Bhagya

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer:
The batswoman played 30 balls. Hence, the total number of trials = 30. If the event that she did not hit a boundary is denoted by A, then. the number of trials when event A occured is 30 – 6 = 24.
∴ p(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{24}{30}\)
= \(\frac{4}{5}\)
Thus, the probability that she did not hit a boundary is \(\frac{4}{5}\).

Question 2.
1500 families with 2 children were selected randomly, and the following data were s recorded:

Compute the probability of a family, chosen at random, having
(i) 2 girls
(ii) 1 girl
(iii) No girl. Also check whether the sum of these probabilities is 1.
Answer:
Here, the total number of families is 1500.
Hence, the total number of trials = 1500

(i) Let event A denote the event that the family chosen at random is having 2 girls.
Then, the number of trials when event A occured is 475.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{475}{1500}\)
= \(\frac{19}{60}\)

(ii) Let event B denote the event that the family chosen at random is having 1 girl.
Then, the number of trials when event B occured is 814.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{814}{1500}\)
= \(\frac{407}{750}\)

(iii) Let event C denote the event that the family chosen at random Is having no girl.
Then, the number of trials when event C occured is 211.
∴ p(C) = \(\frac{\text { No. of trials in which event } \mathrm{C} \text { occured }}{\text { The total number of trials }}\)
= \(\frac{211}{1500}\)
Now,
P(A) + P(B) + P(C) = \(\frac{19}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to sum no. 5 of “Sums to Enrich ‘Remember’” in chapter 14. Find the probability that a student of the class was born in August.
Answer:
From the Bar graph in the sum which is referred here, we get the following information:

Total number of students = 40 and the number of students born in August = 6.
Hence, if event A denotes the event that a student of the class is born in August, then the number of trials when event A occured is 6 and the total number of trials is 40.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{6}{40}\)
= \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer:
Here, the total number of trials = 200. If event A denotes the event that 2 heads come up, then the number of trials when event A occured is 72.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{72}{200}\)
= \(\frac{9}{25}\)

Question 5.
An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below :

Suppose a family is chosen. Find the probability that the family chosen is ( i ) earning ? 10000- ? 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not .own any vehicle.
(iv) earning ₹ 13000 – ₹ 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer:
Here, the total number of families is 2400. Hence, the total number of trials = 2400

(i) Let event A denote the event that the family is earning ₹ 10000 – ₹ 13000 per month and owning exactly 2 vehicles.
Then, the number of trials when event A occured = 29.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{29}{2400}\)

(ii) Let event B denote the event that the family is earning ₹ 16000 or more per month and owning exactly 1 vehicle.
Then, the number of trials when even B occured = 579.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{579}{2400}\)
= = \(\frac{193}{800}\)

(iii) Let event C denote the event that the family is earning less than ₹ 7000 per month and does not own any vehicle.
Then, the number of trials when event C occured = 10.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{10}{2400}\)
= = \(\frac{1}{240}\)

(iv) Let event D denote the event that the family is earning ? 13000 -? 16000 per month and is owning more than 2 vehicles. Then, the number of trials when event D occured = 25.
∴ P(D) = \(\frac{\text { No. of trials in which event D occured }}{\text { The total number of trials }}\)
= \(\frac{25}{2400}\)
= \(\frac{1}{96}\)

(v) Let event E denote the event that the family is owning not more than 1 vehicle, i.e., 1 vehicle or no vehicle.
Then, the number of trials when event E occured.
= 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
∴ P(E) = \(\frac{\text { No. of trials in which event E occured }}{\text { The total number of trials }}\)
= \(\frac{2062}{2400}\)
= \(\frac{1031}{1200}\)

Question 6.
Refer to table 7 of sum no. 7 in “Sums to Enrich ‘Remember’” in chapter 14.
(i) Find the probability that a student obtained less than 20 marks in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer:
According to the table referred here, the total number of students = 90.
Hence, the total number of trials = 90.
(i) According to the same table, the number of students who obtained less than 20 marks in the mathematics test is 7. So, if the event that a student obtained less than 20 marks in mathematics test is called event A, then the number of trials when event A occured is 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{90}\)

(ii) Let event B denote the event that a student obtained 60 or more marks. Then, , according to the same table, the number of trials when event B occured = 15 + 8 = 23.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table:

Opinion	Number of students
Like	135
Dislike	65

Find the probability that a student chosen at random
(i) Likes statistics,
(ii) Does not like it.
Answer:
Here, the total number of students = 200.
Hence, the total number of trials = 200.

(i) Let event A denote the event that a student likes statistics.
Then, the number of trials when event A occured = 135
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{135}{200}\)
= \(\frac{27}{40}\)

(ii) Let event B denote the event that a student does not like statistics. Then, the number of trials when event B occured = 65.
∴ P(B) = \(\frac{\text { No. of trials in which event B occured }}{\text { The total number of trials }}\)
= \(\frac{65}{200}\)
= \(\frac{13}{40}\)

Question 8.
Refer to sum no. 2, Exercise 14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work ?
(ii) more than or equal to 7km from her place of work ?
(iii) within \(\frac{1}{2}\)km from her place to work?
Answer:
The total number of observations in the question referred here is 40.
Hence, the total number of trials = 40.

(i) Let event A denote the event that the distance between her residence and the place of work is less than 7 km. Then there are 9 such observations, viz., 5, 3, 2, 3, 6, 5, 6, 2, 3.
Hence, the number of trials when event A occured = 9.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{9}{40}\)

(ii) Let event B denote the event that the said distance is 7 km or more than 7 km. Then, all the remaining 31(40-9) observations refer to event B.
Hence, the number of trials when event B occured = 31
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{31}{40}\)

(iii) Let event C denote the event that the engineer lives within \(\frac{1}{2}\) km from her place of work. There is no observation which is \(\frac{1}{2}\) or less than \(\frac{1}{2}\).
Hence, the number of trials when event C occured = 0.
∴ P(C) = \(\frac{\text { No. of trials in which event C occured }}{\text { The total number of trials }}\)
= \(\frac{0}{40}\)
= 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Note: Students should do this Activity themselves.

Question 10.
Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her / him is divisible by 3 ? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer:
Note: Students should do this Activity themselves.

Question 11.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg) :
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer:
The total number of bags = 11.
Hence, the total number of trials = 11.
Let event A denote the event that a bag contains more than 5 kg of flour.
There are 7 bags weighing more than 5 kg.
Their weights (in kg) are 5.05, 5.08, 5.03, 5.06, 5.08, 5.04 and 5.07. Hence, the number of trials when event A occured = 7.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{7}{11}\)

Question 12.
In sum no. 5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Answer:
In sum no. 5, Exercise 14.2, total number of days is 30.
Hence, the total number of trials = 30.
In the table prepared there, we see that the frequency of class 0.12 – 0.16 is 2.
Hence, during 2 days the concentration of sulphur dioxide (in ppm) was in the interval 0.12 – 0.16.
Let event A denote the event that the concentration of sulphur dioxide (in ppm) is in the interval 0.12 – 0.16.
Hence, the number of trials when event A occured = 2.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{2}{30}\)
= \(\frac{1}{15}\)

Question 13.
In sum no. 1, Exercise 14.2, you were asked) to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer:
In sum no. 1, Exercise 14.2, the total number of students is 30.
Hence, the total number of trials = 30.
Let event A denote the event that a student has blood group AB. The number of students having blood group AB is 3.
Hence, the number of trials when event A occured = 3.
∴ P(A) = \(\frac{\text { No. of trials in which event A occured }}{\text { The total number of trials }}\)
= \(\frac{3}{30}\)
= \(\frac{1}{10}\)

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 14 Statistics MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The marks scored by Kavya in 10 tests of Mathematics are 35, 18, 41, 24, 45, 10, 28, 32, 40, 15. Then, the range of the data is …………….. .
A. 45
B. 10
C. 35
D. 28.8
Answer:
C. 35

Question 2.
The average of the observations 3, 4, 5, 8, 12, 10, 13, 16, 18, 11 is …………………. .
A. 100
B. 10
C. 18
D. 3
Answer:
B. 10

Question 3.
The mean of first five odd natural numbers is ……………….. .
A. 3
B. 5
C. 4
D. 25
Answer:
B. 5

Question 4.
The mean of first four even natural numbers is ……………….. .
A. 5
B. 10
C. 20
D. 4
Answer:
A. 5

Question 5.
The mean of first five prime numbers is
A. 28
B. 2.8
C. 5.6
D. 1.4
Answer:
C. 5.6

Question 6.
If the mean of 2x, 5, 3x, 12, 5x, 17 and 6 is 20, then x = ………………….. .
A. 10
B. 20
C. 15
D. 40
Answer:
A. 10

Question 7.
The mean of the following distribution is ………………. .

A. 3.9
B. 7.8
C. 78
D. 39
Answer:
A. 3.9

Question 8.
If the mean of 12, 13, x, 17, 18 and 20 is 16, then x = ………………. .
A. 8
B. 4
C. 16
D. 32
Answer:
C. 16

Question 9.
For a given frequency distribution, n = 20 and Σf_ix_i = 140, then X̄ = ………………… .
A. 20
B. 14
C. 7
D. 28
Answer:
C. 7

Question 10.
The mean of \(\frac{2}{5},\), \(\frac{5}{7},\), \(\frac{3}{5},\) and \(\frac{2}{7},\) is ……………… .
A. \(\frac{1}{2},\)
B. \(\frac{3}{5},\)
C. \(\frac{5}{7},\)
D. 2
Answer:
A. \(\frac{1}{2},\)

Question 11.
The median of 14, 6, 2, 13, 9, 15 and 12 is …………………. .
A. 12
B. 10
C. 2
D. 15
Answer:
A. 12

Question 12.
The median of 21, 17, 13, 33, 19, 23 is ………………… .
A. 21
B. 20
C. 33
D. 19
Answer:
B. 20

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches:
2, 3. 4, 5, 0. 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer:
Here, n = 10.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\frac{2+3+4+5+0+1+3+3+4+3}{10}\)
= \(\frac{28}{10}\)
= 2.8
Thus, the mean of the given scores is 2.8 goals.

Arranging the observations in the ascending order, we get:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Since n = 10 is an even number, \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6.

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
= \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
= \(\frac{3+3}{2}\) = 3
Thus, the median of the given scores is 3 goals.
In the given data, observation 3 occurs most frequently (4 times). Hence, the mode of the data is 3 goals.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer:
Here, n = 15.
Mean X̄ = \(\frac{\Sigma x_{i}}{n}\)
= \(\begin{gathered}
41+39+48+52+46+62+54+40 \\
+96+52+98+40+42+52+60 \\
\hline 15
\end{gathered}\)
= \(\frac{822}{15}\) = 54.8
Thus, the mean of the data is 54.8 marks.
Arranging the observations in the ascending order, we get:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 is an odd number.
Median M = \(\left(\frac{n+1}{2}\right)\)th observation
= \(\left(\frac{15+1}{2}\right)\)th observation
= 8 th observation
= 52
Thus, the median of the data is 52 marks.
In the given data, observation 52 occurs most frequently (3 times). Hence, the mode of the data is 52 marks.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer:
Here, the median = 63 and n = 10.
∴ \(\frac{n}{2}\) = 5 and \(\frac{n}{2}\) + 1 = 6

Median M
= \(\frac{\left(\frac{n}{2}\right) \text { th observation }+\left(\frac{n}{2}+1\right) \text { th observation }}{2}\)
∴ 63 = \(\frac{5 \text { th } \text { observation }+6 \text { th } \text { observation }}{2}\)
∴ 63 = \(\frac{(x)+(x+2)}{2}\)
∴63 × 2 = x + x + 12
∴126 = 2x + 2
∴ 2x = 124
∴ x = 62

Question 4.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer:
Here, just by simple observation, it is clearly seen that observation 14 occurs most frequently, i.e., 4 times.
Hence, the mode of the data is 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table:

Answer:

Mean X̄ = \(\frac{\Sigma f_{i} x_{i}}{n}\)
= \(\) = \(\frac{3,05,000}{60}\) = 5083.33
Thus, the mean salary is ₹ 5083.33.

Question 6.
Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
Answer:
For the students studying in the same class, usually their level of knowledge and understanding would be more or less equal. There would be a few student having this level low and there would be a few students having this level high. Their level of knowledge and understanding would be reflected in the marks scored by them at an examination. Hence, the mean of marks scored by them at an examination is an appropriate measure of central tendency.

(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
If we consider the monthly income of the people of certain region, the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %):

Causes	Female fatality rate (%)
1. Reproductive health conditions	31.8
2. Neuropsychiatric conditions	25.4
3. Injuries	12.4
4. Cardiovascular conditions	4.3
5. Respiratory conditions	4.1
6. Other causes	22.0

(i) Represent the information given above graphically.
Answer:

(ii) Which condition is the major cause of women’s ill health and death worldwide?
Answer:
‘Reproductive health conditions’ is the major cause of womens ill health and death worldwide.

(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Answer:
‘Malnutrition’ and ‘Lack of necessary medical facilities’ can be considered as two other factors which play a major role in female fatality.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

Section	Number of girls per thousand bays
Scheduled Caste (SC)	940
Scheduled Tribe (ST)	970
Non-SC/ST	920
Backward districts	950
Non-backward districts	920
Rural	930
Urban	910

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Answer:

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.
Answer:
Seats won by different political parties

(ii) Which political party won the maximum number of seats?
Answer:
Political party: A won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	.5
163-171	4
172-180	2

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous.]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Answer:
Making the class intervals continuous, we get the following table:

Length (in mm)	Number of leaves
117.5-126.5	3
126.5- 135.5	5
135.5-144.5	9
144.5-153.5	12
153.5- 162.5	5
162.5-171.5	4
171.5-180.5	2

(i) Length of leaves in millimetre

(ii) Yes. The given data can also be represented by ‘Frequency polygon’.

(iii) It is not correct to conclude that the maximum number of leaves are 153 mm long, because even if the frequency of class 145-153 is 12, we do not have the information about the length of each of those 12 leaves individually.

Question 5.
The following table gives the life times of 400 neon lamps:

Life time (in hours)	Number of lamps
300 – 400	14
400 – 500	56
500 – 600	60
600 – 700	86
700 – 800	74
800 – 900	62
900 – 1000	48

(i) Represent the given information with the help of a histogram.
Answer:

(ii) How many lamps have a life time of 700 hours or more than 700 hours ?
Answer:
The-frequencies of classes 700-800, 800-900 and 900-1000 are 74, 62 and 48 respectively.
Hence, the life time of 184 (74 + 62 + 48) lamps is 700 hours or more than 700 hours.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Answer:
To draw the frequency polygons of both the sections, we find the class marks of each class and prepare the following tables:

Section A

Marks	Class mark	Frequency
0-10	5	3
10-20	15	9
20-30	25	17
30-40	35	12
40-50	45	9

Section B

Marks	Class mark	Frequency
0-10	5	5
10-20	15	19
20-30	25	15
30-40	35	10
40-50	45	1

Comparing the performance of both the sections from the frequency polygons, we observe that the performance of students of section A is better than the performance of students of section B.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Number of balls	Team A	Team B
1-6	2	5
7-12	1	6
13-18	8	2
19-24	9	10
25-30	4	5
31-36	5	6
37-42	6	3
43-48	10	4
49-54	6	8
55-60	2	10

Represent the data of both the teams on the same graph by frequency polygons.
[Hint: First make the class intervals continuous.]
Answer:

Number of runs made by Team A and Team B in first 60 balls.

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.
Answer:

Children of various age groups playing in a park

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	Number of surnames
1-4	6
4-6	30
6-8	44
8-12	16
12-20	4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Answer:

(i) Information regarding the number of surnames having given number of letters
Answer:

(ii) Write the class interval in which the maximum number of surnames lie.
Answer:
The maximum number of surnames lie in the class interval 6-8.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see the given figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Answer:
Outer faces to be polished:

• One face on back side of the bookshelf, measuring 110 cm × 85 cm.
• Two faces on the sides, each of those measuring 110 cm × 25 cm.
• The top and the base, each of those measuring 85 cm × 25 cm.
• Two vertical strips on the front side, each of those measuring 110 cm × 5 cm.
• Four horizontal strips on the front side, each of those measuring 75 cm × 5 cm.

Thus, total area of region to be polished
= [(110 × 85) + 2(110 × 25) + 2 (85 × 25) + 2(110 × 5) + 4(75 × 5)] cm²
= (9350 + 5500 + 4250 + 1100+ 1500) cm²
= 21700 cm²
20 paise per cm² = ₹ 0.20 per cm²
Cost of polishing 1 cm2 region = ₹ 0.20
∴ Cost of polishing 21700 cm² region
= ₹ (21700 × 0.20)
= ₹ 4340

Inner faces to be painted:

• Two faces on the sides each of those measuring 90 cm × 20 cm.
• Two faces each of two shelves, the top face and the bottom face, in all six face, each of those measuring 75 cm × 20 cm.
• Face on the back side, measuring 90 cm × 75 cm.

Thus, total area of the region to be painted
= [2 (90 × 20) + 6 (75 × 20) + (90 × 75)] cm²
= (3600 + 9000 + 6750) cm²
= 19350 cm²
10 paise per cm² = ₹0.10 per cm²
Cost of painting 1 cm² region = ₹ 0.10
∴ Cost of painting 19350 cm² region = ₹ (19350 × 0.10) = ₹ 1935
Then, the total expense of polishing and painting = ₹ 4340 + ₹ 1935 = ₹ 6275

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be ‘ painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Answer:
For each wooden sphere,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{21}{2}\) cm
Curved surface area of 1 sphere
= 4πr²
= 4 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{21}{2}\) cm²
= 1386 cm²
For each cylindrical support, radius r = 1.5 cm and height h = 7 cm.
Area of top of cylindrical support
= πr²
= \(\frac{22}{7}\) × 1.5 × 1.5 cm²
= 7.07 cm² (approx.)
Hence, the area of each sphere to be painted silver = 1386 cm² – 7.07 cm² = 1378.93 cm²
∴ Total area of eight spheres to be painted silver = 1378.93 cm² × 8 = 11031.44 cm²
25 paise per cm² = ₹ 0.25 per cm²
Cost of painting silver in 1 cm² region = ₹ 0.25
∴ Cost of painting silver in 11031.44 cm² region
= ₹ (11031.44 x 0.25)
= ₹ 2757.86 (approx.)
Curved surface area of 1 cylindrical support
= 2πrh
= 2 × \(\frac{22}{7}\) × 1.5 × 7 cm
= 66 cm²
∴ Total area of eight cylindrical supports to be painted black = 66 cm² × 8 = 528 cm²
5 paise per cm² = ₹ 0.05 per cm²
Cost of painting black in 1 cm² region = ₹ 0.05
∴ Cost of painting black in 528 cm² region = ₹ (528 × 0.05)
= ₹ 26.40
Thus, the total cost of painting = ₹ 2757.86 + ₹ 26.40
= ₹ 2784.26 (approx.)

Question 3.
The diameter of a sphere is decreased by 25 %. By what per cent does its curved surface area decrease?
Answer:
Suppose, the initial diameter of the sphere is d units and radius is r units.
∴ d = 2r
Original curved surface area of the sphere
= 4πr²
= π (4r²)
= π (2r)²
= πd² unit²
Now, the diameter of the sphere is reduced by 25 %. Hence, the new diameter of the sphere is 0.75d units.
New curved surface area of the sphere
= π (diameter)
= π (0.75d)² unit²
= 0.5625 πd² unit²
∴ The decrease in the curved surface area of the sphere = πd² – 0.5625 πd²
= 0.4375 πd² unit²
∴Percentage decrease in the curved surface area of the sphere = \(\frac{0.4375 \pi d^{2}}{\pi d^{2}}\) × 100 = 43.75 %
Thus, when the diameter of a sphere is decreased by 25 %, its curved surface area decreases by 43.75 %.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan² 45° + cos² 30° – sin² 60°

(iii)

(iv)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
= \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\)
= \(\frac{3}{4}+\frac{1}{4}\) = 1.

(ii) 2 tan² 45° + cos² 30° – sin² 60° = 2 (tan 45°)² + (cos 30°)² – (sin 60°)²
= 2 (1)² + (\(\frac{\sqrt{3}}{2}\))² – (\(\frac{\sqrt{3}}{2}\))² = 2.

(iii)
= \(\frac{\frac{1}{\sqrt{2}}}{\left(\frac{2}{\sqrt{3}}\right)+(2)}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)

= \(\frac{1}{\sqrt{2}}: \frac{\sqrt{3}}{2+2 \sqrt{3}}=\frac{\sqrt{3}}{2 \sqrt{2}(\sqrt{3}+1)}\)

= \(\frac{\sqrt{3}(\sqrt{3}-1)}{2 \sqrt{2}(\sqrt{3}+1)(\sqrt{3}-1)}\)

= \(\frac{\sqrt{2} \times \sqrt{3} \times(\sqrt{3}-1)}{4(3-1)}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\).

(iv)

= \(\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{3}{2}}\)

= \(\frac{3 \sqrt{3}-4}{4+3 \sqrt{3}}\)

= \(\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}\)

= \(\frac{27+16-24 \sqrt{3}}{27-16}\)

= \(\frac{43-24 \sqrt{3}}{11}\)

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)

= \(\begin{array}{r}
5\left(\cos 60^{\circ}\right)^{2}+4\left(\sec 30^{\circ}\right)^{2} \\
\frac{-\left(\tan 45^{\circ}\right)^{2}}{\left(\sin 30^{\circ}\right)^{2}+\left(\cos 30^{\circ}\right)^{2}}
\end{array}\)

= \(\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)

= \(\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{5}{4}+\frac{1}{3}-1}{\frac{1}{4}+\frac{3}{4}}\)

= \(\frac{5}{4}+\frac{16}{3}-1=\frac{15+64-12}{12}=\frac{67}{12}\).

Question 2.
Choose the correct option and justify your choice.

(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan 45^{\circ}}\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0.

(iii) sin 2A = 2 sin A is true when
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°.

Solution:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

\(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}\) = sin 60°.
So, correct anwer is (A).

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\frac{1-(1)^{2}}{1+(1)^{2}}\) = 0
So, correct anwer is (D).

(iii) Here when A = 0°
L.H.S. = sin 2A = sin 0° = 0
and R.H.S. = 2 sin A = 2 sin 0°
= 2 × 0 = 0
∴ Option (A) is correct.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

= \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}\)

= tan 60°
∴ Option (C) is correct.

Question 3.
If tan (A + B) = \(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° ∠A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = \(\sqrt{3}\). Given
tan (A + B) = tan 60°
⇒ A + B = 60° ……………..(1)
tan (A – B) = \(\frac{1}{\sqrt{3}}\) (Given)
or tan (A – B) = tan 30°
⇒ A – B = 30° …………….(2)
On adding (1) and (2),

A = 45°

Pu value of A = 45° in (1)
45° + B = 60°
B = 60° – 45°
B = 15°
Hence A = 45° and B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin q increases as q increases.
(iii) The value of cos q Increases as q increases
(iv) sin q = cos q for all value of q.
(v) cot A is not defined for A = 0°.
Solution:
(i) False.
When A = 60°, B = 30°
L.H.S. = sin (A + B) = sin (60° + 30°) = sin 90° = 1
R.H.S. = sin A + sin B
= sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\) ≠ 1
i.e., L.H.S. ≠ R.H.S.

(ii) True, sin 30° = \(\frac{1}{2}\) = 0.5,
Note that sin 0° = 0,
sin 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 (approx.)
sin 60° = \(\frac{\sqrt{3}}{2}\) = 0.87 (approx.)
and sin 90° = 1
i.e., value of sin θ increases as θ increases from 0° to 90°.

(iii) False.
Note that cos 0° = 1,
cos 30° = \(\frac{\sqrt{3}}{2}\) = 0.87(approx.)
cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7.(approx.)
cos 60° = \(\frac{1}{2}\) = 0.5
and cos 90° = 0.
Hence, value of θ decreases as θ increases from 0° to 90°.

(iv) False
Since sin 30° = \(\frac{1}{2}\)
and cos 30° = \(\frac{\sqrt{3}}{2}\)
or sin 30° ≠ cos 30°
Only we have: sin 45° = cos 45°.
\(\frac{1}{\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)

(v) True.
cot 0° = \(\frac{1}{\tan 0^{\circ}}=\frac{1}{0}\), or not defined.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers Ex 1.1Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1

1. Represent these numbers on the number line:

Question (i).
\(\frac {7}{4}\)
Solution:
To represent \(\frac {7}{4}\), make 7 markings each of a distance equal to \(\frac {1}{4}\) on the right side of 0. The 7th point represents the rational number \(\frac {7}{4}\).

The point A is \(\frac {7}{4}\).

Question (ii).
\(\frac {-5}{6}\)
Solution:
To represent (\(\frac {-5}{6}\)) on the number line, make 5 markings each of a distance equal to on the left side of 0. The 5th point represents the rational number (\(\frac {-5}{6}\)).

The point B is (\(\frac {-5}{6}\))

2. Represent \(\frac{-2}{11}, \frac{-5}{11}, \frac{-9}{11}\) on the number line.
Solution:
To represent the given rational numbers on a number line, make 11 markings each being equal to distance \(\frac {1}{11}\) on the left of 0.

Here, the point A is (\(\frac {-2}{11}\)).
the point B is (\(\frac {-5}{11}\)).
the point C is (\(\frac {-9}{11}\)).

3. Write five rational numbers which are smaller than 2.
Solution:
There are infinite rational numbers below 2, positive as well as negative.
Five of them are 1, \(\frac {1}{3}\), \(\frac {1}{4}\), 0, – 1.

4. Find ten rational numbers between \(\frac {-2}{5}\) and \(\frac {1}{2}\).
Solution:
First, convert \(\frac {-2}{5}\) and \(\frac {1}{2}\) having the same denominator, such that the difference between the numerators is more than 10.
\(\frac{-2}{5}=\frac{-2}{5} \times \frac{4}{4}=\frac{-8}{20}\);
\(\frac{1}{2}=\frac{1}{2} \times \frac{10}{10}=\frac{10}{20}\)
∴ The ten rational numbers between \(\frac {-8}{20}\) and \(\frac {10}{20}\) are
\(\frac{-7}{20}, \frac{-6}{20}, \frac{-5}{20}, \frac{-4}{20}, \frac{-3}{20}, \ldots, 0, \frac{1}{20}, \ldots, \frac{9}{20} .\)
(There can be many more such rational numbers.)

5. Find five rational numbers between

Question (i).
\(\frac {2}{3}\) and \(\frac {4}{5}\)
Solution:
First, convert \(\frac {2}{3}\) and \(\frac {4}{5}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{2}{3}=\frac{2}{3} \times \frac{20}{20}=\frac{40}{60}\);
\(\frac{4}{5}=\frac{4}{5} \times \frac{12}{12}=\frac{48}{60}\)
∴ The five rational numbers between \(\frac {2}{3}\) and \(\frac {4}{5}\) are \(\frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60}, \frac{46}{60}\).

Question (ii).
\(\frac {-3}{2}\) and \(\frac {5}{3}\)
Solution:
First, convert \(\frac {-3}{2}\) and \(\frac {5}{3}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{-3}{2}=\frac{-3}{2} \times \frac{3}{3}=\frac{-9}{6}\);
\(\frac{5}{3}=\frac{5}{3} \times \frac{2}{2}=\frac{10}{6}\)
∴ The five rational numbers between \(\frac {-3}{2}\) and \(\frac {5}{3}\) are \(\frac{-8}{6}, \frac{-7}{6}, 0, \frac{7}{6}, \frac{8}{6}\).

Question (iii).
\(\frac {1}{4}\) and \(\frac {1}{2}\)
Solution:
First, convert \(\frac {1}{4}\) and \(\frac {1}{2}\) having the same denominator, such that the difference between the numerators is more than 5.
\(\frac{1}{4}=\frac{1}{4} \times \frac{8}{8}=\frac{8}{32}\);
\(\frac{1}{2}=\frac{1}{2} \times \frac{16}{16}=\frac{16}{32}\)
∴ The five rational numbers between \(\frac {1}{4}\) and \(\frac {1}{2}\) are \(\frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}, \frac{14}{32}\).
(There can be many more such rational numbers.)
[Note : You can write rational numbers of your choice.]

6. Write five rational numbers greater than -2.
Solution:
There can be many rational numbers greater than – 2. Five of them are \(\frac{-3}{2}, \frac{-1}{4}, 0, \frac{1}{2}, \frac{1}{5}\).

7. Find ten rational numbers between \(\frac {3}{5}\) and \(\frac {3}{4}\).
Solution:
First, convert \(\frac {3}{5}\) and \(\frac {3}{4}\) having the same denominator, such that the difference between the numerators is more than 10.
\(\frac{3}{5}=\frac{3}{5} \times \frac{20}{20}=\frac{60}{100}\);
\(\frac{3}{4}=\frac{3}{4} \times \frac{25}{25}=\frac{75}{100}\)
∴ The ten rational numbers between \(\frac {3}{5}\) and \(\frac {3}{4}\) are \(\frac{61}{100}, \frac{62}{100}, \frac{63}{100}, \frac{64}{100}, \frac{65}{100}, \frac{66}{100}, \frac{67}{100}, \frac{68}{100},\)\(\frac{69}{100}, \frac{70}{100}\)

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1

Question 1.
In ∆ABC, right angled at B, AB = 24 cm; BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C.
Solution:
(i) We are to find sin A .cos A AB = 24 cm; BC = 7 cm
By using Pythagoras Theorem,

AC² = AB² + BC²
AC² = (24)² + (7)²
AC² = 576 + 49
AC² = 625
AC = \(\sqrt{625}\)
AC = 25 cm.
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

sin A = \(\frac{7 \mathrm{~cm}}{25 \mathrm{~cm}}=\frac{7}{25}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24 \mathrm{~cm}}{25 \mathrm{~cm}}\)

cos A = \(\frac{24}{25}\)

Hence sin A = \([latex]\frac{7}{25}\)[/latex] and cos A = \([latex]\frac{24}{25}\)[/latex].

(ii) sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24 \mathrm{~cm}}{25 \mathrm{~cm}}\)

sin C = \(\frac{24}{25}\)

cos C = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7 \mathrm{~cm}}{25 \mathrm{~cm}}\)

cos C = \(\frac{7}{25}\)

Hence sin C = \(\frac{24}{25}\) and cos C = \(\frac{7}{25}\).

Question 2
In fig., find tan P – cot R.

Solution:
Hyp. PR = 13 cm

By using Pythagoras Theorem,
PR² = PQ² + QR²
or (13)² = (12)² + QR²
or 169 = 144 + (QR)²
or 169 – 144 = (QR)²
or 25 = (QR)²
or QR = ± \(\sqrt{25}\)
or QR = 5, – 5.
But QR = 5 cm.
[QR ≠ – 5, because side cannot be negative]
tan P = \(\frac{R Q}{Q P}=\frac{5}{12}\)

cot R = \(\frac{R Q}{P Q}=\frac{5}{12}\)

∴ tan P – cot R = \(\frac{5}{12}-\frac{5}{12}\) = 0
Hence tan P – cot R = 0.

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:
Let ABC be any triangle with right angle at B.

sin A = \(\frac{3}{4}\)
But sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) [From figure]
∴ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}\)
But \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3}{4}\) = K
where K, is constant of proportionality.
⇒ BC = 3K, AC = 4K
By using Pythagoras Theorem,
AC² = AB² + BC²
or (4K)² = (AB)² + (3K)²
or 16K² = AB² + 9K²
or 16K² – 9K² = AB²
or 7K² = AB²
or AB = ± \(\sqrt{7 K^{2}}\)
or AB = ± \(\sqrt{7} \mathrm{~K}\)
[AB ≠ \(\sqrt{7 K}\) because side of a triangle cannot be negative]

⇒ AB = \(\sqrt{7} \mathrm{~K}\)
cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
cos A = \(\frac{\sqrt{7} K}{4 K}=\frac{\sqrt{7}}{4}\)
tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 \mathrm{~K}}{\sqrt{7} \mathrm{~K}}=\frac{3}{\sqrt{7}}\)

Hence cos A = \(\frac{\sqrt{7}}{4}\) and tan A = \(\frac{3}{\sqrt{7}}\).

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
Let ABC be any right angled triangle where A is an acute angle with right angle at B.

15 cot A = 8
cot A = \(\frac{8}{15}\)
But cot A = \(\frac{A B}{B C}\) (fromfig.)
⇒ \(\frac{A B}{B C}=\frac{8}{15}\) = K
where K is constant of proportionality.
AB = 8 K, BC = 15 K
By using Pythagoras Theorem.
AC² = (AB)² + (BC)²
(AC)² = (8 K)² + (15 K)²
(AC)² = 64K² + 225 K²
(AC)² = 289 K²
AC = ± \(\sqrt{289 K^{2}}\)
AC = ± 17 K
⇒ AC = 17K
[AC = – 17 K, Because side cannot be negative]
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{15 \mathrm{~K}}{17 \mathrm{~K}}=\frac{15}{17}\)

sin A = \(\frac{15}{17}\)

sec A = \(\frac{\mathrm{AC}}{\mathrm{AB}}\)

sec A = \(\frac{17 \mathrm{~K}}{8 \mathrm{~K}}=\frac{17}{8}\)

sec A = \(\frac{17}{8}\)

Hence, sin A = \(\frac{15}{17}\) and sec A = \(\frac{17}{8}\).

Question 5.
Given sec θ = \(\frac{13}{2}\), calculate all other trigonometric ratios.
Solution:
Let ABC be any right angled triangle with right angle at B.
Let ∠BAC = θ

sec θ = \(\frac{13}{12}\)

But sec θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) ……….[from fig.]

\(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}\)

But \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{13}{12}\) = k where k is constant of proportionality.
AC = 13 k and AB = 12 k
By using Pythagoras Theorem,
AC² = (AB)² + (BC)²
or (13k)² = (12k)² + (BC)²
or 169k² = 144k² + (BC)²
or 169k² – 144k² = (BC)
or (BC)² = 25k²
or BC = ± \(\sqrt{25 k^{2}}\)
or BC = ± 5k
or BC = 5k.
[BC ≠ – 5k because side cannot be negative]

sin θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5 k}{13 k}=\frac{5}{13}\)
cos θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12 k}{13 k}=\frac{12}{13}\)
tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5 k}{12 k}=\frac{5}{12}\)
cosec θ = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{13 k}{5 k}=\frac{13}{5}\)
cot θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}\).

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, show that LA = LB.
Solution:
Let ABC be any triangle, where ∠A and ∠B are acute angles. To find cos A and cos B.

Draw CM ⊥ AB
∠AMC = ∠BMC = 90°
In right angled ∆AMC,
\(\frac{\mathrm{AM}}{\mathrm{AC}}\) = cos A ……………(1)
In right angled ∆BMC,
\(\frac{\mathrm{BM}}{\mathrm{BC}}\) = cos B ……………(2)
But cos A = cos B [given] ………..(3)
From (1), (2) and (3),
\(\frac{\mathrm{AM}}{\mathrm{AC}}=\frac{\mathrm{BM}}{\mathrm{BC}}\)
\(\frac{\mathrm{AM}}{\mathrm{BM}}=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{CM}}{\mathrm{CM}}\)
∴ ∆AMC = ∆BMC [By SSS similarity]
⇒ ∠A = ∠B [∵ Corresponding angles of two similar triangles are equal].

Question 7.
If cot θ = \(\frac{7}{8}\) evaluate
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot² θ.
Solution:
(i) ∠ABC = θ.
In right angled triangle ABC with right angle at C.

Given that, cot θ = \(\frac{7}{8}\)
But cot θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) [From fig.]
⇒ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\)
Let \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\) = k
where k is constant of proportionality.
⇒ BC = 7k, AC = 8k
By using Pythagoras Theorem,
AB² = (BC)² + (AC)²
or (AB)² = (7k)² + (8k)²
or (AB)² = 49k² + 64k²
or (AB)² = 113 k²
or AB = ± \(\)
AB = \(\sqrt{113 k^{2}}\) k
AB = \(\sqrt{113}\) k
[AB ≠ \(\sqrt{113}\) k because side cannot be negative]

sin θ = \(\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{8 k}{\sqrt{113} k}\)
sin θ = \(\frac{8}{\sqrt{113}}\)
cos θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{7 k}{\sqrt{113} k}=\frac{7}{\sqrt{113}}\)
cos θ = \(\frac{7}{\sqrt{113}}\)

(1 + sin θ) (1 – sin θ) = (1 + \(\frac{8}{\sqrt{113}}\)) (1 – \(\frac{8}{\sqrt{113}}\))
= (1)² – (\(\frac{8}{\sqrt{113}}\))²
[By using formula (a + b) (a – b) = a² – b²]
= 1 – \(\frac{64}{113}\)
(1 + sin θ) (1 – sin θ) = \(\frac{113-64}{113}=\frac{49}{113}\)
(1 + sin θ)(1 – sin θ) = \(\frac{49}{113}\) ……………..(1)

(1 + cos θ) (1 – cos θ) = (1 + \(\frac{8}{\sqrt{113}}\)) (1 – \(\frac{8}{\sqrt{113}}\))
(1)² – (\(\frac{7}{\sqrt{113}}\))²
[By using formula(a + b) (a – b) = a² – b²]
= 1 – \(\frac{49}{113}\) = \(\frac{113-49}{113}\)
(1 + cos θ) (1 – cos θ) = \(\frac{64}{113}\) ……….(2)

Consider, \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{\frac{49}{113}}{\frac{64}{113}}\) [From (1) and (2)]

Hence \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}\)

(ii) cot θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{8}\)
cot² θ = (cot θ)²
cot² θ= (\(\frac{7}{8}\))²
⇒ cot² θ = \(\frac{49}{64}\).

Question 8.
If 3 cot A = 4 check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos² A – sin² A or not.
Solution:
Let ABC be a right angled triangle with right angled at B.

It is given that 3 cot A = 4
cot A = \(\frac{4}{3}\)
But cot A = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) [From fig.]
⇒ \(\frac{A B}{B C}=\frac{4}{3}\)
But \(\frac{A B}{B C}=\frac{4}{3}\) = k
⇒ AB = 4k, BC = 3k
By using Pythagoras Theorem,
(AC)² = (AB)² + (BC)²
(AC)² = (4k)² + (3k)²
(AC)² = 16 k² + 9 k²
(AC)² = 25 k²
AC=± \(\sqrt{25 k^{2}}\)
AC = ± 5k

But AC = 5k.
[AC ≠ – 5k. because side cannot be negative]
sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5}\)

tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{3 k}{4 k}=\frac{3}{4}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5}\)

L.H.S. = \(\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}\)

∴ cos² A – sin² A = \(\frac{7}{25}\) ………..(2)

From (1) and (2),
L.H.S = R.H.S
Hence, \(\frac{1-\tan ^{2} \mathrm{~A}}{1+\tan ^{2} \mathrm{~A}}\) = cos² A – sin² A.

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\). Find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C.
Solution:
(i) Given: ABC with right angled at B

tan A = \(\frac{1}{\sqrt{3}}\) ……………..(1)
But tan A = \(\frac{B C}{A B}\) ……………(2)
From (1) and (2),
\(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}\)
Let \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}\) = k
BC = k, AB = k
where k is constant of proportionality.
In right angled triangle ABC,
By using Pythagoras Theorem,
(AC)² = (AB)² + (BC)²
or (AC)² = (Jk)² + (k)²
or AC² = 3k² + k²
or AC² = 4k²
or AC = ± \(\sqrt{4 k^{2}}\)
AC = ± 2k.
where AC = 2k
[AC ≠ – 2k side cannot be negative]

[sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)

cos C = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{k}{2 k}=\frac{1}{2}\)

cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)

sin C = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}\)] …………….(3)

sin A cos C = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)
cos A sin C = \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{3}{4}\)
sin A cos C + cos A sin C = \(\frac{1}{4}+\frac{3}{4}\)
= \(\frac{1+3}{4}\)
= \(\frac{4}{4}\) = 1
∴ sin A cos C + cos A sin C = 1.

(ii) cos A cos C = \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{4}\) [From (3)]
sin A sin C = \(\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) [From (3)]

cos A cos C – sin A sin C = \(\left(\frac{\sqrt{3}}{4}\right)-\left(\frac{\sqrt{3}}{4}\right)\) = 0.

Question 10.
In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given: ∆PQR, right angled at Q

PR + QR = 25 cm
PQ = 5 cm
In right angled triangle PQR
By using Pythagoras Theorem,
(PR)² = (PQ)² + (RQ)²
or (PR)² = (5)² + (RQ)²
[∴ PR + QR = 25, QR = 25 – PR]
or (PR)² = 25 + [25 – PR]²
or (PR)² = 25 + (25)² + (PR)² – 2 × 25 × PR
or (PR)² = 25 + 625 + (PR)² – 50
or (PR)² – (PR)² + 50 PR = 650
or 50 PR = 650
or PR = \(\frac{650}{50}\)
or PR = 13 cm
QR = 25 – PR
QR = (25 – 13) cm
or QR = 12 cm.

sin P = \(\frac{Q R}{P R}=\frac{12}{13}\)

cos P = \(\frac{P Q}{P R}=\frac{5}{13}\)

tan P = \(\frac{Q R}{P Q}=\frac{12}{5}\)

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1
(ii) sec A = \(\frac{12}{5}\) for some value of angle A.
(iii) cos A is abbreviation used for cosecant of angle A.
(iv) cot A is product of cot and A.
(v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ tan 60° = √3 = 1.732 > 1.

(ii) True; sec A = \(\frac{12}{5}\) = 240 > 1
∵ Sec A is always greater than 1.

(iii) False.
Because cos A is used for cosine A.

(iv) False.
Because cot A is cotangent of the angle A not the product of cot and A.

(v) False; sin θ = \(\frac{4}{3}\) = 1.666 > 1
Because sin θ is always less than 1.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 1 Rational Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 1 Rational Numbers Ex 1.1

1. Using appropriate properties find.

Question (i).
\(-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)
Solution:
\(-\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}\)
= \(-\frac{2}{3} \times \frac{3}{5}-\frac{3}{5} \times \frac{1}{6}+\frac{5}{2}\) (Commutative)
= \(\frac{3}{5} \times\left[-\frac{2}{3}-\frac{1}{6}\right]+\frac{5}{2}\) (Distributive)
= \(\frac{3}{5}\left[\frac{-4-1}{6}\right]+\frac{5}{2}\)
= \(\frac{3}{5}\left[\frac{-5}{6}\right]+\frac{5}{2}\)
= \(\frac{3}{5} \times \frac{-5}{6}+\frac{5}{2}\)
= \(-\frac{1}{2}+\frac{5}{2}\)
= \(\frac{-1+5}{2}\)
= \(\frac {4}{2}\)
= 2

Question (ii).
\(\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}\)
Solution:
\(\frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}\)
= \(\frac{2}{5} \times\left(\frac{-3}{7}\right)+\frac{1}{14} \times \frac{2}{5}-\frac{1}{6} \times \frac{3}{2}\) (Commutative)
= \(\frac{2}{5} \times\left(\frac{-3}{7}+\frac{1}{14}\right)-\frac{1}{6} \times \frac{3}{2}\) (Distributive)
= \(\frac{2}{5} \times\left[\frac{-6+1}{14}\right]-\frac{1}{4}\)
= \(\frac{2}{5} \times \frac{-5}{14}-\frac{1}{4}\)
= \(-\frac{1}{7}-\frac{1}{4}=\frac{-4-7}{28}\)
= \(\frac{-11}{28}\)

2. Write the additive inverse of each of the following:

Question (i).
\(\frac{2}{8}\)
Solution:
Additive inverse of \(\frac{2}{8}\) = \(\frac{-2}{8}\)

Question (ii).
\(\frac{-5}{9}\)
Solution:
Additive inverse of \(\frac{-5}{9}\) = \(\frac{5}{9}\)

Question (iii).
\(\frac{-6}{-5}\)
Solution:
Additive inverse of \(\frac{-6}{-5}\) means \(\frac{6}{5}\) = \(\frac{-6}{5}\)

Question (iv).
\(\frac{2}{-9}\)
Solution:
Additive inverse of \(\frac{2}{-9}\) = \(\frac{2}{9}\)

Question (v).
\(\frac{19}{-6}\)
Solution:
Additive inverse of \(\frac{19}{-6}\) = \(\frac{19}{6}\)

3. Verify that – (- x) = x for

(i) x = \(\frac {11}{15}\)
Solution:
x = \(\frac {11}{15}\)
∴ (-x) = \(\left(\frac{-11}{15}\right)\)
-(-x) = –\(\left(\frac{-11}{15}\right)\)
= \(\frac {11}{15}\) = x
∴ -(-x) = x

(ii) x = \(\frac {-13}{17}\)
Solution:
x = \(\frac {-13}{17}\)
∴ (-x) = \(\left(\frac{-13}{17}\right)\)
= \(\frac {13}{17}\)
-(-x) = –\(\left(\frac{-13}{17}\right)\)
= \(\frac {-13}{17}\) = x
∴ -(-x) = x

4. Find the multiplicative inverse of the following:

Question (i).
-13
Solution:
Multiplicative inverse of -13 = \(\frac {-1}{13}\)

Question (ii).
\(\frac {-13}{19}\)
Solution:
Multiplicative inverse of \(\frac {-13}{19}\) \(\frac {-19}{13}\)

Question (iii).
\(\frac {1}{5}\)
Solution:
Multiplicative inverse of \(\frac {1}{5}\) = 5

Question (iv).
\(\frac{-5}{8} \times \frac{-3}{7}\)
Solution:
\(\left(\frac{-5}{8}\right) \times\left(\frac{-3}{7}\right)\)
= \(\frac{(-5 \times-3)}{8 \times 7}\)
= \(\frac {15}{56}\)
Multiplicative inverse of \(\frac {15}{56}\) = \(\frac {56}{15}\)

Question (v) .
1 × \(\frac {-2}{5}\)
Solution:
-1 × \(\frac {-2}{5}\) = \(\frac{(-1 \times-2)}{5}\)
= \(\frac {2}{5}\)
Multiplicative inverse of \(\frac {2}{5}\) = \(\frac {5}{2}\)

Question (vi).
-1
Solution:
Multiplicative inverse of -1 = (-1)
(∵ \(\frac{1}{(-1)}\) = (-1))

5. Name the property under multiplication used in each of the following:

Question (i).
\(\frac{-4}{5} \times 1=1 \times \frac{-4}{5}=-\frac{4}{5}\)
Solution:
1 is the multiplicative identity.

Question (ii).
\(-\frac{13}{17} \times \frac{-2}{7}=\frac{-2}{7} \times \frac{-13}{17}\)
Solution:
Commutative property of multiplication.

Question (iii).
\(\frac{-19}{29} \times \frac{29}{-19}=1\)
Solution:
Existence of multiplicative inverse.

6. Multiply \(\frac {6}{13}\) by the reciprocal of \(\frac {-7}{16}\).
Solution:
Reciprocal of \(\frac{-7}{16}=\frac{-16}{7}\)
∴ \(\frac{6}{13} \times \frac{-16}{7}\)
= \(\frac{6 \times(-16)}{13 \times 7}\)
= \(\frac {-96}{91}\)

7. Tell what property allows you to compute.
\(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)\) as \(\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}\)
Solution:
In computing
\(\frac{1}{3} \times\left(6 \times \frac{4}{3}\right)\) as \(\left(\frac{1}{3} \times 6\right) \times \frac{4}{3}\)
we use the associativity.

8. Is \(\frac {8}{9}\) the multiplicative inverse of – 1 \(\frac {1}{8}\) ? Why or why not ?
Solution:
\(-1 \frac{1}{8}=\frac{-9}{8}\)
\(\frac{8}{9} \times \frac{-9}{8}\) = (-1)
∴ \(\frac {8}{9}\) is is not the multiplicative inverse of -1 \(\frac {1}{8}\) as product of two multiplicative inverse is always 1.

9. Is 0.3 the multiplicative inverse of 3 \(\frac {1}{3}\) ? Why or why not?
Solution:
0.3 = \(\frac {3}{10}\) and 3 \(\frac {1}{3}\) = \(\frac {10}{3}\)
\(\frac{3}{10} \times \frac{10}{3}\) = 1
∴ the multiplicative inverse of 3 \(\frac {1}{3}\) is 0.3.

10. Write:

Question (i).
The rational number that does not have a reciprocal.
Solution:
The rational number that does not have a reciprocal is 0.

Question (ii).
The rational numbers that are equal to their reciprocals.
Solution:
The rational numbers that are equal to their reciprocals are 1 and (-1).

Question (iii).
The rational number that is equal to its negative.
Solution:
The rational number that is equal to its negative is zero (0).

11. Fill in the blanks:

Question (i).
Zero has ……………. reciprocal.
Solution:
Zero has no reciprocal.

Question (ii).
The numbers ……………. and ……………. are their own reciprocals.
Solution:
The numbers 1 and -1 are their own reciprocals.

Question (iii).
The reciprocal of – 5 is …………….
Solution:
The reciprocal of – 5 is \(\frac {-1}{5}\)

Question (iv).
Reciprocal of \(\frac{1}{x}\), where x ≠ 0 is …………….
Solution:
Reciprocal of \(\frac{1}{x}\), where x ≠ 0 is x

Question (v) .
The product of two rational numbers is always a …………….
Solution:
The product of two rational numbers is always a rational number.

Question (vi).
The reciprocal of a positive rational number is …………….
Solution:
The reciprocal of a positive rational number is positive.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line it + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).
Solution:
Let line 2x + y – 4 = 0 divides the line segment joining the points A (2,- 2) and B(3, 7) at C (x, y) in the ratio k : 1

∴ Coordinates of C are x = \(\frac{3 k+2 \times 1}{k+1}=\frac{3 k+2}{k+1}\) and y = \(\frac{7 k+(-2) \times 1}{k+1}=\frac{7 k-2}{k+1}\)
∴ C \(\left[\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right]\). must lie on the line 2x + y – 4 = 0

i.e., 2\(\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)\) – 4 = 0
or \(\frac{6 k+4+7 k-2-4 k-4}{k+1}\) = 0
or 9k – 2 = 0
or 9k = 2
or k = \(\frac{2}{9}\).
∴ ratio k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9.
Hence required ratio is 2 : 9.

Question 2.
Find a relation between x and y if (x, y) ; (1, 2) and (7, 0) are collinear.
Solution:
Let given points are A (x, y); B (1, 2) and C (7, 0).
Here x₁ = x, x₂ = 1, x₃ = 7
y₁ = y, y₂ = 2, y₃ = 0
∵ Three points are collinear
iff \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) x (2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0 is the required relation.

Question 3.
Find the centre of a cirçle passing through the points (6, —6); (3, —7) and (3,3).
Solution:
Let O (x, y) be the required centre of the circle which passes through points P(6, – 6); Q(3, – 7) and R (3, 3).
∴ radii of circle are equal.

∴ OP = OQ = OR
or (OP)² = (OQ)² = (OR)²
Now, (OP)² = (OQ)²
(x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
or x² + 36 – 12x + y² + 36 + 12y = x² + 9 – 6x + y² + 49 + 14y
or – 12x + 12y + 72 = – 6x + 14y + 58
or – 6x – 2y + 14 = 0
or 3x + y – 7 = 0 ………………(1)
Also, (OQ)² = (OR)²
or (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
or (y + 7)² = (y – 3)²
or y² + 49 + 14y = y² + 9 – 6y
or 20y = – 40
y = \(\frac{-40}{20}\) = – 2
Substitute this value of)’ in (1), we get
3x – 2 – 7 = 0
or 3x – 9 = 0
or 3x = 9
or x = \(\frac{9}{3}\) = 3
∴ Required centre is (3, – 2).

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.
Solution:
Let two opposite vertices of a square ACBD are A (- 1, 2) and B (3, 2) and coordinates of C are (x, y)
∵ Length of each sides of square are equal.
∴ AC = BC
or (AC)² = (BC)²
or (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²

or (x + 1)² = (x – 3)²
or x² + 1 + 2x = x² + 9 – 6x
or 8x = 8
or x = \(\frac{8}{8}\) = 1
Now, in rt ∠d ∆ACB,
Using Pythagoras Theorem,
(AC)² + (BC)² = (AB)²
(x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
or x² + 1 + 2x + y² + 4 – 4y + x² + 9 – 6x + y² + 4 – 4y = 16
or 2x² + 2y² – 4x – 8y + 2 = 0
or x² + y² – 2x – 4y + 1 = 0
Putting the value of x = 1 in (1), we get
(1)² + y² – 2 (1) – 4y + 1 = 0
or y² – 4y = 0
or y (y – 4) = 0
Either y = 0 or y – 4 = 0
Either y = 0 or y = 4
∴ y = 0, 4
∴ Required points are (1. 0) and (1.4).

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There ¡s a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of A PQR if C is the origin? Also calculate the areas of the triangles In these cases. What do you observe?
Solution:
Case I:
When taking A as origin then AD is X-axis and AB is Y-axis.
∴ Coordinates of triangular grassy Lawn
PQR are P (4, 6); Q (3, 2) and R(6, 5).
Here x₁ = 4, x₂ = 3, x₃ = 6
y₁ = 6, y₂ = 2, y₃ = 50
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= \(\frac{1}{2}\) [- 12 – 3 + 24] = \(\frac{9}{2}\)
= 4.5 sq. units.

Case II: When taking C as origin then CB is X – axis and CD is Y – axis.
∴ Coordinates of triangular grassy lawn PQR
are P(12, 2); Q (13,6) and R (10, 3)
Here x₁ = 12, x₂ = 13, x₃ = 10
y₁ = 2, y₂ = 6, y₃ = 3
Now, area of ∆PQR = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)]
= \(\frac{1}{2}\) [36 + 13 – 40]
= \(\frac{9}{2}\) = 4.5 sq. units.
From above two cases, it is clear that area of triangular grassy lawn is same.

Question 6.
The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. (Recall Theorem 6.2 and Theorem 6.6).
Solution:
The vertices of ∆ABC are A (4, 6); B (1, 5) and C (7, 2)

A line is drawn to intersect sides AB and AC at D (x₁, y₁) and E (x₂, y₂) respectively such that \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\).

∴ D and E divides AB and AC in the ratio 1 : 3.
∴ Coordinates of D are
x₁ = \(\frac{1(1)+3(4)}{1+3}=\frac{1+12}{4}=\frac{13}{4}\) and y₁ = \(\frac{1(5)+3(6)}{1+3}=\frac{5+18}{4}=\frac{23}{4}\)

∴ Coordinates of D are (\(\frac{13}{4}\), \(\frac{23}{4}\))
Now, coordinates of E are
x₂ = \(\frac{1(7)+3(4)}{1+3}=\frac{7+12}{4}=\frac{19}{4}\) and y₂ = \(\frac{1(2)+3(6)}{1+3}=\frac{2+18}{4}=\frac{20}{4}=5\)

∴ Coordinates of E are (\(\frac{19}{4}\), 5).

In ∆ADE
x₁ = 4, x₂ = \(\frac{13}{4}\), x₃ = \(\frac{19}{4}\)
y₂ = 6, y₂ = \(\frac{23}{4}\), y₃ = 5
area of ∆ADE = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= \(\frac{1}{2}\left[4\left(\frac{23}{4}-5\right)+\frac{13}{4}(5-6)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]\)

= \(\frac{1}{2}\left[4\left(\frac{23-20}{4}\right)+\frac{13}{4}(-1)+\frac{19}{4}\left(\frac{24-23}{4}\right)\right]\)

= \(\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]\)

= \(\frac{1}{2}\left[\frac{48-52+19}{16}=\frac{15}{16}\right]\)
= \(\frac{15}{32}\) sq. units.

In ∆ABC
x₁ = 4, x₂ = 1, x₃ = 7
y₂ = 6, y₂ = 5, y₃ = 2
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

= \(\frac{1}{2}\) [4 (5 – 2) + 1 (2 – 6) + 7 (6 – 5)]
= \(\frac{1}{2}\) [12 – 4 + 7] = \(\frac{15}{2}\) sq.units.

Now, \(\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\frac{15}{32}}{\frac{15}{2}}=\frac{15}{32} \times \frac{2^{1}}{16_{1}}\)

= \(\frac{1}{16}=\left(\frac{1}{4}\right)^{2}\)

= \(\left(\frac{A D}{A B}\right)^{2} \text { or }\left(\frac{A E}{A C}\right)^{2}\).

Question 7.
Let (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the potnt P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians ¡s called centroid and this point divides each median in the ratio 2: 1]
(v) if A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.
Solution:
Given that vertices of ∆ABC are A (4, 2); B (6, 5) and C (1, 4).
(i) AD is the median from the vertex A.
∴ D is the mid point of BC.

then x = \(\frac{6+1}{2}=\frac{7}{2}\) and y = \(\frac{5+4}{2}=\frac{9}{2}\)
Hence, coordinates of D is (\(\frac{7}{2}\), \(\frac{9}{2}\)).

(ii) Let P(x, y) be point on AD such that AP : PD = 2 : 1

then x = \(\frac{2\left(\frac{7}{2}\right)+1(4)}{2+1}\)
= \(\frac{7+4}{3}=\frac{11}{3}\)

and y = \(\frac{2\left(\frac{9}{2}\right)+1(2)}{2+1}\)
= \(\frac{9+2}{3}=\frac{11}{3}\)

Hence, Coordinates of P is (\(\frac{11}{3}\), \(\frac{11}{3}\)).

(iii) Le BE and CF are the medians of ∆ABC to AC and AB respectively.
∴ E and F are mid points of AC and AB respectively.

Coordinate of E are
x₁ = \(\frac{4+1}{2}=\frac{5}{2}\)
and y₁ = \(\frac{4+2}{2}=\frac{6}{2}\) = 3
Coordinate of E are (\(\frac{5}{2}\), 3)
Coordinate of F are
x₂ = \(\frac{4+6}{2}=\frac{10}{2}\) = 5
and y₂ = \(\frac{5+2}{2}=\frac{7}{2}\)
∴ Coordinate of F are (5, \(\frac{7}{2}\))
Now, Q divides BE such that BQ : QE = 2: 1

∴ Coordinate of Q are \(\left(\frac{2\left(\frac{5}{2}\right)+6(1)}{2+1}, \frac{2(3)+1(5)}{2+1}\right)\)

= \(\left(\frac{5+6}{3}, \frac{6+5}{3}\right)\) = \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

Also, R divides CF such that CR : RF = 2 : 1

∴ Coordinate of R are = \(\left(\frac{2(5)+1(1)}{2+1}, \frac{2\left(\frac{7}{2}\right)+(4)}{2+1}\right)\)

= \(\left(\frac{10+1}{3}, \frac{7+4}{3}\right)\)

= \(\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2: 1.

(v) The vertices of given ∆ABC are
A (x₁, y₁); B (x₂, y₂) and C (x₃, y₃).

Let AD is median of E, ∆ABC.
∴ D is the mid point of BC then coordinates of D are \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

Now, G be the centroid of ABC, which divides the median AD in the ratio 2: 1
∴ Coordinates of G are [using (iv)]

= \(\left[\frac{2\left(\frac{x_{2}+x_{3}}{2}\right)+1\left(x_{1}\right)}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right)+1\left(y_{1}\right)}{2+1}\right]\)

= \(\left[\frac{x_{2}+x_{3}+x_{1}}{3}, \frac{y_{2}+y_{3}+y_{1}}{3}\right]\)

= \(\left[\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right]\).

Question 8.
ABCD is a rectangle formed by the points A (- 1, – 1), B (- 1, 4), C (5, 4) and D (5, – 1). P, Q R and S are the mid points
of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle? or a rhombus ? Justify your answer.
Solution:
Given: The vertices ot’ given rectangle ABCD are
A(- 1, – 1); B(- 1, 4); C(5, 4) and D (5, – 1).

∵ P is the mid point of AB.
∴ Coordinates of P are \(\left(\frac{-1-1}{2}, \frac{-1+4}{2}\right)=\left(-1, \frac{3}{2}\right)\)
∵ Q is the mid point of BC.
∴ Co-ordinates of Q are \(\left(\frac{-5+5}{2}, \frac{4+4}{2}\right)\) = (2, 4)
∵ R is the mid point of CD.
∴ Coordinates of R are \(\left(\frac{5+5}{2}, \frac{4+1}{2}\right)=\left(5, \frac{3}{2}\right)\)

∵ S is the mid point of AD.
∴ Co-ordinates of S are \(\left(\frac{5-1}{2}, \frac{-1-1}{2}\right)\) = (2, -1)

PQ = \(\sqrt{(2+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9 \times \frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

PQ = \(\sqrt{\frac{61}{4}}\)

QR = \(\sqrt{(5-2)^{2}+\left(\frac{3}{2}-4\right)^{2}}\)

= \(\sqrt{(3)^{2}+\left(\frac{3-8}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

QR = \(\sqrt{\frac{61}{4}}\)

RS = \(\sqrt{(2-5)^{2}+\left(-1-\frac{3}{2}\right)^{2}}\)

= \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\)

RS = \(\sqrt{\frac{61}{4}}\)

and SP = \(\sqrt{(-1-2)^{2}+\left(\frac{3}{2}+1\right)^{2}}\)

SP = \(\sqrt{9+\frac{25}{4}}=\sqrt{\frac{61}{4}}\)

Also PR = \(\sqrt{(5+1)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}}\)
PR = \(\sqrt{36+0}=\sqrt{36}\) = 6
QS = \(\sqrt{(2-2)^{2}+(4+1)^{2}}\)
= \(\sqrt{0+25}=\sqrt{25}\) = 5.

Form above discussion it is clear that PQ = QR = RS = SP.
Also, PR ≠ QS.
⇒ All sides of quad. PQRS are equal but their diagonals are not equal.
Quad. PQRS is a rhombus.