Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3

1. Carry out the following divisions:

Question (i)
28x⁴ ÷ 56x
Solution:

Question (ii)
– 36y³ ÷ 9y²
Solution:

Question (iii)
66pq²r³ ÷ 11qr²
Solution:

Question (iv)
34x³y³z³ ÷ 51 xy²z³
Solution:

Question (v)
12a⁸b⁸ ÷ (- 6a⁶b⁴)
Solution:

2. Divide the given polynomial by the given monomial:

Question (i)
(5x² – 6x) ÷ 3x
Solution:

Question (ii)
(3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
Solution:

Question (iii)
8 (x³y²z² + x²y³z² ÷ 4 x²y²z²)
Solution:

Question (iv)
(x³ + 2x² + 3x) ÷ 2x
Solution:

Question (v)
(P³ q⁶ – p⁶ q³) ÷ p³ q³
Solution:

3. Work out the following divisions:

Question (i)
(10x – 25) ÷ 5
Solution:
= \(\frac{10 x-25}{5}\)
= \(\frac{5(2 x-5)}{5}\)
= 2x – 5

Question (ii)
(10x-25) ÷ (2x – 5)
Solution:
= \(\frac{10 x-25}{2 x-5}\)
= \(\frac{5(2 x-5)}{(2 x-5)}\)
= 5

Question (iii)
10y (6y + 21) ÷ 5 (2y + 7)
Solution:
= \(\frac{10 y(6 y+21)}{5(2 y+7)}\)
= \(\frac{2 \times 5 \times y \times 3 \times(2 y+7)}{5(2 y+7)}\)
= 2 × y × 3
= 6y

Question (iv)
9x²y²(3z – 24) ÷ 27xy (z – 8)
Solution:

Question (v)
96 abc (3a – 12) (5b – 30) ÷ 144 (a – 4)(b – 6)
Solution:
= \(\frac{96 a b c(3 a-12)(5 b-30)}{144(a-4)(b-6)}\)
= \(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 3 \times a \times b \times c \times 3 \times(a-4) \times 5 \times(b-6)}{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times(a-4) \times(b-6)}\)
= 2 × 5 × a × b × c
= 10 abc

4. Divide as directed:

Question (i)
5 (2x + 1) (3x + 5) ÷ (2x + 1)
Solution:
= \(\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\)
= \(\frac{5 \times(3 x+5)}{1}\)
= 5(3x + 5)

Question (ii)
26xy (x + 5) (y – 4) ÷ 13x (y – 4)
Solution:
= \(\frac{26 x y(x+5)(y-4)}{13 x(y-4)}\)
= \(\frac{2 \times 13 \times x \times y(x+5)(y-4)}{13 x(y-4)}\)
= 2y(x + 5)

Question (iii)
52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
Solution:
= \(\frac{52 p q r(p+q)(q+r)(r+p)}{104 p q(q+r)(r+p)}\)
= \(\frac{52 \times p \times q \times r \times(p+q)(q+r)(r+p)}{2 \times 52 \times p \times q \times(q+r)(r+p)}\)
= \(\frac{r \times(p+q)}{2}\)
= \(\frac {1}{2}\)r (p + q)

Question (iv)
20 (y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
Solution:
= \(\frac{20(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= \(\frac{2 \times 2 \times 5 \times(y+4)\left(y^{2}+5 y+3\right)}{5(y+4)}\)
= 2 × 2 × (y² + 5y + 3)
= 4(y² + 5y + 3)

Question (v)
x (x + 1) (x + 2) (x + 3) ÷ x (x + 1)
Solution:
= \(\frac{x(x+1)(x+2)(x+3)}{x(x+1)}\)
= \(\frac{(x+2)(x+3)}{1}\)
= (x + 2)(x + 3)

5. Factorise the expressions and divide them as directed:

Question (i)
(y² + 7y + 10) ÷ (y + 5)
Solution:
First we factorise
y² + 7y + 10
= y² + 5y + 2y + 10
= y (y + 5) + 2 (y + 5)
= (y + 5) (y + 2)
∴ (y² + 7y + 10) ÷ (y + 5)
= \(\frac{(y+5)(y+2)}{(y+5)}\)
= y + 2

Question (ii)
(m² – 14m – 32) ÷ (m + 2)
Solution:
First we factorise
m² – 14m – 32
= m² – 16m + 2m – 32
= m (m – 16) + 2 (m – 16)
= (m – 16) (m + 2)
∴ (m² – 14m – 32) ÷ (m + 2)
= \(\frac{(m-16)(m+2)}{m+2}\)
= m – 16

Question (iii)
(5p² – 25p + 20) ÷ (p – 1)
Solution:
First we factorise
5p² – 25p + 20
= 5 (p² – 5p + 4)
= 5 (p² – 4p – p + 4)
= 5 [p (p – 4) – 1 (p – 4)]
= 5 (p – 4) (p – 1)
∴ (5p² – 25p + 20) ÷ (p – 1)
= \(\frac{5(p-4)(p-1)}{p-1}\)
= 5 (p – 4)

Question (iv)
4yz (z² + 6z – 16) ÷ 2y (z + 8)
Solution:
= \(\frac{4 y z\left(z^{2}+6 z-16\right)}{2 y(z+8)}\)
= \(\frac{2 z\left(z^{2}+6 z-16\right)}{z+8}\)
Now, factorise
z² + 6z – 16
= z² + 8z – 2z – 16
= z (z + 8) – 2 (z + 8)
= (z + 8) (z – 2)
∴ 4yz (z² + 6z – 16) ÷ 2y (z + 8)
= \(\frac{2 z(z+8)(z-2)}{z+8}\)
= 2z (z – 2)

Question (v)
5pq (p² – q²) ÷ 2p(p + q)
Solution:

Question (vi)
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
Solution:
= \(\frac{12 x y\left(9 x^{2}-16 y^{2}\right)}{4 x y(3 x+4 y)}\)
= \(\frac{12 x y(3 x-4 y)(3 x+4 y)}{4 x y(3 x+4 y)}\)
= 3 (3x – 4y)

Question (vii)
39y³ (50y² – 98) ÷ 26y² (5y + 7)
Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2

1. Factorise the following expressions:

Question (i)
a² + 8a + 16
Solution:
= (a)² + 2 (a)(4) + (4)²
= (a + 4)²

Question (ii)
p² – 10p + 25
Solution:
= (p)² – 2 (p)(5) + (5)²
= (P – 5)²

Question (iii)
25m² + 30m + 9
Solution:
= (5m)² + 2 (5m) (3) + (3)²
= (5m + 3)²

Question (iv)
49y² + 84yz + 36z²
Solution:
= (7y)² + 2 (7y)(6z) + (6z)²
= (7y + 6z)²

Question (v)
4x² – 8x + 4
Solution:
= 4(x² – 2x + 1)
= 4 [(x)² – 2 (x)(1) + (1)²]
= 4 (x – 1)²

Question (vi)
121b² – 88bc + 16c²
Solution:
= (11b)² – 2 (11b)(4c) + (4c)²
= (11b – 4c)²

Question (vii)
(l + m)² – 4lm [Hint: Expand (1 + m)² first]
Solution:
= l² + 2lm + m² – 4lm
= l² + 2lm – 4lm + m²
= l² – 2lm + m²
= (l)² – 2 (l) (m) + (m)²
= (l – m)²

Question (viii)
a⁴ + 2a²b² + b⁴
Solution:
= (a²)² + 2 (a²)(b²) + (b²)²
= (a² + b²)²

2. Factorise:

Question (i)
4p² – 9q²
Solution:
= (2p)² – (3q)²
= (2p – 3q) (2p + 3q)

Question (ii)
63a² – 112b²
Solution:
= 7 (9a² – 16b²)
= 7 [(3a)² -(4b)²]
= 7 (3a – 4b) (3a + 4b)

Question (iii)
49x² – 36
Solution:
= (7x)² – (6)²
= (7x – 6) (7x + 6)

Question (iv)
16x⁵ – 144x³
Solution:
= 16x³(x² – 9)
= 16x³ [(x)² – (3)²]
= 16x³ (x-3) (x + 3)

Question (v)
(l + m)² – (l – m)²
Solution:
=[(l + m) + (l – m)] [(l + m) – (l – m)]
= [l + m + l – m] [l + m – l + m]
= (2l) (2m)
= 4lm

Question (vi)
9x²y² – 16
Solution:
= (3xy)² – (4)²
= (3xy – 4) (3xy + 4)

Question (vii)
(x² – 2xy + y²) – z²
Solution:
= (x – y)² – (z)²
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)

Question (viii)
25a² – 4b² + 28bc – 49c²
Solution:
= (25a²) – (4b² – 28bc + 49c²)
= (5a)² – (2b – 7c)²
= [(5a) – (2b – 7c)] [(5a) + (2b – 7c)]
= (5a – 2b + 7c) (5a + 2b – 7c)

3. Factorise the expressions:

Question (i)
ax² + bx
Solution:
= x (ax + b)

Question (ii)
7p² + 21q²
Solution:
= 7 (p² + 3q²)

Question (iii)
2x³ + 2xy² + 2xz²
Solution:
= 2x(x² + y² + z²)

Question (iv)
am² + bm² + bn² + an²
Solution:
= am² + bm² + an² + bn²
= m² (a + b) + n²(a + b)
= (a + b) (m² + n²)

Question (v)
(lm + l) + m + 1
Solution:
= l (m + 1) + 1 (m + 1)
= (m + 1) (l + 1)

Question (vi)
y(y + z) + 9(y + z)
Solution:
= (y + z)(y + 9)

Question (vii)
5y² – 20y – 8z + 2yz
Solution:
= 5y² – 20y + 2yz – 8z
= 5y (y – 4) + 2z (y – 4)
= (y- 4) (5y + 2z)

Question (viii)
10ab + 4a + 5b + 2
Solution:
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

Question (ix)
6xy – 4y + 6 – 9x
Solution:
= 6xy – 4y – 9x + 6
= 2y (3x-2)-3(3x-2)
= (3x-2) (2y – 3)

4. Factorise:

Question (i)
a⁴ – b⁴
Solution:
= (a²)² – (b²)²
= (a² – b²) (a² + b²)
= ((a)² – (b²)] (a² + b²)
= (a – b) (a + b) (a² + b²)

Question (ii)
p⁴ – 81
Solution:
= (p²)² – (9)²
= (p² – 9) (p² + 9)
= ((p)² – (3)²] (p² + 9)
= (p – 3)(p + 3)(p² + 9)

Question (iii)
x⁴ – (y + z)⁴
Solution:
= (x²)² – (a²)² (∵ y + z = a)
= (x² – a²) (x² + a²)
= (x – a) (x + a) (x² + a²)
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²] (∵ a = y + z)
= (x – y – z) (x + y + z) [x² + (y + z)²]

Question (iv)
x⁴ – (x – z)⁴
Solution:
= (x²)² – [(x – z)²]²
= [x² – (x – z)²] [x² + (x – z)²]
= [x² – (x² – 2xz + z²)] [x² + (x² – 2xz + z²)]
= (x² – x² + 2xz – z²) (x² + x² – 2xz + z²)
= (2xz – z²) (2x² – 2xz + z²)
= z (2x – z) (2x² – 2xz + z²)

Question (v)
a⁴ – 2a²b² + b⁴
Solution:
= (a²)² – 2(a²)(b²) + (b²)²
= (a² – b²)²
= (a² – b²) (a² – b²)
= (a – b) (a + b) (a – b) (a + b)

5. Factorise the following expressions:

Question (i)
p² + 6p + 8
Solution:
= p² + 6p + 9 – 1
= (p² + 6p + 9) – (1)
= (p + 3)² – (1)²
= (p + 3 + 1) (p + 3 – 1)
= (P + 4) (p + 2)
Here, last term is 8.
∴ 9 – 1 = 8.

OR
p² + 6p + 8
Here, ab = 8 and a + b = 6
On solving equations, a = 4, b = 2
Now, p² + 6p + 8
= p² + 4p + 2p + 8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question (ii)
q² – 10q + 21
Solution:
= q² – 10q + 25 – 4
= (q² – 10q + 25) – (4)
= (q – 5)² – (2)²
= (q – 5 + 2) (q – 5 – 2)
= (q – 3) (q – 7)
Here, last term is 21.
∴ 25 – 4 = 21.

OR
q² – 10q + 21
Here, ab = 21 and a + b = (- 10)
Possible values of a = 7 or (-7)
b = 3 or (- 3)
Let us check, 7 + 3 = 10 ≠ (- 10)
∴ a = – 7, b = – 3
Now, q² – 10q + 21
= q² – 7q – 3q + 21
= q (q – 7) – 3 (q – 7)
= (q – 7) (q – 3)

Question (iii)
p² + 6p – 16
Solution:
= p² + 6p + 9 – 25
= (P² + 6p + 9) – (25)
= (p + 3)² – (5)²
= (p + 3 – 5) (p + 3 + 5)
= (p – 2) (p + 8)
Here, last term is (-16).
∴ (-25) + 9 = (-16)

OR

p² + 6p – 16
Here, ab = – 16 and a + b = 6
Possible values of a = 8 or (-8) b = 2 or (-2)
Let us check, 8 + 2 = 10 ≠ 6
(- 8) + 2 = (-6) ≠ 6
8 + (-2) = 8 – 2 = 6
Now, p² + 6p – 16
= p² + 8p – 2p – 16
= P (P + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1

1. Find the common factors of the given terms:

Question (i)
12x, 36
Solution:
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Common factors of 12x and 36 = 2 × 2 × 3
= 12

Question (ii)
2y, 22xy
Solution:
2y = 2 xy
22xy = 2 × 11 × x × y
∴ Common factors of 2y and 22xy
= 2 × y = 2y

Question (iii)
14pq, 28p²q²
Solution:
14pq = 2 × 7 × p × q
28p²q² = 2 × 2 × 7 × p × p × q × q
∴ Common factors of 14pq and 28p²q²
= 2 × 7 × p × q = 14pq

Question (iv)
2x, 3x², 4
Solution:
2x = 2 × x
3x² = 3 × x × x
4 = 2 × 2
∴ Common factors of 2x, 3x² and 4 = 1 [Note: 1 is a factor of each term.]

Question (v)
6abc, 24ab², 12a²b
Solution:
6abc = 2 × 3 × a × b × c
24ab² = 2 × 2 × 2 × 3 × a × b × b
12a²b = 2 × 2 × 3 × a × a × b
∴ Common factors of 6abc, 24ab² and 12a²b
= 2 × 3 × a × b
= 6ab

Question (vi)
16x³, – 4x², 32x
Solution:
16x³ = 2 × 2 × 2 × 2 × x × x × x
– 4x² = (-1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ Common factors of 16x³, – 4x² and 32x = 2 × 2 × x = 4x

Question (vii)
10pq, 20qr, 30rp
Solution:
10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ Common factors of 10pq, 20qr and 30rp = 2 × 5 = 10

Question (viii)
3x²y³, 10x³y², 6x²y²z
Solution:
3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ Common factors of 3x²y³, 10x³y² and 6x²y²z
= x × x × y × y = x²y²

2. Factorise the following expressions:

Question (i)
7x – 42
Solution:
7x = 7 × x
42 = 2 × 3 × 7
∴ 7 is common in both terms.
7x – 42 = 7 (x – 6)

Question (ii)
6p – 12q
Solution:
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
∴ 2 × 3 = 6 is common in both terms.
6p – 12q = 6 (p – 2q)

Question (iii)
7a² + 14a
Solution:
7a² = 7 × a × a
14a = 2 × 7 × a
∴ 7 × a = 7a is common in both terms.
∴ 7a² + 14a = 7a (a + 2)

Question (iv)
– 16z + 20z³
Solution:
– 16z = -2 × 2 × 2 × 2 × z
20z³ = 2 × 2 × 5 × z × z × z
∴ 2 × 2 × z = 4z is common in both terms.
∴ – 16z + 20z³ = 4z (- 4 + 5z²)

Question (v)
20l²m + 30alm
Solution:
= 2 × 2 × 5 × l × l × m + 2 × 3 × 5 × a × l × m
= 10lm (2l + 3a)

Question (vi)
5x²y – 15xy²
Solution:
= 5xy (x – 3y)

Question (vii)
10a² – 15b² + 20c²
Solution:
= 5 (2a² – 3b² + 4c²)

Question (viii)
– 4a² + 4ab – 4ca
Solution:
= 4a (- a + b – c)

Question (ix)
x²yz + xy²z + xyz²
Solution:
= xyz (x + y + z)

Question (x)
ax²y + bxy² + cxyz
Solution:
= xy (ax + by + cz)

3. Factorise:

Question (i)
x² + xy + 8x + 8y
Solution:
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

Question (ii)
15xy – 6x + 5y – 2
Solution:
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

Question (iii)
ax + bx- ay – by
Solution:
= x (a + b) – y (a + b)
= (a+ b) (x- y)

Question (iv)
15pq + 15 + 9q + 25p
Solution:
= 15pq + 9q + 25p + 15
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

Question (v)
z – 7 + 7xy – xyz
Solution:
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7) (1 – xy)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 15 Introduction to Graphs InText Questions

Try These : [Textbook Page No. 244]

1. In the above example, use the graph to find how much petrol can be purchased for ₹ 800.

Solution:
We can find the quantity of petrol to be got for ₹ 800. For this take a point on the Y-axis (0, 800). Now, draw a line parallel to X-axis to meet the graph at the point B. Now, from the point B, draw a line parallel to Y-axis, which intersect X-axis in the point C. Coordinate of the point C : (16, 0).
Hence, 16 litres of petrol can be purchased for ₹ 800.

Think, Discuss and Write : [Textbook Page No. 243]

1. The number of litres of petrol you buy to fill a car’s petrol tank will decide the amount you have to pay. Which is the independent variable here? Think about it.
Solution:
Here, we clearly understand that graph of quantity of petrol (litre) and amount to pay (₹) should be a line.
Both quantities are in direct proportion. If we fill more litres of petrol, we have to pay more amount and vice versa.
∴ Petrol is the independent variable.

Try These : [Textbook Page No. 245]

1. Is Example 7, a case of direct variation?
Solution:
Yes, Example 7 given on page 245 (Textbook), is a case of direct variation. As the principal increases, the simple interest on it also increases proportionately.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3

1. Draw the graphs for the following tables of values, with suitable scales on the axes.

Question (a)
Cost of apples

Number of apples	1	2	3	4	5
Cost (in ₹)	5	10	15	20	25

Solution:

1. Draw 2 lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale :
On X-axis : 1 cm = 1 apple
On Y-axis 1 cm = ₹ 5
3. Plot the points (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25) on graph paper.
4. Join these points and extend line.

(b) Distance travelled by a car

Time (in hours)	6 a.m.	7 a.m.	8 a.m.	9 a.m.
Distance (in km)	40	80	120	160

Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = 1 unit (Time in hours.)
On Y-axis : 1 cm = 10 km
3. Plot the points (6, 40), (7, 80), (8, 120) and (9, 160) on graph paper.
4. Join these points and extend line.

Question (i)
How much distance did the car cover during the period 7:30 a.m. to 8:00 a.m.?
Solution:
In the graph, draw a perpendicular at the point indicating 7:30 a.m. on the X-axis such that it meets the graph at P.
From P draw a line parallel to X-axis to meet Y-axis at 100 km.
∴ Distance travelled between 7:30 am and 8:00 am.
= (120 – 100) km
= 20 km

Question (ii)
What was the time when the car had covered a distance of 100 km since it’s start?
Solution:
When the car had covered a distance of 100 km, the time was 7 : 30 am.

(c) Interest on deposits for a year.

Deposit (in ₹)	1000	2000	3000	4000	5000
Simple Interest (in ₹)	80	160	240	320	400

Solution :
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = ₹ 1000 (deposit)
On Y-axis : 1 cm = ₹ 40 (simple interest)
3. Plot the points (1000, 80), (2000, 160), (3000, 240), (4000, 320) and (5000, 400) on graph paper.
4. Join these points and extend line.

Question (i)
Does the graph pass through the origin?
Solution:
Yes, it passes through the origin.

Question (ii)
Use the graph to find the interest on ₹ 2500 for a year.
Solution:
From the graph, the interest on ₹ 2500 for a year is ₹ 200.

Question (iii)
To get an interest of ₹ 280 per year, how much money should be deposited?
Solution:
From the graph an interest of ₹ 280 can be got by depositing ₹ 3500.

2. Draw a graph for the following:

Question (i)

Side of square (in cm)	2	3	3.5	5	6
Perimeter (in cm)	8	12	14	20	24

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm = 4 cm (Perimeter of a square) )
3. Plot the points (2, 8), (3, 12), (3.5, 14), (5, 20) and (6, 24) on graph paper.
4. Join these points and extend line.

Yes, it is a linear graph.

Question (ii)

Side of square (in cm)	2	3	4	5	6
Area (in cm2)	4	9	16	25	36

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm 5 cm (Area of a square)
3. Plot the points (2, 4), (3, 9), (4, 16), (5, 25) and (6, 36) on graph paper.
4. Join these points.

No, this graph is not a straight line. So it is not a linear graph.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral LIFT.
LI = 4 cm,
IF = 3 cm,
TL = 2.5 cm,
LF = 4.5 cm,
IT = 4 cm.
Solution:

Steps of construction:

• Draw a line segment LI = 4 cm.
• With L as centre and radius = 2.5 cm, draw an arc.
• With I as centre and radius = 4 cm, draw an arc to intersect the previous arc at T.
• With L as centre and radius = 4.5 cm draw an arc.
• With I as centre and radius 3 cm, draw an arc to intersect the previous, arc at F.
• Draw \(\overline{\mathrm{LT}}, \overline{\mathrm{IF}}, \overline{\mathrm{FT}}, \overline{\mathrm{LF}}\) and \(\overline{\mathrm{IT}}\).

Thus, LIFT is the required quadrilateral.

Question (ii).
Quadrilateral GOLD
OL = 7.5 cm,
GL = 6 cm,
GD = 6 cm,
LD = 5 cm,
OD = 10 cm.
Solution:

Steps of construction:

• Draw a line segment LD = 5 cm.
• With L as centre and radius = 7.5 cm, draw an arc.
• With D as centre and radius = 10 cm, draw another arc to intersect the previous arc at O.
• With L as centre and radius = 6 cm, draw an arc.
• With D as centre and radius = 6 cm, draw another arc to intersect previous arc at G.
• Draw \(\overline{\mathrm{LO}}, \overline{\mathrm{GO}}, \overline{\mathrm{DG}}, \overline{\mathrm{LG}}\) and \(\overline{\mathrm{DO}}\).

Thus, GOLD is the required quadrilateral.

Question (iii).
Rhombus BEND
BN = 5.6 cm,
DE = 6.5 cm.
Solution:

[Note: Diagonals of a rhombus are perpendicular bisectors of each another. Here, diagonals of □ BEND \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{BN}}\) intersect at A. So AN = 2.8 cm and AB = 2.8 cm (BN = 5.6 cm, \(\frac {1}{2}\) BN = AN and AB)]
Steps of construction:

• Draw a line segment DE = 6.5 cm.
• Draw perpendicular bisector \(\overleftrightarrow{X Y}\) of \(\overline{\mathrm{DE}}\), which intersects \(\overline{\mathrm{DE}}\) at A.
• With centre at A and radius = 5.6 × \(\frac {1}{2}\) = 2.8 cm, draw two arcs intersecting \(\overleftrightarrow{X Y}\) in points B and N.
• Draw \(\overline{\mathrm{DN}}, \overline{\mathrm{EN}}, \overline{\mathrm{EB}}\) and \(\overline{\mathrm{DB}}\).

Thus, BEND is the required quadrilateral.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.2

1. Plot the following points on a graph sheet. Verify if they lie on a line:

Question (a)
A(4, 0), B (4, 2), C(4, 6), D(4, 2.5)
Solution:

Plotting the given points and then l joining them we find that they all S lie on the same line.

Question (b)
P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
Solution:

Plotting the given points and then joining them we find that they all lie on the same line.

Question (c)
K(2, 3), L(5, 3), M(5, 5), N(2, 5)
Solution:

Plotting the given points and then joining them we find that all of them do not lie on the same line.

2. Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Solution:

Plot the given points and join them to make a line. When you extend this line, it meets the X-axis at C (5, 0) and the Y-axis at D (0, 5).

3. Write the coordinates of the vertices of each of these adjoining figures:

Solution:
Figure:
(i) The coordinates of the vertices of quadrilateral OABC:
O are (0, 0)
A are (2, 0)
B are (2, 3)
C are (0, 3)

(ii) The coordinates of the vertices of quadrilateral PQRS:
P are (4, 3)
Q are (6, 1)
R are (6, 5)
S are (4, 7)

(iii) The coordinates of the vertices of triangle KLM:
K are (10, 5)
L are (7, 7)
M are (10, 8)

4. State whether True or False. Correct that are false:

Question (i)
A point whose x-coordinate is zero and y- coordinate is non-zero will lie on y-axis.
Solution:
True

Question (ii)
A point whose y-coordinate is zero and x-coordinate is 5 will lie on y- axis.
Solution:
False
Correct statement: A point whose y-coordinate is 0 and x-coordinate is 5 will lie on X-axis at a distance 5 units from origin.

Question (iii)
The coordinates of the origin are (0, 0).
Solution:
True

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral ABCD.
AB = 4.5 cm,
BC = 5.5 cm,
CD = 4 cm,
AD = 6 cm,
AC = 7 cm.
Solution :

Steps of construction:

• Draw a line segment AB = 4.5 cm.
• With A as centre and radius = 7 cm, draw an arc.
• With B as centre and radius = 5.5 cm, draw another arc to intersect previous arc at C.
• With A as centre and radius 6 cm draw an arc.
• With centre at C and radius 4 cm, draw another arc to intersect previous arc at D.
• Draw \(\overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{AD}}\) and \(\overline{\mathrm{AC}}\).

Thus, ABCD is the required quadrilateral.

Question (ii).
Quadrilateral JUMP
JU = 3.5 cm,
UM = 4 cm,
MP = 5 cm,
PJ = 4.5 cm,
PU = 6.5 cm.
Solution:

Steps of construction :

• Draw a line segment JU = 3.5 cm.
• With J as centre and radius = 4.5 cm, draw an arc.
• With U as centre and radius = 6.5 cm, draw another arc to intersect previous arc at P
• With U as centre and radius = 4 cm draw an arc.
• With P as centre and radius 5 cm, draw an arc which intersects previous arc at M.
• Draw \(\overline{\mathrm{JP}}, \overline{\mathrm{UM}}, \overline{\mathrm{MP}}\) and \(\overline{\mathrm{UP}}\).

Thus, JUMP is the required quadrilateral.

Question (iii).
Parallelogram MORE ?
OR = 6 cm,
RE = 4.5 cm,
EO = 7.5 cm.
Solution:

[Note: □ MORE is a parallelogram. So length of opposite sides are equal. ]
∴ RE = MO = 4.5 cm; OR = ME = 6 cm
Steps of construction:

• Draw a line segment MO = 4.5 cm.
• With M as centre and radius = 6 cm, draw an arc.
• With O as centre and radius = 7.5 cm, draw another arc to intersect the previous arc at E.
• With O as centre and radius = 6 cm, draw an arc.
• With E as centre and radius = 4.5 cm, draw another arc to intersect the previous arc at R.
• Draw \(\overline{\mathrm{ME}}, \overline{\mathrm{OR}}, \overline{\mathrm{RE}}\) and \(\overline{\mathrm{OE}}\).

Thus, MORE is the required parallelogram.

Question (iv).
Rhombus BEST
BE = 4.5 cm,
ET = 6 cm.
Solution:

[Note : BEST is a rhombus. So length of all four sides are equal.]
BE = 4.5 cm
∴ ES = ST = TB = 4.5 cm
ET = 6 cm (Given)
Steps of construction:

• Draw a line segment BE = 4.5 cm.
• With B as centre and radius = 4.5 cm, draw an arc.
• With E as centre and radius = 6 cm, draw another arc to intersect the previous arc at T.
• With E as centre and radius = 4.5 cm, draw an arc.
• With T as centre and radius = 4.5 cm, draw another arc to intersect previous arc at S.
• Draw \(\overline{\mathrm{BT}}, \overline{\mathrm{ES}}, \overline{\mathrm{ST}}\) and \(\overline{\mathrm{ET}}\).

Thus, BEST is the required quadrilateral.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1

1. The following graph shows the temperature of a patient in a hospital, recorded every hour:

Question (a)
What was the patient’s temperature at 1 p.m.?
Solution:
The patient’s temperature at 1 p.m. was 36.5 °C.

Question (b)
When was the patient’s temperature 38.5 °C?
Solution:
The patient’s temperature was 38.5 °C at 12 noon.

Question (c)
The patient’s temperature was the same two times during the period given. What were these two times?
Solution:
The patient’s temperature was same (36.5 °C) at 1 p.m. and 2 p.m.

Question (d)
What was the temperature at 1:30 p.m.? How did you arrive at your answer?
Solution:
The patient’s temperature at 1:30 p.m. was 36.5 °C.
(The temperature did not change during interval of 1 p.m. and 2 p.m. So the temperature did not show any change and it was 36.5 °C at 1:30 p.m.)

Question (e)
During which periods did the patients’ temperature showed an upward trend?
Solution:
The patient’s temperature showed an upward trend during the periods 9 a.m. to 10 a.m., 10 a.m. to 11a.m. and 2 p.m. to 3 p.m., because the temperature increased during these intervals.

2. The following line graph shows the yearly sales figures for a manufacturing company:

Question (a)
What were the sales in (i) 2002 (ii) 2006?
Solution:
1. The sales in the year 2002 was ₹ 4 crores.
2. The sales in the year 2006 was ₹ 8 crores.

Question (b)
What were the sales in (i) 2003 (ii) 2005?
Solution:
1. The sales in the year 2003 was ₹ 7 crores.
2. The sales in the year 2005 was ₹ 10 crores.

Question (c)
Compute the difference between the sales in 2002 and 2006.
Solution:
The difference between the sales in 2002 and 2006 = ₹ (8 – 4) crore
= ₹ 4 crores

Question (d)
In which year was there the greatest difference between the sales as compared to its previous year?
Solution:
In year 2005, there was the greatest difference between the sales as compared to its previous year.

3. For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph:

Question (a)
How high was Plant A after
1. 2 weeks
2. 3 weeks?
Solution:
1. The plant A was 7 cm high after 2 weeks.
2. The plant A was 9 cm high after 3 weeks.

Question (b)
How high was Plant B after
1. 2 weeks
2. 3 weeks?
Solution:
1. The plant B was 7 cm high after 2 weeks.
2. The plant B was 10 cm high after 3 weeks.

Question (c)
How much did Plant A grow during the 3rd week?
Solution:
Plant A grew (9 cm – 7 cm) = 2 cm during 3rd week.

Question (d)
How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
Solution:
The plant B grew (10cm-7cm) = 3 cm from the end of 2nd week to the end of 3rd week.

Question (e)
During which week did Plant A grow most?
Solution:
The growth of the plant A During the 1st week = 2 cm – 0 cm = 2 cm
During the 2nd week = 7 cm – 2 cm = 5 cm
During the 3rd week = 9 cm – 7 cm = 2 cm
Thus, during the 2nd week, the plant A grew the most.

Question (f)
During which week did Plant B grow least?
Solution:
The growth of the plant B.
During the 1st week = 1cm – 0 cm
= 1 cm
During the 2nd week = 7 cm – 1 cm
= 6 cm
During the 3rd week = 10 cm-7 cm
= 3 cm
Thus, the plant B grew the least in the first week.

Question (g)
Were the two plants of the same height during any week shown here? Specify.
Solution:
At the end of 2nd week, both the plants were of the same height, that is 7 cm.

4. The following graph shows the temperature forecast and the actual temperature for each day of a week.

Question (a)
On which days was the forecast temperature the same as the actual temperature?
Solution:
The forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday.

Question (b)
What was the maximum forecast temperature during the week?
Solution:
The maximum forecast temperature during the week was 35 °C.

Question (c)
What was the minimum actual temperature during the week?
Solution:
The minimum actual temperature during the week was 15 °C.

Question (d)
On which day did the actual temperature differ the most from the forecast temperature?
Solution:
On Thursday, the actual temperature differed the most from the forecast temperature (7.5 °C).

Difference of temperature:

• Monday : 17.5 °C – 15 °C = 2.5 °C
• Tuesday : 20 °C – 20 °C = o°c
• Wednesday : 30 °C – 25 °C = 5°C
• Thursday : 22.5 °C – 15 °C = 7.5 °C
• Friday : 15 °C – 15 °C = o°c
• Saturday : 30 °C – 25 °C = 5°C
• Sunday : 35 °C – 35 °C = o°c

5. Use the tables below to draw linear graphs:

Question (a)
The number of days a hillside city received snow in different years:

Year	2003	2004	2005	2006
Days	8	10	5	12

Solution:
Linear graph to show snowfall in different years:

Question (b)
Population (in thousands) of men and women in a village in different years:

Year	2003	2004	2005	2006	2007
Number of Men	12	12.5	13	13.2	13.5
Number of Women	11.3	11.9	13	13.6	12.8

Solution:
Draw two perpendicular lines on the graph paper. Take year along X-axis (horizontal line) and population (in thousand) along Y-axis (vertical line).
For men: Mark the points (2003, 12), (2004, 12.5); (2005, 13); (2006, 13.2) and (2007, 13.5) and join them.
For women: Mark the points (2003, 11.3); (2004, 11.9); (2005, 13); (2006, 13.6) and (2007, 12.8) and join them.

6. A courier cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph:

Question (a)
What is the scale taken for the time axis?
Solution:
The time is taken along the X-axis. The scale along X-axis is 4 units = 1 hour.

Question (b)
How much time did the person take for the travel?
Solution:
Total travel time taken by a courier : = 8:00 am to 11:30 am = 3\(\frac {1}{2}\) hours

Question (c)
How far is the place of the merchant from the town?
Solution:
Distance of the merchant from the town is 22 km.

Question (d)
Did the person stop on his way? Explain.
Solution:
Yes, the stopage time = 10:00 am to 10:30 am. This is indicated by the horizontal part of the graph.

Question (e)
During which period did he ride fastest?
Solution:
He rode fastest between 8:00 am and 9:00 am.

7. Can there be a time-temperature graph as follows? Justify your answer.

Solution:
In case of (iii), the graph shows different number of temperatures at the same time which is not possible.
∴ Case (iii) does not represent a time-temperature graph.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals InText Questions

Try These (Textbook Page No. 43)

Take a regular hexagon Fig 3.10.

Question 1.
What is the sum of the measures of its exterior angles x, y, z, p, q r?
Solution:
∠x + ∠y + ∠z + ∠p + ∠q + ∠r = 360°
(∵ Sum of exterior angles of a polygon = 360°)

Question 2.
Is x = y = z = p = q = r ? Why?
Solution:
Since, all the sides of the polygon are equal, it is a regular hexagon. So its interior angles are equal.
∴ x = (180° – a), y = (180° – a),
z = (180° – a), p = (180° – a),
q = (180° – a), r = (180° – a)
∴ x = y = z = p = q = r

Question 3.
What is the measure of each ?
(i) exterior angle
(ii) interior angle
Solution:
(i) x + y + z + p + q + r = 360°
(∵ Sum of exterior angles = 360°)
All angles are equal.
∴ Measure of each exterior angle = \(\frac{360^{\circ}}{6}\) = 60°

(ii) Exterior angle = 60°
∴ Interior angle = 180° – 60° = 120°.

Question 4.
Repeat this activity for the cases of:
(i) a regular octagon
(ii) a regular 20-gon
Solution:
(i) In a regular octagon, number of sides (n) = 8.
∴ Each exterior angle = \(\frac{360^{\circ}}{8}\) = 45°
∴ Each interior angle = 180° – 45° = 135°

(ii) For a regular 20-gon, the number of sides (n) = 20.
∴ Each exterior angle = \(\frac{360^{\circ}}{20}\) = 18°
∴ Each interior angle = 180° – 18° = 162°

Try These (Textbook Page No. 47)

Question 1.
Take two identical set squares with angles 30°-60°-90° and place them adjacently to form a parallelogram as shown in figure. Does this help you to verify the above property ?

Solution:
Yes, the given figure helps us to verify that opposite sides of a parallelogram are equal.

Try These (Textbook Page No. 48)

Question 1.

Solution:
Yes, this figure also helps us to confirm that opposite angles of a parallelogram are equal.

Think, Discuss and Write (Textbook Page No. 50)

Question 1.
After showing m∠R = m∠N = 70°, can you find m∠I and m∠G by any other method ?
Solution:

Yes, without using the property of parallelogram, we can find m∠I and m∠G.
m∠R = m∠N = 70° (Given)
RG || IN, the transversal RI intersecting them,
∴ m∠R + m∠I = 180° (Sum of interior angles is 180°)
∴ 70° + m∠I = 180° (∵ m∠R = m∠N = 70°)
∴ m∠I = 180° – 70°
∴ m∠I = 110°
Similarly, m∠G = 110°

Think, Discuss and Write (Textbook Page No. 56)

Question 1.
A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?
Solution :
He can make sure that it is rectangular using the following different ways :

• By making opposite sides of equal length.
• By keeping each angle at the corners as 90°.
• By keeping the diagonals of equal length.
• By making opposite sides parallel.

Question 2.
A square was defined as a rectangle with all sides equal. Can we define it as rhombus with equal angles? Explore this idea.
Solution:
Yes, because a rhombus becomes a square if its all angles are equal.

Question 3.
Can a trapezium have all angles equal ?
Can it have all sides equal ? Explain.
Solution:
Yes, a trapezium can have all angles equal. In this case, it becomes a square or rectangle.
Yes, it can have all sides equal. In this case, it becomes a rhombus or square.