\hline 2 & 3872 \\
\hline 2 & 1936 \\
\hline 2 & 968 \\
\hline 2 & 484 \\
\hline 2 & 242 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11
= 2² × 2² × 2² × 11²
∴ \(\sqrt{7744}\) = 2 × 2 × 2 × 11
= 88

Question (vi).
9604
Solution:
9604
\(\begin{array}{l|r}
2 & 9604 \\
\hline 2 & 4802 \\
\hline 7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
9604 = 2 × 2 × 7 × 7 × 7 × 7
= 2² × 7² × 7²
∴ \(\sqrt{9604}\) =2 × 7 × 7
= 98

Question (vii).
5929
Solution:
5929
\(\begin{array}{r|r}
7 & 5929 \\
\hline 7 & 847 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
5929 = 7 × 7 × 11 × 11
= 7² × 11²
∴ \(\sqrt{5929}\) = 7 × 11
= 77

Question (viii).
9216
Solution:
9216
\(\begin{array}{r|r}
2 & 9216 \\
\hline 2 & 4608 \\
\hline 2 & 2304 \\
\hline 2 & 1152 \\
\hline 2 & 576 \\
\hline 2 & 288 \\
\hline 2 & 144 \\
\hline 2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{9216}\) = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question (ix).
529
Solution:
529
\(\begin{array}{l|r}
23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
529 = 23 × 23
= 23²
∴ \(\sqrt{529}\) = 23

Question (x).
8100
Solution:
8100
\(\begin{array}{l|r}
2 & 8100 \\
\hline 2 & 4050 \\
\hline 3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5
= 2² × 3² × 3² × 5²
∴ \(\sqrt{8100}\) = 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:

Question (i).
252
Solution:
252
\(\begin{array}{l|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
The prime factor 7 is unpaired.
∴ [252] × 7 = [2 × 2 × 3 × 3 × 7] × 7
1764 = 2 × 2 × 3 × 3 × 7 × 7
= 2² × 3² × 7²
∴ \(\sqrt{1764}\) = 2 × 3 × 7 = 42
Thus, 252 should be multiplied by smallest whole number 7 to get a perfect square.

Question (ii).
180
Solution:
180
\(\begin{array}{l|r}
2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
180 = 2 × 2 × 3 × 3 × 5
Here, the prime factor 5 is unpaired.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2 × 2 × 3 × 3 × 5 × 5
= 2² × 3² × 5²
∴ \(\sqrt{900}\) = 2 × 3 × 5 = 30
Thus, 180 should be multiplied by smallest whole number 5 to get a perfect square.

Question (iii).
1008
Solution:
1008
\(\begin{array}{l|r}
2 & 1008 \\
\hline 2 & 504 \\
\hline 2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired.
∴ [1008] × 7 = [2 × 2 × 2 × 2 × 3 × 3 × 7] × 7
∴ 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
= 2² × 2² × 3² × 7²
∴ \(\sqrt{7056}\) = 2 × 2 × 3 × 7 = 84
Thus, 1008 should be multiplied by smallest whole number 7 to get a perfect square.

Question (iv).
2028
Solution:
2028
\(\begin{array}{r|r}
2 & 2028 \\
\hline 2 & 1014 \\
\hline 3 & 507 \\
\hline 13 & 169 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2028 = 2 × 2 × 3 × 13 × 13
Here, the prime factor 3 is unpaired.
∴ [2028] × 3 = [2 × 2 × 3 × 13 × 13] × 3
∴ 6084 = 2 × 2 × 3 × 3 × 13 × 13
= 2² × 3² × 13²
∴ \(\sqrt{6084}\) = 2 × 3 × 13 = 78
Thus, 2028 should be multiplied by smallest whole number 3 to get a perfect square.

Question (v).
1458
Solution:
1458
\(\begin{array}{l|r}
2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, the prime factor 2 is unpaired.
∴ [1458] × 2 = [2 × 3 × 3 × 3 × 3 × 3 × 3] × 2
∴ 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 2² × 3² × 3² × 3²
∴ \(\sqrt{2916}\) = 2 × 3 × 3 × 3 = 54
Thus, 1458 should be multiplied by smallest whole number 2 to get a perfect square.

Question (vi).
768
Solution:
768
\(\begin{array}{l|r}
2 & 768 \\
\hline 2 & 384 \\
\hline 2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, the prime factor 3 is unpaired.
∴ [768] × 3 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3] × 3
∴ 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2² × 2² × 2² × 2² × 3²
∴ \(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3 = 48
Thus, 768 should be multiplied by smallest whole number 3 to get a perfect square.

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:

Question (i).
252
Solution:
252
\(\begin{array}{r|r}
2 & 252 \\
\hline 2 & 126 \\
\hline 3 & 63 \\
\hline 3 & 21 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
252 = 2 × 2 × 3 × 3 × 7
Here, the prime factor 7 is unpaired. So, given number should be divided by 7.
∴ [252] ÷ 7 = [2 × 2 × 3 × 3 × 7] ÷ 7
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 252 should be divided by smallest whole number 7 to get a perfect square number.

Question (ii).
2925
Solution:
2925
\(\begin{array}{r|r}
3 & 2925 \\
\hline 3 & 975 \\
\hline 5 & 325 \\
\hline 5 & 65 \\
\hline 13 & 13 \\
\hline & 1
\end{array}\)
2925 = 3 × 3 × 5 × 5 × 13
Here, the prime factor 13 is unpaired. So, given number should be divided by 13.
∴ [2925] ÷ 13 = [3 × 3 × 5 × 5 × 13] ÷ 13
∴ 225 = 3 × 3 × 5 × 5
= 3² × 5²
∴ \(\sqrt{225}\) = 3 × 5 = 15
Thus, 2925 should be divided by smallest whole number 13 to get a perfect square number.

Question (iii).
396
Solution:
396
\(\begin{array}{r|r}
2 & 396 \\
\hline 2 & 198 \\
\hline 3 & 99 \\
\hline 3 & 33 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
396 = 2 × 2 × 3 × 3 × 11
Here, the prime factor 11 is unpaired. So, given number should be divided by 11.
∴ [396] ÷ 11 = [2 × 2 × 3 × 3 × 11] ÷ 11
∴ 36 = 2 × 2 × 3 × 3
= 2² × 3²
∴ \(\sqrt{36}\) = 2 × 3 = 6
Thus, 396 should be divided by smallest whole number 11 to get a perfect square number.

Question (iv).
2645
Solution:
2645
\(\begin{array}{r|r}
5 & 2645 \\
\hline 23 & 529 \\
\hline 23 & 23 \\
\hline & 1
\end{array}\)
2645 = 5 × 23 × 23
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [2645] ÷ 5 = [5 × 23 × 23] ÷ 5
∴ 529 = 23 × 23 = 23²
∴ \(\sqrt{529}\) = 23
Thus, 2645 should be divided by smallest whole number 5 to get a perfect square number.

Question (v).
2800
Solution:
2800
\(\begin{array}{l|r}
2 & 2800 \\
\hline 2 & 1400 \\
\hline 2 & 700 \\
\hline 2 & 350 \\
\hline 5 & 175 \\
\hline 5 & 35 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
Here, the prime number 7 is unpaired. So, given number should be divided by 7.
∴ [2800] ÷ 7 = [2 × 2 × 2 × 2 × 5 × 5 × 7] ÷ 7
∴ 400 = 2 × 2 × 2 × 2 × 5 × 5
= 2² × 2² × 5²
∴ \(\sqrt{400}\) = 2 × 2 × 5 = 20
Thus, 2800 should be divided by smallest whole number 7 to get a perfect square number.

Question (vi).
1620
Solution:
1620
\(\begin{array}{l|r}
2 & 1620 \\
\hline 2 & 810 \\
\hline 3 & 405 \\
\hline 3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
Here, the prime factor 5 is unpaired. So, given number should be divided by 5.
∴ [1620] ÷ 5 = [2 × 2 × 3 × 3 × 3 × 3 × 5] ÷ 5
∴ 324 = 2 × 2 × 3 × 3 × 3 × 3
= 2² × 3² × 3²
∴ \(\sqrt{324}\) = 2 × 3 × 3 = 18
Thus, 1620 should be divided by 5 to get a perfect square number.

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:
Let the number of students be x.
Amount each student donated = Number of students in the class.
So, amount donated by each student = ₹ x
Total amount donated by class = ₹ x × x = x²
\(\begin{array}{l|r}
7 & 2401 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
∴ x² = 2401
∴ \(\sqrt{x^{2}}=\sqrt{2401}\)
∴ x = \(\sqrt{7 \times 7 \times 7 \times 7}\)
= \(\sqrt{7^{2} \times 7^{2}}\)
∴ x = 7 × 7 = 49
Hence, number of students in the class is 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:
Let the number of rows be x.
Number of rows = Number of plants in each row
So, number of plants in a row = x
∴ Number of plants to be planted in a garden = x × x = x²
\(\begin{array}{l|r}
3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
∴ x² = 2025
∴ \(\sqrt{x^{2}}=\sqrt{2025}\)
∴ x = \(\sqrt{3 \times 3 \times 3 \times 3 \times 5 \times 5}\)
∴ \(\sqrt{3^{2} \times 3^{2} \times 5^{2}}\)
∴ x = 3 × 3 × 5 = 45
Hence, the number of rows is 45 and the number of plants in each row is 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.
Solution :
[Note: LCM is the number, which is divided by all factors of it without leaving remainder. ]
Here, the smallest square number divisible by each one of 4, 9 and 10 is equal to some multiple of the LCM of 4, 9 and 10.
\(\begin{array}{l|ll}
2 & 4, & 9, & 10 \\
\hline 2 & 2, & 9, & 5 \\
\hline 3 & 1, & 9, & 5 \\
\hline 3 & 1, & 3, & 5 \\
\hline 5 & 1, & 1, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 4, 9 and 10 = 2 × 2 × 3 × 3 × 5 = 180
The prime factor 5 is unpaired.
So, 180 must be multiplied by 5.
∴ [180] × 5 = [2 × 2 × 3 × 3 × 5] × 5
∴ 900 = 2² × 3² × 5²
Hence, 900 is the required perfect square number.

10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.
Solution:
[Note: LCM is the number, which is divided by all factors of it without leaving remainder.]
Here, the smallest square number divisible by each of 8, 15 and 20 is equal to some multiple of the LCM of 8, 15 and 20.
\(\begin{array}{r|rrr}
2 & 8, & 15, & 20 \\
\hline 2 & 4, & 15, & 10 \\
\hline 2 & 2, & 15, & 5 \\
\hline 3 & 1, & 15, & 5 \\
\hline 5 & 1, & 5, & 5 \\
\hline & 1, & 1, & 1
\end{array}\)
LCM of 8, 15 and 20 = 2 × 2 × 2 × 3 × 5 = 120
The prime factors 2, 3 and 5 are unpaired.
So, 120 should be multiplied by 2 × 3 × 5 = 30.
∴ [120] × 2 × 3 × 5 = [2 × 2 × 2 × 3 × 5] × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5
∴ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
Hence, 3600 is the required perfect square number.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1

1. Identify the terms, their coefficients for each of the following expressions:

(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:

Terms	Coefficient of terms
(i) 5 xyz2 -3zy	5 – 3
(ii) 1 x x2 ‘	1 1 1
(iii) 4x2y2 – 4x2y2z2 z2	4 -4 1
(iv) 3 – pq qr – rp	3 – 1 1 – 1
(v) \(\frac {x}{2}\) \(\frac {y}{2}\) – xy	\(\frac {1}{2}\) \(\frac {1}{2}\) – 1
(vi) 0.3 a – 0.6ab 0.5b	0.3 – 0.6 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q.
Solution:

Monomials	Binomials	Trinomials
1000 pqr	x + y 2y – 3y2 4z – 15y2 p2q +pq2 2p + 2q	7 + y + 5x 2y – 3y2 + 4y3 5x – 4y + 3xy

Following polynomials do not fit in any catagories:
x + x² + x³ + x⁴ [∵ Polynomial has 4 terms] ab + bc + cd + da [∵ Polynomial has 4 terms]

3. Add the following:

Question (i)
ab – bc, bc – ca, ca – ab
Solution:
To add, let us arrange like terms one below the other.

Question (ii)
a – b + ab, b – c + bc, c – a + ac
Solution:

Question (iii)
2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
Solution:

Question (iv)
l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Solution:

4.

Question (a)
Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
Solution:

Question (b)
Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
Solution:

Question (c)
Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution: