PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4 1
Steps of construction:

  • Draw a line segment DE = 4 cm.
  • At E, draw \(\overrightarrow{\mathrm{EM}}\) such that ∠DEM = 60°.
  • With E as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EM}}\) at A.
  • At A, draw \(\overrightarrow{\mathrm{AN}}\) such that ∠EAN = 90°.
  • With A as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{AN}}\) at R.
  • Draw \(\overline{\mathrm{DR}}\).

Thus, DEAR is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Question (ii).
Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.4 2
Steps of construction :

  • Draw a line segment TR = 3.5 cm.
  • At R, draw a \(\overrightarrow{\mathrm{RM}}\) such that ∠TRM = 75°.
  • With R as centre and radius = 3 cm, draw an arc intersecting \(\overrightarrow{\mathrm{RM}}\) at U.
  • At U, draw \(\overrightarrow{\mathrm{UN}}\) such that ∠RUN = 120°.
  • With U as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{UN}}\) at E.
  • Draw \(\overline{\mathrm{ET}}\).

Thus, TRUE is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 1

  • Draw a line segment MO = 6 cm.
  • At M, draw \(\overrightarrow{\mathrm{MA}}\), such that ∠OMA = 60°
  • At O, draw \(\overrightarrow{\mathrm{OB}}\) such that ∠MOB = 105°.
  • With O as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OB}}\) at R.
  • At R, draw \(\overrightarrow{\mathrm{RC}}\) such that ∠ORC = 105°.
  • Locate E at intersection of \(\overrightarrow{\mathrm{RC}}\) and \(\overrightarrow{\mathrm{MA}}\).

Thus, MORE is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (ii).
Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 2
[Note : In □ PLAN, m∠P = 90°, m∠A = 110° and m∠N = 85°)
∴ m∠L = 360°- (m∠P + m∠A + m∠N)
= 360° – (90° + 110° + 85°)
= 360° – 285°
= 75°

Steps of construction:

  • Draw a line segment AL = 6.5 cm.
  • At A, draw \(\overrightarrow{\mathrm{AX}}\) such that ∠XAL = 110°. (Use protractor)
  • At L, draw \(\overrightarrow{\mathrm{LY}}\) such that ∠YLA = 75°. (Use protractor)
  • With L as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{LY}}\) at P.
  • At P, draw \(\overrightarrow{\mathrm{PZ}}\) such that ∠ZPL = 90°. (∵ ∠ ZPY = 90°)
  • Locate N at intersection of \(\overrightarrow{\mathrm{AX}}\) and \(\overrightarrow{\mathrm{PZ}}\).

Thus, PLAN is the required quadrilateral.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (iii).
Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 3
[Note : □ HEAR is a parallelogram.
Opposite sides of parallelogram are of equal lengths.]
HE = 5 cm, ∴ AR = 5 cm, EA = 6 cm, ∴ HR = 6 cm
Adjacent angles of a parallelogram arc supplementary.
m∠R = 85° (given)
∴ m∠H = 180° – 85° = 95°
Opposite angles of a parallelogram are of equal measures,
m ∠ R = 85°
∴ m ∠ E = 85°

Steps of construction:

  • Draw a line segment HE = 5 cm.
  • At H, draw \(\overrightarrow{\mathrm{HX}}\), such that ∠ XHE = 95°. (Use protractor)
  • With H as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{HX}}\) at R.
  • At E, draw \(\overrightarrow{\mathrm{EY}}\) such that ∠ HEY = 85°. (Use protractor)
  • With E as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EY}}\) at A.
  • Draw \(\overline{\mathrm{AR}}\).

Thus, HEAR is the required parallelogram.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question (iv).
Rectangle OKAY
OK = 7 cm
KA = 5 cm
Solution:
PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry Ex 4.3 4
[Note: Here, OKAY is a rectangle. Opposite sides of a rectangle are of equal lengths.]
OK = 7 cm, ∴ AY = 7 cm and KA = 5 cm, ∴ OY = 5 cm
Moreover, all angles of a rectangle are right angles.

Steps of construction:

  • Draw a line segment OK = 7 cm.
  • At O, draw \(\overrightarrow{\mathrm{OM}}\) such that ∠ MOK = 90°.
  • With O as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OM}}\) at Y.
  • At K, draw \(\overrightarrow{\mathrm{KN}}\) such that ∠ NKO = 90°.
  • With K as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{KN}}\) at A.
  • Draw \(\overline{\mathrm{AY}}\).

Thus, OKAY is the required rectangle.

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1

Question 1.
For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station.
Give reasons for each.
Solution:
We represent those data by a histogram which can be grouped into class intervals. Obviously, for (b) and (d), the data can be represented by histograms.

Question 2.
The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution:
The frequency distribution table for the above data can be done as follows :
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 1
We can represent the above data by a bar graph as given below:
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 2

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Question 3.
The weekly wages (in ₹) of 30 workers in a factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on.
Solution:
The lowest observation = 804
The highest observation = 898
The classes are 800-810, 810-820, etc.
∴ The frequency distribution table is
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 3

Question 4.
Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions:
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more ?
(iii) How many workers earn less than ₹ 850 ?
Solution:
The histogram from the above frequency table is given below.
Here we have represented the class intervals along X-axis and frequencies of the class intervals along the Y-axis.
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 4
(i) The group 830 – 840 has the maximum number of workers.
(ii) Number of workers earning ₹ 850 or more = 1 + 3 + 1 + 1 – 1 – 4 = 10
(iii) Number of workers earning less than ₹ 850 = 3 + 2 + 1 + 9 + 5 = 20

PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1

Question 5.
The number of hours for which students of a particular class watched television during holidays is shown through the given graph.
Answer the following:
(i ) For how many hours did the maximum number of students watch TV ?
(ii) How many students watched TV for less than 4 hours ?
(iii) How many students spent more than 5 hours in watching TV ?
PSEB 8th Class Maths Solutions Chapter 5 Data Handling Ex 5.1 5
Solution:
(i) The maximum number of students watched TV for 4 to 5 hours.
(ii) 34 students watched TV for less than 4 hours (4 + 8 + 22 = 34).
(iii) 14 students spent more than 5 hours in watching TV (8 + 6 = 14).

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 204]

1. Observe the following tables and find if x and y are directly proportional.

Question (i)

X 20 17 14 11 8 5 2
y 40 34 28 22 16 10 4

Solution:
\(\begin{aligned}
&\frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2}, \frac{8}{16}=\frac{1}{2} \\
&\frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}
\end{aligned}\)
The value of \(\frac{\text {x}}{\text {y}}\) is same for different values of x and y. So these values x and y are directly proportional.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Question (ii)

X 6 10 14 18 22 26 30
y 4 8 12 16 20 24 28

Solution:
\(\begin{aligned}
&\frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8}, \frac{22}{20}=\frac{11}{10}, \\
&\frac{26}{24}=\frac{13}{12}, \frac{30}{28}=\frac{15}{14}
\end{aligned}\)
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively. So these values of x and y are not directly proportional.

Question (iii)

X 5 8 12 15 18 20
y 15 24 36 60 72 100

Solution:
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively.
So these values of x and y are not directly proportional.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

2. Principal = ₹ 1000, Rate = 8 % per annum. Fill in the following table and find which type of interest (simple or compound) changes in direct proportion with time period.

Time Period 1 year 2 years 3 years
Simple Interest (in ₹)
Compound Interest (in ₹)

Solution:
Simple interest : SI = \(\frac{P \times R \times T}{100}\)
For calculation:
P = ₹ 1000, R = 8 %, T = …………….

Time (T) → 1 year: T = 1 2 years : T = 2 3 years : T = 3
Simple interest SI = \(\frac{P \times R \times T}{100}\) ₹ \(\frac{1000 \times 8 \times 1}{100}\)
= ₹ 80
₹ \(\frac{1000 \times 8 \times 2}{100}\)
= ₹ 160
₹ \(\frac{1000 \times 8 \times 3}{100}\)
= ₹ 240
\(\frac{\text { SI }}{\text { T }}\) \(\frac {80}{1}\) = 80 \(\frac {160}{2}\) = 80 \(\frac {240}{3}\) = 80

Here, the ratio of simple interest with time period is same for every year.
Hence, simple interest changes in direct proportion with time period.
Compound interest:
For calculation:
P = ₹ 1000, R = 8 %, T = ……………

Time → 1 year : n = 1
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))1
= 1000 × \(\frac {108}{100}\) = 1080
∴ CI = 1080 – 1000 = ₹ 80
\(\frac{\text { CI }}{\text { T }}\) \(\frac {80}{1}\)
Time → 2 years : n = 2
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))2
= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1166.40
∴ CI = ₹ 1166.40 – 1000 = ₹ 166.40
\(\frac{\text { CI }}{\text { T }}\) \(\frac {166.40}{2}\)
Time → 3 years : n = 3
A = P(1 + \(\frac {R}{100}\))n
CI = A – P
A = 1000(1 + \(\frac {8}{100}\))3
= 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1259.712
∴ CI = ₹ 1259.712 – ₹ 1000 = ₹ 259.712
\(\frac{\text { CI }}{\text { T }}\) \(\frac {259.712}{3}\)

Here, the ratio of CI and T is not same.
Thus, compound interest is not proportional with time period.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Think, Discuss and Write: [Textbook Page No. 204]

1. If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution:
Time period (T) and rate of interest (R) are fixed, then
Simple interest = \(\frac {PRT}{100}\) = P × Constant
So simple interest depends on principal. Simple interest changes proportionally with principal.
Now, compound interest = P(1 + \(\frac {R}{100}\))T – P
i.e., A – P
= P [(1 + \(\frac {R}{100}\)T – 1]
= P × Constant
So compound interest depends on principal.
If principal increases or decreases, then compound interest will also increases or decreases.
Thus, compound interest changes with principal.

Think, Discuss and Write : [Textbook Page No. 209]

1. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by ‘unitary method’?
Solution:
Yes, each problem can be solved by unitary method.
e.g. Question 4 of Exercise: 13.1
Number of bottles filled in 6 hours = 840
∴ The number of bottles filled in 1 hour = \(\frac {840}{6}\) = 140
The number of bottles filled in 5 hours = 140 × 5 = 700
Thus, 700 bottles will it fill in five hours.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 211]

1. Observe the following tables and find which pair of variables (here x and y) are in inverse proportion.

Question (i)

X 50 40 30 20
y 5 6 7 8

Solution:
x1 = 50 and y1 = 5
∴ x1y1 = 50 × 5
∴ x1y1 = 250

x2 = 40 and y2 = 6
∴ x2y2 = 40 × 6
∴ x2y2 = 240

x3 = 30 and y3 = 7
∴ x3y3 = 30 × 7
∴ x3y3 = 210

x4 = 20 and y4 = 8
∴ x4y4 = 20 × 8
∴ x4y4 = 160
Now 250 ≠ 240 ≠ 210 ≠ 160
∴ x1y1 ≠ x2y2 ≠ x3y3 ≠ x4y4
∴ x and y are not in inverse proportion.

Question (ii)

X 100 200 300 400
y 60 30 20 15

Solution:
x1 = 100 and y1 = 60
∴ x1y1 = 100 × 60
∴ x1y1 = 6000

x2 = 200 and y2 = 30
∴ x2y2 = 200 × 30
∴ x2y2 = 6000

x3 = 300 and y3 = 20
∴ x3y3 = 300 × 20
∴ x3y3 = 6000

x4 = 400 and y4 = 15
∴ x4y4 = 400 × 15
∴ x4y4 = 6000
Now x1y1 = x2y2 = x3y3 = x4y4
∴ x and y are in inverse proportion.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Question (iii)

X 90 60 45 30 20 5
y 10 15 20 25 30 35

Solution:
x1 = 90 and y1 = 10
∴ x1y1 = 90 × 10
∴ x1y1 = 900

x2 = 60 and y2 = 15
∴ x2y2 = 60 × 15
∴ x2y2 = 900

x3 = 45 and y3 = 20
∴ x3y3 = 45 × 20
∴ x3y3 = 900

x4 = 30 and y4 = 25
∴ x4y4 = 30 × 25
∴ x4y4 = 750

x5 = 20 and y5 = 30
∴ x5y5 = 20 × 30
∴ x5y5 = 600

x6 = 30 and y6 = 35
∴ x6y6 = 5 × 35
∴ x6y6 = 175

Now x1y1 = x2y2 = x3y3 ≠ x4y4 ≠ x5y5 ≠ x6y6
∴ x and y are in inverse proportion.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Try These : [Textbook Page No. 219]

1. Factorise:

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

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

Question (ii)
22y – 33z
Solution:
22y = 2 × 11 × y and 33z = 3 × 11 × z
Common factor =11
∴ 22y – 33z = (2 × 11 × y) – (3 × 11 × z)
= (11) × (2 × y – 3 × z)
= 11 (2y – 3z)

Question (iii)
14pq + 35pqr
Solution:
14pq = 2 × 7 × p × q and
35pqr = 7 × 5 × p × q × r
Common factor = 7pq
∴ 14pq + 35pqr = (2 × 7 × p × q) + (7 × 5 × p × q × r)
= 7 × p × q (2 + 5 × r)
= 7pq (2 + 5r)

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

Try These : [Textbook Page No. 225]

1. Divide:

Question (i)
24xy2z3 by 6yz2
Solution:
= \(\frac{24 x y^{2} z^{3}}{6 y z^{2}}\)
= \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)
= \(\frac{2 \times 2 \times x \times y \times z}{1}\) = 4xyz
∴ 24xy2z3 ÷ 6yz2
= 4xyz

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

Question (ii)
63a2b4c6 by 7a2b2c3
Solution:
= \(\frac{63 a^{2} b^{4} c^{6}}{7 a^{2} b^{2} c^{3}}\)
= \(\frac{3 \times 3 \times 7 \times a^{2} \times b^{4} \times c^{6}}{7 \times a^{2} \times b^{2} \times c^{3}}\)
= 3 × 3 × \(\frac{a^{2}}{a^{2}} \times \frac{b^{4}}{b^{2}} \times \frac{c^{6}}{c^{3}}\)
= 9 × a2-2 × b4-2 × c6-3
= 9 × a0 × b2 × c3
= 9 × 1 × b2 × c3
= 9b2c3
∴ 63a2b4c6 ÷ 7a2b2c3
= 9b2c3

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

1. Which of the following are in inverse proportion?

Question (i)
The number of workers on a job and the time to complete the job.
Solution:
If the number of workers on a job increases, then time to complete the job decreases. So, it is the case of inverse proportion.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

Question (ii)
The time taken for a journey and the distance travelled in a uniform speed.
Solution:
Here, the speed is uniform.
∴ The time taken for a journey is directly proportional to the speed. So, it is not the case of inverse proportion.

Question (iii)
Area of cultivated land and the crop harvested.
Solution:
For more area of cultivated land, more crops would be harvested. So, it is not the case of inverse proportion.

Question (iv)
The time taken for a fixed journey and the speed of the vehicle.
Solution:
If speed of vehicle is more, then time to cover the fixed journey would be less. So, it is the case of inverse proportion.

Question (v)
The population of a country and the area of land per person.
Solution:
For more population, less area per person would be in the country. So, it is a case of inverse proportion.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

2. In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners 1 2 4 5 8 10 20
Prize for each winner (in ₹) 1,00,000 50,000

Solution:
Here, more the number of winners, less is the prize money for each winner.
∴ This is a case of inverse proportion.
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 1
Now, the table is as follows:

Number of winners 1 2 4 5 8 10 20
Prize for each winner (in ₹) 1,00,000 50,000 25,000 20,000 12,500 10,000 5,000

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table:
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 2

Number of spokes 4 6 8 10 12
Angle between a pair of consecutive spokes 90° 60°

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

Question (i)
Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
Solution:
Here, more the number of spokes, less the measure of angle between a pair of consecutive spokes.
∴ This is a case of inverse proportion.
Here,
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 3.1
Now, the table is as follows:

Number of spokes 4 6 8 10 12
Angle between a pair of consecutive spokes
90° 60° 45° 36° 30°

Question (ii)
Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
Solution:
Let the measure of angle be x°.
∴ 15 × x° = 4 × 90°
∴ x° = \(\frac{4 \times 90^{\circ}}{15}\) = 24°
Hence, this angle should be 24°.

Question (iii)
How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
Let the number of spokes be n.
∴ n × 40° = 4 × 90°
∴ n = \(\frac{4 \times 90^{\circ}}{40^{\circ}}\) = 9
Thus, 9 spokes would be needed.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:

Number of children x Number of sweets y
x1 = 24 y1 = 5
x2 = 24 – 4 = 20 y2 = (?)

Here, if the number of children decreases, then the number of sweets received by each child will increase.
∴ This is a case of inverse proportion.
x1 × y1 = x2 × y2
∴ 24 × 5 = 20 × y2
∴ y2 = \(\frac{24 \times 5}{20}\) = 6
Thus, each child will get 6 sweets.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

Number of animals x Number of days y
x1 = 20 y1 = 6
x2 = 20 + 10 = 20 y2 = (?)

Here, the number of animals increases, so the number of days to feed them will decrease.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 20 × 6 = 30 × y2
∴ y2 = \(\frac{20 \times 6}{30}\) = 4
Thus, the food will last for 4 days.

6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:

Number of persons x Number of days y
x1 = 42 y1 = 63
x2 = (?) y1 = 63

Here, more the number of persons, less will be the time required to complete the job.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 3 × 4 = 4 × y2
∴ y2 = \(\frac{3 \times 4}{4}\) = 3
Thus, 3 days will be required to complete the job.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 4.1
Solution:

Number of bottles in a box x Number of boxes y
x1 = 12 y1 = 25
x2 = 20 y2 = (?)

Here, more the number of bottles in a box, less would be the number of boxes.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 3 × 4 = 4 × y2
∴ y2 = \(\frac{3 \times 4}{4}\) = 3
Thus, 15 boxes would be filled.

8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:

Number of machines x Number of days y
x1 = 42 y1 = 63
x2 = (?) y2 = 54

Here, if the number of days will be less, the number of machines required will be more.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 42 × 63 = x2 × 54
∴ x2 = \(\frac{42 \times 63}{54}\) = 49
Thus, 49 machines would be required.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:

Speed (km/h) x Number of hours y
x1 = 60 y1 = 2
x2 = 80 y2 = (?)

Here, if the speed of car increases, then the time taken to cover the same distance will decrease.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 60 × 2 = 80 × y2
∴ y2 = \(\frac{60 \times 2}{80}=\frac{3}{2}=1 \frac{1}{2}\) h
Thus, car would take 1\(\frac {1}{2}\) hours.

10. Two persons could fit new windows in a house in 3 days.

Question (i)
One of the persons fell ill before the work started. How long would the job take now?
Solution:

Number of persons x Number of days y
x1 = 2 y1 = 3
x2 = 2 – 1 = 1 y2 = (?)

Here, less the number of persons, more would be the number of days to complete the job.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 2 × 2 = 1 × y2
∴ y2 = \(\frac{2 \times 3}{1}\) = 6
Thus, it would take 6 days to complete the job.

Question (ii)
How many persons would be needed to fit the windows in one day?
Solution:

Number of days x Number of persons y
x1 = 3 y1 = 2
x2 = 1 y2 = (?)

Here, less the number of days, more will be the number of persons needed.
∴ This is a case of inverse proportion.
∴ y2 = ? and x2 = 1
∴ x1 × y1 = x2 × y2
∴ 3 × 2 = 1 × y2
∴ y2 = \(\frac{3 \times 2}{1}\) = 6
Thus, 6 persons would be needed.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:

Number of periods x Length of each period (in minute) y
x1 = 8 y1 = 45
x2 = 9 y2 = (?)

Here, the number of periods is more, then the length of each period will be less.
∴ This is a case of inverse proportion.
∴ x1 × y1 = x2 × y2
∴ 8 × 45 = 9 × y2
∴ y2 = \(\frac{8 \times 45}{9}\) = 40
Thus, each period would be of 40 minutes.

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

1. Find and correct the errors in the following mathematical statements:

Question 1.
4 (x – 5) = 4x – 5
Solution:
Error: 4 × – 5 = (- 20) and not (- 5)
Correct statement: 4 (x – 5) = 4x – 20

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 2.
x (3x + 2) = 3x2 + 2
Solution:
Error: x × 2 = 2x
Correct statement: x (3x + 2) = 3x2 + 2x

Question 3.
2x + 3y = 5xy
Solution:
Error: 2x and 3y are unlike terms.
So their sum is not possible.
Correct statement: 2x + 3y = 2x + 3y

Question 4.
x + 2x + 3x = 5x
Solution:
Error: x, 2x and 3x are like terms. So sum of their coefficient =1 + 2 + 3 = 6.
Correct statement: x + 2x + 3x = 6x

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 5.
5y + 2y + y – 7y = 0
Solution:
Error : 5y, 2y, y and – 7y all are like terms here. So sum of their coefficient = 5 + 2 + 1 – 7 = 1.
Correct statement: 5y + 2y + y – 7y = y

Question 6.
3x + 2x = 5x2
Solution:
Error: When like terms are added or subtracted their exponents do not change.
Correct statement: 3x + 2x = 5x

Question 7.
(2x)2 + 4 (2x) + 7 = 2x2 + 8x + 7
Solution:
Error: (2x)2 = 2x × 2x = 4x2
Correct statement:
(2x)2 + 4 (2x) + 7 = 4x2 + 8x + 7

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 8.
(2x)2 + 5x = 4x + 5x = 9x
Solution:
Error : (2x)2 = (2x × 2x) = 4x2
Correct statement: (2x)2 + 5x = 4x2 + 5x

Question 9.
(3x + 2)2 = 3x2 + 6x + 4
Solution:
Error : (3x + 2)2
= (3x)2 + 2 (3x)(2) + (2)2
= 9x2 + 12x + 4
Correct statement:
(3x + 2)2 = 9x2 + 12x + 4

10. Substituting x = – 3 in

Question (a)
x2 + 5x + 4 gives (- 3)2 + 5 (- 3) + 4 = 9 + 2 + 4 = 15
Solution:
Error : 5 (- 3) = – 15 and not 2
Correct statement:
Substituting x = (- 3) in, x2 + 5x + 4
= (- 3)2 + 5 (-3) + 4
= 9 – 15 + 4 = 9 + 4 – 15
= 13 – 15
= (-2)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question (b)
x2 – 5x + 4 gives (- 3)2 – 5 (- 3) + 4 = 9 – 15 + 4 = -2
Solution:
Error: -5 (-3) = + 15 and not (-15)
Correct statement:
Substituting x = (- 3) in,
x2 – 5x + 4
= (- 3)2 – 5 (- 3) + 4
= 9 + 15 + 4
= 28

Question (c)
x2 + 5x gives (- 3)2 + 5 (- 3) = – 9 – 15 = – 24
Solution:
Error : (- 3)2 = + 9
Correct statement:
Substituting x = (- 3) in, x2 + 5x
= (- 3)2 + 5 (- 3)
= 9 – 15 = – 6

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 11.
(y – 3)2 = y2 – 9
Solution:
Error : (y – 3)2
= (y)2 – 2 (y)(3) + (- 3)2
= y2 – 6y + 9
Correct statement: (y – 3)2 = y2 – 6y + 9.

Question 12.
(z + 5)2 = z2 + 25
Solution:
Error : (z + 5)2
= (z)2 + 2 (z)(5) + (5)2
= z2 + 10 z + 25
Correct statement: (z + 5)2
= z2 + 10z + 25

Question 13.
(2a + 3b) (a – b) = 2a2 – 3b2
Solution:
Error : (2a + 3b) (a – b)
= 2a (a-b) + 3b (a-b)
= 2a2 – 2ab + 3ab – 3b2
= 2a2 + ab – 3b2
Correct statement: (2a + 3b) (a – b)
= 2a2 + ab – 3b2

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 14.
(a + 4) (a + 2) = a2 + 8
Solution:
Error : (a + 4) (a + 2) = a (a + 2) + 4 (a + 2)
= a2 + 2a + 4a + 8
= a2 + 6a + 8
Correct statement: (a + 4) (a + 2)
= a2 + 6a + 8

Question 15.
(a – 4) (a – 2) = a2 – 8
Solution:
Error : (a – 4) (a – 2) = a (a – 2) – 4 (a – 2)
= a2 – 2a – 4a + 8
= a2 – 6a + 8
Correct statement: (a – 4) (a – 2)
= a2 – 6a + 8

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 16.
\(\frac{3 x^{2}}{3 x^{2}}\) = 0
Solution:
Error: Numerator and denominator, both are same. So their division is 1.
Correct statement:\(\frac{3 x^{2}}{3 x^{2}}\) = 1

Question 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
Solution:
Error: \(\frac{3 x^{2}+1}{3 x^{2}}=\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}\)
= 1 + \(\frac{1}{3 x^{2}}\)
Correct statement: \(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + \(\frac{1}{3 x^{2}}\)

Question 18.
\(\frac{3 x}{3 x+2}=\frac{1}{2}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}\)

Question 19.
\(\frac{3}{4 x+3}=\frac{1}{4 x}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

PSEB 8th Class Maths Solutions Chapter 14 Factorization Ex 14.4

Question 20.
\(\frac{4 x+5}{4 x}\) = 5
Solution:
Error: \(\frac{4 x+5}{4 x}\)
= \(\frac{4 x}{4 x}+\frac{5}{4 x}\)
= 1 + \(\frac{5}{4 x}\)
Correct statement: \(\frac{4 x+5}{4 x}\) = 1 + \(\frac{5}{4 x}\)

Question 21.
\(\frac{7 x+5}{5 x}\) = 7x
Error: \(\frac{7 x+5}{5 x}\)
= \(\frac{7 x}{5}+\frac{5}{5}\)
= \(\frac{7 x}{5}\) + 1
Correct statement: \(\frac{7 x+5}{5}=\frac{7 x}{5}\) + 1

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used:
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 1
(i) AD = ……………
(ii) ∠DCB = ……………
(iii) OC = ……………
(iv) m∠DAB + m∠CDA = ……………
Solution:
(i) AD = BC
(∵ Opposite sides are equal)

(ii) ∠DCB = ∠DAB
(∵ Opposite angles are equal)

(iii) OC = OA
(∵ Diagonals bisect each other)

(iv) m∠DAB + m∠CDA = 180°
(∵ Adjacent angles are supplementary)

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
(i)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 2
Solution:
□ ABCD is a parallelogram.
∴ ZB = ZD
∴ y = 100°
(∵ Opposite angles)
Now, y + z = 180° (∵ Adjacent singles are supplementary)
∴ 100° + z = 180°
∴ z = 180° – 100°
∴ z = 80°
Now, x = z (∵ Opposite angles)
∴ x = 80°
Thus, x = 80°, y = 100° and z = 80°.

(ii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 3
Solution:
It is a parallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 4
∴ m∠P + m∠S = 180°
(∵ Adjacent angles are supplementary)
∴ x + 50° = 180°
∴ x = 180° – 50°
∴ x = 130°
x = y (∵ Opposite angles)
∴ y = 130°
Now, m∠Q = 50° (∵ ∠S and ∠Q are opposite angles)
m∠Q + z = 180° (∵ Linear pair of angles)
∴ 50° + z = 180°
∴ z = 180° – 50°
∴ z = 130°
Thus, x = 130°, y = 130° and z = 130°.

(iii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 5
Solution:
Vertically opposite angles are equal.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 6
∴ m∠AMD = m∠BMC = 90°
∴ x = 90°
In ΔBMC
y + 90° + 30° = 180°
(Angle sum property of a triangle)
∴ y + 120° = 180°
∴ y = 180° – 120° = 60°
Here, ABCD is a parallelogram
∴ \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) is a transversal.
∴ y = z (∵ Alternate angles)
∴ z = 60° (∵ y = 60°)
Thus, x = 90°, y = 60° and z = 60°

(iv)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 7
Solution:
□ ABCD is a Dparallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 8
∠D = ∠B (∵ Opposite angles)
∴ y = 80°
m∠A + m∠D = 180° (∵ Adjacent angles are supplementary.)
∴ x + y = 180°
∴ x + 80° = 180°
∴ x = 180° – 80°
∴ x = 100°
m∠A = m∠BCD (∵ Opposite angles are equal)
∴ 100° = m∠BCD
Now, z + m∠BCD = 180° (∵ Linear pair of angles)
∴ z + 100° = 180°
∴ z = 180° – 100°
∴ z = 80°
Thus, x = 100°, y = 80° and z = 80°

(v)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 9
Solution:
□ ABCD is a parallelogram.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 10
∴ m∠B = m∠D (∵ Opposite angles)
∴ y = 112°
m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ (40° + z) + 112° = 180°
∴ 40° + z + 112° = 180°
∴ z + 152° = 180°
∴ z = 180° – 152°
∴ z = 28°
Now, \(\overline{\mathrm{DC}} \| \overline{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{AC}}\) is their transversal.
∴ z = x
∴ x = 28° (∵ z = 28°)
Thus, x = 28°, y = 112° and z = 28°

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
Solution:
In a quadrilateral ABCD,
∠D + ∠B = 180° (Given)
In parallelogram, the sum of measures of adjacent angles is 180°.
But here, the sum of measures of opposite angles is 180°.
∴ The quadrilateral may be a parallelogram.

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
Solution:
In a quadrilateral ABDC,
AB = DC = 8 cm
AD = 4 cm
BC = 4.4 cm
∵ Opposite sides AD and BC are not equal.
∴ It cannot be a parallelogram.

(iii) ∠A = 70° and ∠C = 65° ?
Solution:
In a quadrilateral ABCD, m∠A = 70° and m∠C = 65°
Opposite angles ∠A ≠ ∠C
∴ It cannot be a parallelogram.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 11
Here, ∠B = ∠D (see figure)
Yet, it is not a parallelogram.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 12
Adjacent angles are supplementary.
∴ m∠A + m∠B = 180°
∴ 3x + 2x – 180°
∴ 5x = 180°
∴ \(\frac{180^{\circ}}{5}\)
∴ m∠A = 3x = 3 × 36° = 108°
m∠B = 2x = 2 × 36° = 72°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 108° and m∠D = 72°
Thus, ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 13
∴ m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ m∠A + m∠A = 180° (m∠B = m∠A)
∴ 2m∠A = 180°
∴ m∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ m∠B = 90°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 90° and m∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 14
Solution:
∠POA is the exterior angle of ΔHOE
∴ y + z = 70° (Exterior)
∴ m∠HOP = 180° – 70° = 110°
∠HEP = ∠HOP = 110° (Opposite angles of a parallelogram)
∴ x = 110°
∵ \(\overline{\mathrm{EH}} \| \overline{\mathrm{PO}}\), \(\overleftrightarrow{\mathrm{PH}}\) is a transversal.
∴ y = 40° (∵ Alternate angles are equal)
Now, y + z = 70°
∴ 40° + z = 70°
∴ z = 70° – 40° = 30°
Thus, x = 110°, y = 40° and z = 30°

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 15
Solution :
□ GUNS is a parallelogram.
∴ GS = NU and SN = GU (Y Opposite sides)
∴ 3x = 18 and 26 = 3y – 1
∴ x = \(\frac {18}{3}\)
∴ x = 6

∴ 3y – 1 = 26
∴ 3y = 26 + 1
∴ 3y = 27
∴ y = \(\frac {27}{3}\) = 9
Thus, x = 6 cm and y = 9 cm

(ii)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 16
Solution:
□ RUNS is a parallelogram.
∴ Its diagonals bisect each other.
x + y = 16 ……(1)
and y + 7 = 20
y = 20 – 7 = 13 …..(2)
Substituting value of y in (1)
x + y = 16
x + 13 = 16
∴ x = 16 – 13
= 3
Thus, x = 3 cm and y = 13 cm

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 9.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 17
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
□ RISK is a parallelogram.
∴ m∠R + m∠K = 180°
(∵ Adjacent angles are supplementary)
∴ m∠R + 120° = 180°
∴ m∠R = 180° – 120°
∴ m∠R = 60°
∠R and ∠S are opposite angles of parallelogram.
∴ m∠S = 60°
□ CLUE is a parallelogram.
∴ m∠E = m∠L = 70° (∵ Opposite angles)
In ΔESQ
∴ m∠E + m∠S + x = 180° (∵ Angle sum property of triangle)
∴ 70° + 60° + x = 180°
∴ 130° + x = 180°
∴ x = 180° – 130°
∴ x = 50°
Thus, x = 50°

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel ? (Fig3.32)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 18
Solution:
Since, 100° + 80° = 180°
i. e., ∠M and ∠L are supplementary.
∴ \(\overline{\mathrm{NM}} \| \overline{\mathrm{KL}}\)
(∵ Interior angles along on the same side of the transversal are supplementary)
One pair of opposite side of □ LMNK is parallel.
∴ □ LMNK is a trapezium.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 11.
Find m∠C in figure 3.33 if \(\overline{\mathbf{A B}} \| \overline{\mathbf{D C}}\).
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 19
Solution:
In □ ABCD, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) and BC is their transversal,
∴ m∠B + m∠C = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ 120° + m∠C = 180°
∴ m∠C = 180° – 120° = 60°
Thus, m∠C = 60°

Question 12.
Find the measure of ∠P and ∠S if \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) in figure 3.34. (If you find m∠R, is there more than one method to find m∠P ?)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 20
Solution:
In □ PQRS \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
∴ □ PQRS is a trapezium.
\(\overleftrightarrow{\mathrm{PQ}}\) is a transversal of \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
m∠P + m∠Q = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ mZP + 130° = 180°
∴ mZP = 180° – 130°
∴ mZP = 50°
In □ PQRS, ∠R = 90°
\(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\), \(\overleftrightarrow{\mathrm{RS}}\) is their transversal.
∴ m∠S + m∠R = 180°
∴ m∠S + 90° = 180°
∴ m∠S= 180°-90°
∴ m∠S = 90°
Yes, the sum of the measures of all angles of quadrilateral is 360°. ∠P and ∠R can be found.
m∠P + m∠Q + m∠R + m∠S = 360°
∴ m∠P + 130° + 90° + 90° = 360°
∴ m∠P + 310° = 360°
∴ m∠P = 360° – 310°
∴ m∠P = 50°

PSEB 8th Class English Creative Writing

Punjab State Board PSEB 8th Class English Book Solutions English Creative Writing Exercise Questions and Answers, Notes.

PSEB 8th Class English Creative Writing

(A) Picture Composition

Now write some sentences on a bicycle. You can take help from the picture.
1.
PSEB 8th Class English Creative Writing 1
Answers:
I have a bicycle. Its colour is black. It has a shining strong rims with tyres on them. There are spokes in the rims. Its tongs are tightly fitted to the wheels. Its chain is very light and runs smoothly. I sit on the saddle on my bicycle. Catches its handle and moves its paddles. My bicycle runs very fast.

PSEB 8th Class English Creative Writing

2.
PSEB 8th Class English Creative Writing 2

1. This picture shows the protection of forest trees.
2. A little girl has put her loving arms around a tree.
3. He does not want it to be cut down.
4. Her love for trees has inspired many other women too.
5. They all try to stop people from cutting trees.
6. Whenever people come to their village to cut trees they reach there crying ‘Chipko ! Chipko !” Hug the trees.
7. It has now become a big movement called ‘Chipko Movement’.
8. They cling to the trees till the tree fallers go back.
9. They tell them the great benefits of trees.
Trees, for example, give us fruits, firewood, medicines and many other things.
10. They check air pollution and supply us with fresh oxygen.
11. Above all our forest trees along with various birds and animals living in the forests are the beauty of our earth.

3. Family Picnic
PSEB 8th Class English Creative Writing 3

1. The picture depicts (Gericht) a family picnic.
2. Three children with their parents and grand parents have come to enjoy the day.
3. It is bright morning.
4. They have spread a mat near a tree.
5. They have brought some food in g big box. There is another dish in a steel container.
6. The house wife is giving some slikes to the old lady to eat.
7. Her grand-daughter is also giving her something.
8. Father and the son are enjoying the game of chakri/chakti.
9. The grandfather is busy in playing some other game with his grandson.
10. The scene is a feast to the eyes.
11. All look happy and carefree.

4.
PSEB 8th Class English Creative Writing 4

1. It is a kids party-five, kids are celebrating it in a beautiful hall outside.
2. The hall is open and is decoated with colourful balloons.
3. The outside panorama (figga) is adding new colours to the scene.
4. The children are in fancy dresses.
5. They have clown caps on their heads.
6. The have a big table in front of them.
7. The cake with candles on it suggest that it is a birthday party of the girl standing near the table.
8. Other kids have brought different gifts for her.
9. There is an arrangement of cold drinks too.
10. The cake will be cut with clappie and cheers
11. The hall will echo with the merry songs of ‘Happy Birthday to ……….

PSEB 8th Class English Creative Writing

(B) Guided Composition

Write a paragraph with the help of the given, hints:

1. My Best Friends
Hints : True friends rare……….. name……….. classmate……….. tall, smart, hard-working,: intelligent………..family……..one of the best students………. kind, gentle, helpful.

True friends are rare in the world. But I am very lucky to have a true friend. His name is Mohan. He is my classmate. He is tall and smart. He is hard-working and intelligent. He is one of the best students of our school. He comes of a noble family. His father is a doctor. His mother is a teacher. He has many good habits. He gets up early and goes for a walk. He wears neat and clean dress. He is never late for school. He obeys the school rules. He respects his teachers. He is a good player of hockey. He is very kind and gentle. He helps the poor and the needy. He is like a rose among flowers.

Word-Meanings : Rare-दुर्लभ, Classmate-सहपाठी, Habits-आदतें, Needy-ज़रूरतमंद।

2. The Postman:
Hints : Nathu Ram our postman……..to post office……… sorts…….. arranges…….. New bag……. keeps……..letters and parcels……..from house to house………..busy…….. hardworking……….. punctual………..enjoys work………..few holidays……….. brings news………..

Sh. Nathu Ram is our postman. He gets early in the morning. He puts on his dress. He goes to the post office in the morning. He sorts the daak there. He arranges letters and parcels. He has a bag round his shoulder. He keeps letters and parcels in it. He goes from house to house. He gives letters and parcels to the right persons. He is busy from morning to evening. He is very hard-working. He is very regular and punctual. He is never late. He enjoys his work very much. He knows no rest. He has a few holidays. He brings all types of news to us. We wait for him every day.

Word-Meanings : Sorts-छांटता है, Arranges-क्रम में रखता है, Punctual-समय का पाबन्द।

3. My Favourite Teacher
Hints : My teacher……. able and hard-working…….like Sh. K.S. Bedi the most……….tall and handsome………..M.A.; B.Ed……….. noble family………..good habits………..gets up early………. prays to God………..school in time……… well dressed……….polite………. way of teaching………good speaker……….Singer……….. helps the needy.

There are many teachers in our school. They are verable and hard-working. But I like Sh. K.S. Bedi the most. He is my favourite teacher. He is ofr class-in-charge. He is about 42 years old. He is tall and handsome. He is an M.A.; B.Ed. He belongs to a noble family. My teacher has many good habits. He gets up early in the morning He goes out for a walk. He prays to God daily. He always comes to school in time. He is an ays well dressed. He never gets angry.

He is polite to all. He teaches us English. His way of teaching is very simple. He takes part in games. He is a good speaker and fine singer. He is a lover of books. He helps the poor and the needy. All like him very much. May be live long!

Word-Meanings : Handsome – संदर, Belongs-संबंध रखते है, Noble – नेक.

4. Our School Library
Hints : School is a temple of learning………..library an altar……. a big library in my school………..50,000 books………. kept subject-wise………. newspapers and magazines ……….. librarian very helpful and kind……….. really useful.

A school is a temple of learning. A library is an altar ( èÁæ-SÍÜ) in it. My school too has a big library. It is housed in a corner. It has about 50,000 books in it. The books are kept subject-wise. They are kept in almirahs with glass-panes. The library has a number of newspapers and magazines too. They are in Punjabi, Hindi and English. We can borrow books from the library. But no student can keep a book for more than fifteen days. The librarian is very helpful, kind and gentle. But he is very strict. He maintains perfect silence and discipline in the library. We go to the library thrice a week. We read newspapers and magazines. The library is really very useful to all of us.

Word-Meanings : Housed in a corner- एक कोने में स्थित हैं, Magazines- पत्रिकाएं, Perfect silence-पूर्ण खामोशी, Strict-सख्त।

PSEB 8th Class English Creative Writing

5. A Cricket Match
Hints: We played against the KVM School………..won the toss………. decided to bat……….skipper Abdul made 20 runs………..Surinder and Hira Singh………201 runs………KVM team scored 180 runs………..we won………..dancing.

Last Sunday we played a cricket match against the KVM Sen. Sec. School. It was a 50 over match. Both the teams were equally strong. We won the toss. We decided to bat. Our skipper Abdul opened the innings with his partner Hamid. He made twenty runs. Then came Surinder. He delighted the spectators with five sixes. Hira Singh was the next batsman. He stayed there for two hours and scored seventy runs. The whole team was out for 201 runs. It was now our turn to field. Our bowlers bowled very well. Our opponents were in great difficulty. They were not in a good form. They were all out for 180 runs. As a result we won the match. All of us started dancing in the field. Other students also joined us.

Word-Meanings : Equally-एक समान, Delighted-प्रसन्न किया, Spectators-दर्शक, Opponents-विरोधी

6. A Street Quarrel
Hints : A common sight……….. last night………..studying in my room………..loud noise in the street………..two persons………. hot words exchanged………. blows……….both injured……….had started over a trifle………..pacified by people……….. boys playing……….a ball struck another……….. parents joined………..should be avoided.

A street quarrel is a common sight. Mostly, an ordinary quarrel of children takes an ugly shape. Last night I happened to see a street quarrel. I was studying in my room. Suddenly, I heard a loud noise in the street. I saw a big crowd there. Two persons were exchanging hot words. Soon they came to blows. Both got injured and started bleeding. With great difficulty some people pacified them. The quarrel had started over a trifle. Some boys were playing in the open. By chance a ball struck another boy. They started fighting. Someone informed the parents of both the boys. They came out and started fighting. Thus a big scene was created out of a small thing. Such quarrels should be avoided.

Word-Meanings : Quarrel-झगड़ा, Common sight- आम दृश्य, Ugly shape – गंदा रूप, Exchanging hot words-गुस्से में बोल रहे थे, Blows-मुक्के, Bleeding-खून बहना, Pacified – रांत क्रिया, Trifle – तुच्छ बात

7. A Picnic
Hints : Sunday morning….party of fifteen boys…..carried fruit and cold drinks…. reached bank of a river……splashed water at one another….chatted and laughed…….titbits…….. tea and snacks………. took a bus and reached home ………. remember the day.

“It was Sunday morning. My friends and I decided to go for a picnic. We were a party of fifteen boys. We carried fruit, cold drinks and some snacks with us. We went to a river by bus. We reached the bank of the river at 9 a.m. We took some rest. Then we took off our clothes and jumped into the water. We splashed water at one another. After half an hour we came out of the water. We took fruit and cold drinks. We took rest for some time. Once again we jumped into the water. After half an hour we came out. We were in a cheerful mood. We chatted and laughed. Gurvinder related a few titbits. We had tea and snacks. At 6 p.m. we took a bus and reached our homes at 7 p.m. We remember the day even today.

Word-Meanings : Took off-उतरे, Splashed-छुटि उड़ाए, Related- सुनाये, Titbits – चुटकले, Chatted-बतें की.

8. A Journey by Train
Hints : Travelled from Chandigarh to Delhi by train ……… reached before time ……… bought a ticket ………. the train late, arrived, comfortable seat …….. trees, houses running backwards …….. stopped at Ambala ……. took tea ……… reached Delhi ……. sigh of relief.

Last Sunday, I travelled from Chandigarh to Delhi by train. I reached the railway station before the arrival of the train. I bought a ticket and came to the platform. The train was late. At last the train arrived. The platform was crowded with people. It was very difficult to get a seat. But I was lucky. I got a very comfortable seat near the window. The train started moving. I looked out of the window. The trees, the houses and the electric poles seemed running backwards. The train stopped even at small stations. At Ambala it stopped for half an hour. I came down and took tea. After some time the train started again. I reached Delhi at sunset. I heaved a sigh of relief.

Word-Meanings – Arrival-आगमन, Crowded-मश हुआ था, Seemed- लगता था, Comfortable-आरामदायक, Heaved a sigh of relief-चैन की सांस ली।

9. Going to the Market
Hints : There is a………… in our ……….. Men and women………… Some buy fruit, vegetables, some ……… grains open ………… Sunday ……….. I visited ………… last Sunday ………… many shops………. sari ………. my mother, shirt ……….. book shop …………. a story book ………… my younger sister ………… toys ………… no money left ………… back home.

There is a big market near our town. Men, women and children go there. They buy different things. Some buy fruit and vegetables. Some buy grains. Cloth is also sold there. The market is open on Sunday too. I visited it last Sunday. I went to many shops. First of all I went to a cloth shop. I bought a sari for my mother and a shirt for my brother. Then I visited a book shop. It had all types of books. I bought a story book for my younger sister. I saw a beautiful toy at toy shop. I wanted to buy it. But I had spent all the money. So I came back home.

Word-Meanings : Market- बाजर, Different- भिन्न-भिन्न, Grains-अनज, Bought-खरीदी, Types-प्रकार, Spent-खर्च कर दिए.

PSEB 8th Class English Creative Writing

10. A Morning Walk
Hints : A morning walk …………. useful ………… fit and healthy ………… I get up ….. friend’s house ………… to the canal ……….. meet people ………… scene very beautiful …………. birds ……… dew-drops ……….. everything fresh ……….. cool breeze ……….. refreshes mind………… reach canal ………… Some people bathing ………… praying………. return home.

A morning walk is very useful. It is a light exercise. It keeps us fit and healthy. Therefore, I go for a morning walk daily. I get up at 5 a.m. and go to my friend’s house. We go out for a walk together. We go to the canal. We meet many people on the way. They are also going for a walk. The morning scene is very beautiful. The birds are chirping. There are dew-drops on the grass. They shine like pearls. Everything looks fresh. A cool breeze is blowing. It refreshes our mind and body. We feel very happy. We reach the canal at about 6 a.m. We walk along the bank. Many men are there. Some are praying. Others are bathing. We also take our bath in the canal. The sun rises and we come back home.

Word-Meanings : Light Exercise-हल्का व्यायम, Canal-नहर, Dew-drops-ओस की बूंदें, Pearls-मोती, Refreshes-ताज़गी देती है।

11. Our Neighbour
Hints : Mohan Babu our neighbour ………… a shopkeeper ………… goes out for a walk ……. returns ………… vegetable market ………… buys vegetables ………… temple ………… devotee of goddess Lakshmi ……….. back from temple ……….. breakfast ……….. leaves for shop ………… a cloth merchant ……….. clever ………. comes home with money ……….. hearty meal with family sleeps early ………. a happy life.

A neighbour is a person who lives near us. Mohan Babu is our neighbour. He is a happy person. He is a shopkeeper. He gets up early in the morning. He goes out for a walk. He comes back after an hour. Then he goes to the vegetable market. He buys vegetables for the family. He goes to the temple daily. He is a devotee of Lakshmi, the goddess of wealth. He takes his breakfast and leaves for his shop. He is a cloth merchant. He is very clever. He deals with his customers well. He returns home late in the evening. His purse is full. He enjoys his meal with his family. He goes to bed early. Thus he leads a real happy life.

Word-Meanings : Neighbour-पड़ोसी, Vegetable market-सब्जी मंडी, Devotee-उपासक, Wealth-धन दौलत, Customers-ग्रहक

12. My Father-A Farmer Hints : Father a farmer ………… not well-educated ………… hard-working ………. truthful and honest ………… respected in village ………… proud of him ………… an open minded man ………… ready to accept things ………… a pleasant person………… settles village quarrels ……….. the wisest man of the village ………… simple and clean life ………… does not lose temper ………… respected by all.

My father is an able person. He is a farmer. Though he is not well-educated, he knows his work well. He is very hard-working. He is truthful and honest. All the villagers respect him. They greet him respectfully. They often consult him about their private affairs. My father has an open mind. He accepts things easily. Moreover, he settles the quarrels among the people of the village. He is the wisest man of the village. He leads a simple and clean life. He does not lose temper with anybody. He is respected by one and all. I am proud of my father.

Word-Meanings-Well-educated-सुरिक्षित, Hard-working-परिश्मी, Greet-अभिवादन करना, Respectfully-आदरपूर्वक, Consult-सलाह लेना, Accept things-बात मान लेना/समझ जाना, Private affairs-निजी मामले, Quarrels-झगड़े, Lose temper-क्रोध करना।

Completion of Incomplete Paragraphs

1. My Mother-A Nurse My mother is a nurse. She spends her day according to a set time-table. She gets up early in the morning. She prepares bed tea for all of us…………..

My mother is a nurse. She spends her day according to a set time-table. She gets up early in the morning. She prepares bed tea for all of us. Then she goes out for a walk. On coming back, she prepares breakfast. She then changes her clothes. She goes to the hospital for her duty. She takes care of every patient. She looks after them very nicely. She comes back in the evening. She prepares tea. We all sit together and talk for some time. Then she starts preparing dinner for us. Really, my mother is a noble lady.

Word-Meanings : According to-के अनुसार, Prepares-तैयार करती है, Breakfast- नारता, Take care of-देखभाल करना, Patients-मरीज, Together-इकट्ठे, Noble- नेक

2. A Football Match Yesterday, a football match was played between our school and Khalsa High School. It was final match. It was played on our school playground…. ……….

Yesterday a football match was played between our school and Khalsa High School. It was final match. It was played on our school playground. The students of both the schools had come to see the match. The match started at 4 p.m. Both the teams were equally strong. The match was very brisk from the very start. The players were in high spirits. Both the teams tried hard to score a goal. But there was no score till interval. After the interval, our team attacked hard. Our centre forward got the ball and scored the first goal. The other team tried hard to equalize but in vain. After a few minutes, the referee blew a long whistle and the match was over. Our team won the match by one goal to nil.

Word- Meanings: Equally-सामन रूप से, Brisk-तेज/ शेचक
उत्सहित In high spirits-Jnifen, Score a goal-गोल करना, Equalize-बराबर करना।

PSEB 8th Class English Creative Writing

3. An Indian Festival
Or
Diwali

The Diwali is a great Indian festival. It comes off in the month of October or November. It has a story behind it. ……….. .

The Diwali is a great Indian festival. It comes off in the month of October or November. It has a story behind it. Rama came back to Ayodhya after 14 years of exile. This festival is held to mark this day. This festival is held with great pomp and show. People whitewash their houses. They hang pictures on the walls. The Diwali day is a happy day. The bazaars look beautiful. A few years ago the sweets were in great demand on Diwali. But today the taste of the people has changed. They prefer dry fruits to sweets. The Diwali night is a beautiful night. Everywhere there is light. There are rows of candles on the roofs. There is a great noise of crackers. Lakshmi Pooja is held. Some people drink and gamble. It is bad. We should celebrate this festival with purity.

Word-Meanings : Festival-त्योहार, Exile-वनवास, Pomp and show-धूमधाम से, Prefer-अधिक पसन्द करना, Crackers-पटाखे, Gamble-जुआ खेलना, Celebrate-उत्सव मनाना, Purity-fastati

4. A Visit To Shimla

I went to Shimla last year. My uncle was with me. We left by the Kalka Mail. …

I went to Shimla last year. My uncle was with me. We left by the Kalka Mail. It left Jalandhar City at 11 p.m. and reached Kalka at 6 a.m. There we changed the train. The train passed through many tunnels. It snaked its way to Shimla. We stayed there in a good hotel. As we were tired, we slept for three or four hours. In the evening we visited the Mall. There was a great rush at the Mall. The next day we visited Kufri. We also visited Jakhoo hill and other places of interest. We stayed at Shimla for a fortnight. We had a nice time there.

Word-Meanings : Changed the train-रेलगाड़ी बदली, Passed through-गुजरी, Tunnels-सुरंगें, Stayed-ठहरे, Snaked-बल खाते हुए चली, Fortnight-पखवाड़ा (पंद्रह दिन)।

5. The Morning Assembly of Our School

Our school begins at half past eight in the morning. All the students and the teachers reach the school in time. The morning assembly begins……….

Our school begins at half past eight in the morning. All the students and the teachers reach the school in time. The morning assembly begins. All the students stand in rows. They stand facing the school building. Four boys stand facing the assembly. They say prayer. The other students say after them. A student from the seniormost class reads out the main news of the day. A teacher gives the thought for the day and explains it. Sometimes our Headmaster makes some important announcement. Then there is roll-call. The assembly ends with the National Anthem and we go to our class-rooms.

Word-Meanings : Morning Assembly–प्राता: कालीन प्रार्थन साभा, Rows-पंक्तियां, Seniormost सबसे ऊंची, Thought-विचार, Explain-व्याख्या करना, Announcement-घोषणा, Rollcall – हाजिरी

6. A Fire Scene
Or
A House On Fire

It was a summer night. I was sleeping on the roof of my house. I was awakened by the cries of Fire ! Fire ! I got up and reached the spot………….

It was a summer night. I was sleeping on the roof of my house. I was awakened by the cries of Fire ! Fire ! I got up and reached the spot. The house of the Mahajans was on fire. I saw flames rising higher. Many people had gathered there. Some were throwing water on the fire. Others were throwing sand. A few people were taking out goods. The youngest son was sleeping in the upper storey. He was crying loudly. I at once took a ladder and climbed up. The boy was safe. I took him down before the fire reached the upper storey. The boy’s mother was very happy. She thanked and blessed me. Soon the fire-brigade arrived. The firemen struggled hard to put out the fire. After an hour, they brought the fire under control. There was a great loss of property. But thank God there was no loss of life.

Word-Meanings: Awaken- नींद से उठना, Spot-जगह, Flames-लपटें, Sand-रेत, Upper storey-ऊपरी मंज़िल, Ladder-सीढ़ी, Blessed-आशीर्वाद दिया, Under control नियन्त्रण में, Property-सम्मत्ति

C-Picture Description

1.
PSEB 8th Class English Creative Writing 5
A boy was coming back from the school. He was on a bicycle. He was cycling very fast. A stone was lying on the road. He did not notice it. His bicycle hit it and lost its balance. He fell down and injured his leg.

One of his schoolmates was coming after him. She was on foot. She saw her wounded classmate and a clinic on the road side. Unluckily, the clinic was closed. Soon the girl remembered a piece of bandage in her bag. She bandanged the wound of the boy. The blood stopped and the boy started walking with her. Soon the boy’s parents reached there. They thanked the girl for her timely help.

2.
PSEB 8th Class English Creative Writing 6
Some monkeys were playing in a jungle. Suddenly, they heard the loud sound of an elephant’s trumpet. They got afraid. They ran up and down a tree.

Soon the elephant reached very close to the tree. A little monkey looked towards the elephant and smiled.

The elephant picked him up and put him on his back. The little monkey had a joyful ride. The other monkeys saw him and one by one also jumped on to elephant’s back. The elephant started moving. The monkeys had a jolly good elephant ride.

3.
PSEB 8th Class English Creative Writing 7
Once there was a donkey. One day he was wandering in the forest. Suddently he found a lion’s skin lying on the ground. He wore it. Animals saw him and ran away with fear. The donkey was very happy. After some time he heard some donkeys braying. He also stared braying. The animals heard his braying and came to know that he was donkey not a lion. They chased him and drove him away from the forest.

4.
PSEB 8th Class English Creative Writing 8
A man was going to the market with his son and ass (गंधा). They met a couple (दम्मत्ति पति- पत्नी) on the way. ‘Why walk when you have an ass to ride ?” called out the husband ‘Seat (बिठा दो) at the boy on the ass.” “I would like that,” said the boy. “Help me up. Father.”

Soon they met another couple. How shameful; (शर्म की बात हे) of you !” cried the woman. “Let your father ride. Won’t he be tired ?” So, the boy got down and the father ride the ass.

Again they marched on. “Poor boy”, “said the next person they met. “Why should the lazy father ride while his son is walking ?” So, the boy got onto the ass too. As they went on, they met some travellers (यात्री). “How cruel of them! They are up to kill the poor ass” cried one of the travellers. Hearing this the father and the son got down. Now they decided (फैसला किया) to carry the ass on their shoulders. As they did so, the travellers brake into laughter (जोर से हंसने लगे) The laughter frightened (डरा दिया) the ass. It broke free and ran away.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 1.
Given here are some figures.
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 1
Classify each of them on the basis of the following:
(a) Simple curve
(b ) Simple closed curve
(c) Polygon
(d ) Convex polygon
(e) Concave polygon
Solution:
(a) Simple curves : (i), (ii), (v), (vi) and (vii)
(b) Simple closed curves : (i), (ii), (v), (vi) and (vii)
(c) Polygons : (i) and (ii)
(d) Convex polygon : (ii)
(e) Concave polygons : (i)

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 2.
How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle
Solution :
(a) In a convex quadrilateral, number of sides (n) = 4
∴ Number of diagonals = \(\frac{n(n-3)}{2}\)
= \(\frac{4(4-3)}{2}\)
= \(\frac{4 \times 1}{2}\)
= 2

(b) In a regular hexagon, number of sides (n) = 6
∴ Number of diagonals = \(\frac{n(n-3)}{2}\)
= \(\frac{6(6-3)}{2}\)
= \(\frac{6 \times 3}{2}\)
= 9

(c) In a triangle, number of sides (n) = 3
∴ Number of diagonals = \(\frac{n(n-3)}{2}\)
= \(\frac{3(3-3)}{2}\)
= \(\frac{3 \times 0}{2}\)
= 0.

Question 3.
What is the sum of the measures of the angles of a convex quadrilateral ? Will this property hold if the quadrilateral is not convex ? (Make a non-convex quadrilateral and try!)
Solution:
The sum of measures of angles of a convex quadrilateral = 360°
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 2
Yes, this property holds, even if the quadrilateral is not convex.
Here, quadrilateral ABCD is a non-convex quadrilateral.
m∠A + m∠B + m∠C + m∠D
= 40° + 55° + 35° + 230°
= 360°

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 4.
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 3
What can you say about the angle sum of a convex polygon with number of sides ?
(a) 7 (b) 8 (c) 10 (d) n
Solution :
From the above table, we conclude that sum of interior angles of polygon with n-sides = (n – 2) × 180°
(a) When n = 7
Sum of interior angles of a polygon of 7 sides = (n – 2) × 180°
= (7 – 2) × 180°
= 5 × 180°
= 900°

(b) When n = 8
Sum of interior angles of a polygon of 8 sides = (n – 2) × 180°
= (8 – 2) × 180°
= 6 × 180°
= 1080°

(c) When n = 10
Sum of interior angles of a polygon having n-sides = (n – 2) × 180°
= (10 – 2) × 180°
= 8 × 180°
= 1440°

(d) When n = n
Sum of interior angles of a polygon having n-sides = (n – 2) × 180°

Question 5.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides
Solution:
A polygon is said to be a regular polygon if:
(1) the measures of its interior angles are equal.
(2) the lengths of its sides are equal.

The name of a regular polygon having:
(i) 3 sides is a ‘equilateral triangle’.
(ii) 4 sides is a ‘square’.
(iii) 6 sides is a ‘regular hexagon’.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

Question 6.
Find the angle measure x in the following figures.
(a)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 4
Solution:
The sum of the interior angles of a quadrilateral = 360°
∴ x + 120° + 130° + 50° = 360°
∴ x + 300° = 360°
∴ x = 360° – 300°
(Transposing 300° to RHS)
∴ x = 60°

(b)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 5
Solution:
The sum of the interior angles of a quadrilateral = 360°
∴ x + 60° + 70° + 90° = 360°
∴ x + 220° = 360°
∴ x = 360° – 220° (Transposing 220° to RHS)
∴ x = 140°

(c)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 6
Solution:
Interior angles are : 30°, x, x,
(180° – 70°) = 110° (Linear pair) and
(180° – 60°) = 120° (Linear pair)
The given figure is a pentagon.
∴ Sum of interior angles of a pentagon = (n – 2) × 180°
= (5 – 2) × 180°
= 3 × 180° = 540°
∴ 30° + x + x + 110° + 120° = 540°
∴ 2x + 260° = 540°
∴ 2x = 540° – 260° (Transposing 260° to RHS)
∴ 2x = 280°
∴ \(\frac{2 x}{2}=\frac{280^{\circ}}{2}\)
(Dividing both the sides by 2)
∴ x = 140°

(d)
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 7
Solution:
It is a regular pentagon.
Sum of all interior angles of a pentagon
= (5 – 2) × 180°
= 3 × 180°
= 540°
∴ x + x + x + x + x = 540°
∴ 5x = 540°
∴ \(\frac{5 x}{5}=\frac{540^{\circ}}{5}\)
(Dividing both the sides by 5)
∴ x = 108°

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1

7. (a).
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 8
Find x + y + z
Solution:
x + 90° = 180° (Linear pair)
∴ x = 180° – 90°
∴ x = 90°
y = 30° + 90° = 120°
(Exterior angle of a triangle is equal to the sum of interior opposite angles)
z = 180° – 30° (Linear pair) = 150°
Now, x + y + z = 90° + 120° + 150°
= 360°

(b).
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 9
Find x + y + z + ω
Solution:
The sum of interior angles of a quadrilateral = 360°
PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.1 10
∴ a + 60° + 80° + 120° = 360°
∴ a + 260° = 360°
∴ a = 360° – 260° = 100°
Now, x + 120° = 180° (Linear pair)
∴ x = 180°- 120° = 60°
y + 80° = 180° (Linear pair)
∴ y = 180° – 80° = 100°
z + 60° = 180° (Linear pair)
∴ z = 180° – 60° = 120°
ω + 100° = 180°
∴ ω = 180° – 100° = 80°
Thus, x + y + z + ω = 60° + 100° + 120° + 80°
= 360°