Bhagya

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

1. Construct a right angled triangle ABC with ∠C = 90°, AB = 5 cm and BC = 3 cm.
Solution:
Given : Two sides of ΔABC as
AB = 5 cm,
BC = 3 cm
and ∠C = 90°.
To construct : A triangle with these two sides and one right angle.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the measure of these two sides and mark the right angle.

Step 2. Draw BC of length 3 cm.

Step 3. At C, draw CX ⊥ BC. (A should be somewhere on this perpendicular).

Step 4. With B as centre, draw an arc of radius 5 cm. (A must be on this arc since it is at a distance of 5 cm from B).

Step 5. A has to be on the perpendicular line CX as well as on the arc drawn with centre C.
∴ A is the meeting point of these two.
ΔABC is now obtained.

2. Construct an isosceles right angled triangle DEF where ∠E = 90° and EF = 6 cm.
Solution:
Given : An isoscele right angled ΔDEF where ∠E = 90° and EF = 6 cm.
To Construct: A right angled triangle with one side.
Steps of Construction:
Steps 1. Draw a rough sketch of given measures.

Step 2. Draw a line segment EF = 6 cm.

Step 3. With the help of compass taking E as centre, draw a ray EX making an angle of 90° with EF.

Step 4. With E as centre and radius 6 cm (= DE) draw an arc intersecting EX at D.

Step 5. Join D and F. Therefore ΔDEF is required isosceles right triangle.

3. Construct a right-angled triangle PQR in which :
∠Q= 90°, PQ = 3.6 cm and PR = 8.5 cm
Solution:
Given : Right triangle be PQR; right-angled at Q
i. e. ∠Q = 90°
and PQ = 3.6 cm,
PR = 8.5 cm
To construct : A triangle with these two sides and one right angle.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the measure of these two sides and mark the right angle.

Step 2. Draw PQ of length 3.6 cm.

Step 3. At Q, draw QX ⊥ PQ.
(R should be somewhere on this perpendicular).

Step 4. With P as centre, draw arc of radius
(R must be on this arc, since it is at a distance of 8.5 cm from P).

Step 5. R has to be on the perpendicular line QX as well as on the arc drawn with centre P.
∴ R is the meeting point of these two.
ΔPQR is now obtained.

4. Question (i).
Which of the following is a pythagorian triplet ?
(a) 1, 2, 3
(b) 2, 3, 4
(c) 4, 5, 6
(d) 12, 13, 5
Answer:
(d) 12, 13, 5

Question (ii).
Construction of unique triangle is not possible when :
(a) Three sides are given.
(b) Two sides and an included angle are given.
(c) Three angles are given.
(d) Two angles & included side are given.
Answer:
(c) Three angles are given.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 4 Integers MCQ Questions

Multiple Choice Questions

Question 1.
How many integers are between -3 to 3?
(a) 5
(b) 6
(c) 4
(d) 3.
Answer:
(a) 5

Question 2.
Which of the following integer is greater than -3?
(a) -5
(b) -4
(c) 0
(d) -10.
Answer:
(c) 0

Question 3.
Which of the following integers are in ascending order?
(a) -5, -9, -7, -8
(b) -9, -8, -7, -5
(c) -5, -7, -8, -8, -9
(d) -8, -5, -9, -7.
Answer:
(b) -9, -8, -7, -5

Question 4.
Which of the following integers are in descending order?
(a) 3, 0, -2, -5
(b) -5, -2, 0, 3
(c) -5, 3,-2, 0
(d) -2, 0, -5, 3.
Answer:
(a) 3, 0, -2, -5

Question 5.
The given number line represents:

(a) 5 + 1
(b) 1 + 5
(c) 1 + 1 + 1 + 1 + 1
(d) 5 + 5 + 5 + 5 + 5.
Answer:
(c) 1 + 1 + 1 + 1 + 1

Question 6.
3 less than -2 =
(a) -5
(b) -6
(c) 5
(d) 6.
Answer:
(a) -5

Question 7.
(-2) + 8 =
(a) -6
(b) -10
(c) 10
(d) 6.
Answer:
(d) 6.

Question 8.
Which of the following statements is true about the given number line:

(a) Value of A is greater than value of B.
(b) Value of A is greater than value of C.
(c) Value of B is less than value of C.
(d) Value of C is less than value of B.
Answer:
(c) Value of B is less than value of C.

Question 9.
(-7) + (-12) + 11 =
(a) -19
(b) 30
(c) -23
(d) -8.
Answer:
(d) -8.

Question 10.
15 – (-12) + (-27) =
(a) 0
(b) -54
(c) -24
(d) 54.
Answers :
(a) 0

Question 11.
What is the number of integers between -4 and -1?
(a) 3
(b) 4
(c) 5
(d) 6.
Answer:
(b) 4

Question 12.
What is the number of integers between -8 and -2?
(a) 3
(b) 4
(c) 5
(d) 6.
Answer:
(c) 5

Question 13.
Which is the largest integers among -7, -6, -5, -4 and -3?
(a) -6
(b) -5
(c) -4
(d) -3.
Answer:
(d) -3.

Question 14.
Which is the smallest integer among -3, -2, 0 and 1?
(a) -3
(b) -2
(c) 0
(d) 1.
Answer:
(a) -3

Question 15.
The value of (-7) + (-9) + 4 + 16 is:
(a) 36
(b) 22
(c) 4
(d) 27.
Answer:
(c) 4

Fill in the blanks:

Question (i)
The sum of (-9) + (+4) + (-6) + (+3) is …………. .
Answer:
– 8

Question (ii)
The successor of -5 is ……………. .
Answer:
– 4

Question (iii)
-19 + …………. = 0.
Answer:
19

Question (iv)
100 + …………. = 0.
Answer:
– 100

Question (v)
50 + (-50) = ……………. .
Answer:
0

Write True/False:

Question (i)
-8 is to the right of -10 on number line. (True/False)
Answer:
True

Question (ii)
-100 is to the right of -50 on number line. (True/False)
Answer:
False

Question (iii)
Smallest negative integer is -1. (True/False)
Answer:
False

Question (iv)
-26 is larger than -25. (True/False)
Answer:
False

Question (v)
The sum of two integers is always an integer. (True/False)
Answer:
True

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is 15 for the first km and 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs 150 for the first metre and rises by 50 for each msubsequent metre.

(iv) The amount of money in the account every year when 10000 is deposited at compound interest at 8% per annum.

Solution:
(i) Let T_n denotes the taxi fare in n^th km.
According to question,
T₁ = 15 km;
T₂ = 15 + 8 = 23;
T₃ = 23 + 8 = 31
Now, T₃ – T₂ = 31 – 23 = 8
T₂ – T₁ = 23 – 15 = 8
Here, T₃ – T₂ = T₂ – T₁ = 8
∴ given situation form an AP.

(ii) Let T_n denotes amount of air present in a cylinder.
According to question,
T₁ = x;
T₂ = x – \(\frac{1}{4}\)x
= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x
T₃ = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)

= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on
Now, T₃ – T₂ = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x
= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x

T₂ – T₁ = \(\frac{3}{4}\)x – x
= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation donot form an AP.

(iii) Let T_n denotes cost of digging a well for the nth metre,
According to question,
T₁ = ₹ 150; T₂ = (150 + 50) = ₹ 200;
T₃ = ₹ (200 + 5o) = 250 and so on
Now, T₃ – T₂ = ₹ (250 – 200) = 50
T₂ – T₁ = ₹ (200 – 150) = 50
Here, T₃ – T₂ = T₂– T₁ = 50
∴ given situation form an A.P.

(iv) Let T_n denotes amount of money in the nth year.
According to question
T₁ = ₹ 10,000
T₂ = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + ₹ 800 = ₹ 10,800
T₃ = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + ₹ 864
= ₹ 11,640 and so on.
Now, T₃ – T₂ = ₹ (11,640 – 10,800) = ₹ 840
T₂ – T₁ = ₹ (10,800 – 10,000) = ₹ 800
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation do not form an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(1) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac{1}{2}\)
(w) a = -1.25, d = -0.25
Solution:
(i) Given that first term = a = 10
and common difference = d = 10
∴ T₁ = a = 10;
T₂ = a + d = 10 + 10 = 20;
T₃ = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T₄ = a + 3d = 10 + 3 × 10
= 10 + 30 = 40
Hence, first four terms of an A.P. are 10, 20, 30, 40………….

(ii) Given that first term = a = -2
and common iifference = d = 0
∴ T₁ = a = -2;
T₂ = a + d = -2 + 0 = -2
T₃ = a + 2d = -2 + 2 × 0 = -2
T₄ = a + 3d = -2 + 3 × 0 = -2
Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….

(iii) Given that first term = a = 4
and common difference = d = -3
∴ T₁ = a = 4;
T₂= a + d = 4 – 3 = 1
T₃ = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T₄ = a + 3d = 4 + 3(-3) = 4 – 9 = -5
Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….

(iv) Given that first term = a = -1
and common difference = d = \(\frac{1}{2}\)
∴ T₁ = a = -1; T₂ = a + d
= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)
T₃ = a + 2d = -1 + 2(\(\frac{1}{2}\))
= -1 + 1 = 0
T₄ = a + 3d = -1 + 3(\(\frac{1}{2}\))
= \(\frac{-2+3}{2}=\frac{1}{2}\)
Hence, first four terms of an AP are -1, –\(\frac{1}{2}\), 0, \(\frac{1}{2}\), …………..

(v) Given that first term = a = – 1.25
and common difference = d = – 0.25
∴ T₁ = a = – 1.25;
T₂ = a + d = – 1.25 – 0.25 = -1.50
T₃ = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T₄ = a + 3d = – 1.25 + 3(- 0.25)
= – 1.25 – 0.75 = – 2
Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..

Question 3.
For the following APs, wilte the first term and the common difference:
(i) 3, 1, -1, -3, …………
(ii) 5, -1, 3, 7, ………….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\), …………..
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) Given A.P., is 3, 1, -1, -3, ………
Here T₁ = 3, T₂ = 1,
T₃ = -1, T₄ = -3
First term = T₁ = 3
Now, T₂ – T₁ = 1 – 3 = – 2
T₃ – T₂ = – 1 – 1 = -2
T₄ – T₃ = -3 + 1 = -2
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
Hence, common difference = – 2 and first term = 3.

(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T₁ = – 5, T₂ = – 1,
T₃ = 3, T₄ = 7
First term T₁ = -5
Now, T₂ – T₁ = -1 + 5 = 4
T₃– T₂ = 3 + 1 = 4
T₄ – T₃ = 7 – 3 = 4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4
Hence, common difference = 4 and first term = – 5.

(iii) Given AP. is:
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
Here T₁ = \(\frac{1}{3}\), T₂ = \(\frac{5}{3}\),
T₃ = \(\frac{9}{3}\), T₄ = \(\frac{13}{3}\)
First term = T₁ = \(\frac{1}{3}\)
Now, T₂ – T₁ = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)
T₃ – T₂ = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)
T₄ – T₃ = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{4}{3}\)

Hence, common difference = \(\frac{4}{3}\) and first term = \(\frac{1}{3}\).

(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…
Here, T₁ = 0.6, T₂ = 1.7, T₃ = 2.8, T₄ = 3.9
First term = T₁ = 0.6
Now, T₂ – T₁ = 1.7 – 0.6 = 1.1
T₃ – T₂ = 2.8 – 1.7 = 1.1
T₄ – T₃ = 3.9 – 2.8 = 1.1
Hence, common difference = 1.1 and first term = 0.6.

Question 4.
WhIch of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ………….
(iv) – 10, – 6, – 2, 2, ………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ……….
(vi) 0.2, 0.22, 0.222, 0.2222, ………….
(vii) 0, -4, -8, -12, …………..
(viii) \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ………..
(ix) 1, 3, 9, 27 …………….
(x) a, 2a, 3a, 4a, ………………
(xi) a, a², a³, a⁴, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 1², 3², 5², 7², ………..
(xv) 1², 5², 7², 73, ………….

Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T₁ = 2, T₂ = 4, T₃ = 8, T₄ = 16
T₂ – T₁ = 4 – 2 = 2
T₃ – T₂ = 8 – 4 = 4
∵ T₂ – T₁ ≠ T₃ – T₂
Hence, given terms do not form an A.P.

(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
Here T₁ = 2, T₂ = 4, T₃ = 3, T₄ = 16
T₂ – T₁ = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)
T₃ – T₂ = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)
T₄ – T₃ = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{1}{2}\)
∴ Common difference = d = \(\frac{1}{2}\)
Now, T₅ = a + 4d = 2 + 4\(\frac{1}{2}\) = 4

T₆ = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)

T₇ = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.

(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T₁ = – 1.2, T₂ = – 3.2,
T₃ = – 5.2, T₄ = – 7.2
T₂ – T₁ = – 3.2 + 1.2 = – 2
T₃ – T₂ = – 5.2 + 3.2 = – 2
T ₄ – T₃ = – 7.2 + 5.2 = – 2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
∴ Common difference = d = – 2
Now, T₅ = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T₆ = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T₇ = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2

(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T₁ = – 10,T₂ = – 6
T₃ = – 2, T₄=2 .
T₂ – T₁ = – 6 + 10 = 4
T₃ – T₂ = – 2 + 6 =4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁=T₃ – T₂ = T₄ – T₃ = 4 .
∴ Common difference = d = 4
Now, T₅ = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T₆ = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T₇ = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.

(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T₁ = 3, T₂ = 3 + √2,
T₃ = 3 + 2√2, T₄= 3 + 3√2
T₂ – T₁ = 3 + √2 – 3 = √2
T₃ – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T₄ – T₃ = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T₂ -T₁ = T₃ – T₂ = T₄ – T₃ = √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = 3 + 4(√2) = 3 + 4√2
T₆ = a + 5d = 3 + 5√2
T₇ = a + 6d = 3 + 6√2

(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T₁ = 0.2, T₂ = 0.22,
T₃ = 0.222, T₄ = 0.2222.
T₂ – T₁ = 0.22 – 0.2 = 0.02
T₃ – T₂ = 0.222 – 0.22 = 0.002
∵ T₂ – T₁ ≠ T₃ – T₂
∴ given terms do not form an A.P.

(vii) Given terms are 0, -4, -8, -12
Here T₁ = 0, T₂ = -4,
T₃ = -8, T₄ = -12
T₂ – T₁ = – 4 – 0 = -4
T₃ – T₂= – 8 + 4 = -4
T₄ – T₃= – 12 + 8 = -4.
T₂ – T₁ = T₃ – T₂ = T₄ – T₃
∴ Common difference = d = -4
Now, T₅= a + 4d = 0 + 4(-4) = -16
T₆ = a + 5d = 0 + 5(-4) = -20
T₇ = a + 6d = 0 + 6(-4) = -24.

(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….
Here T₁ = \(-\frac{1}{2}\), T₂ = –\(\frac{1}{2}\)
T₃ = \(-\frac{1}{2}\), T₄ = \(-\frac{1}{2}\)
T₂ – T₁ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
T₃ – T₂ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
∵ T₂ – T₁ = T₃ – T₂ = 0
∴ Common difference = d = 0
Now, T₅ = T₆ = T₇ = –\(\frac{1}{2}\)
[∵ a = –\(\frac{1}{2}\), d = 0]

(ix) Given terms are 1, 3, 9, 27
T₁ = 1, T₂ = 3, T₃ = 9, T₄ = 27
T₂ – T₁ = 3 1 = 2
T₃ – T₂ = 9 – 3 = 6.
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(x) Given terms are a, 2a, 3a, 4a, …
T₁ = a, T₂ = 2a, T₃ = 3a, T4 = 4a
T₂ – T₁ = 2a – a = a
T₃ – T₂ = 3a – 2a = a
T₄ – T₃ = 4a – 3a = a
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = a
∴ Common difference = d = a
Now T₅ = a + 4d = a + 4(a) = a + 4a = 5a
T₆ = a + 5d = a + 5a = 6a
T₇ = a + 6d = a + 6a = 7a

(xi) Given terms are a, a², a³, a⁴, …………
T₁ = a, T₂ = a², T3 = a³, T₄ = a⁴
T₂ – T₁ = a² – a
T₃ – T₂ = a³ – a²
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xii) Given terms are √2, √8, √18, √32, …………
T₁ = √2, T₂ = √8, T₃ = √18, T₄ = √32
or T₁ = √2, T₂ = 2√2 T₃ = 3√2, T₄ = 4√2
T₂ – T₁ = 2√2 – √2 = √2
T₃ – T = 3√2 – 2√2 = √2
T₄ – T₃ = 4√2 – 3√2 = √2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃= √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = √2 + 4√2 = 5√2
T₆ = a + 5d = √2 + 5√2 = 6√2
T₇ = a + 6d = √2 + 6√2 = 7√2

(xiii) Given terms are √3, √6, √9, √12, ……………..
T₁ = √3, T₂= √6, T₃= √9, T₄= √12
or T₁ = √3, T₂ = √6, T₃ = 3, T₄ = 2√3
T₄ – T₁ = √6 – √3
T₃ – T₂ = 3 – √6
∵ T₂ – T₁ ≠ T₃ – T₂
∴Given terms do not form an A.P.

(xiv) Given terms are 1², 3², 5², 7², ………..
T₁ = 1², T₂ = 3², T₃ = 5², T₄ = 7²
or T₁ = 1, T₂ = 9, T₃ = 25, T₄ = 49
T₄ – T₁ = 9 – 1 = 8
T₃ – T₂ = 25 – 9 = 16
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xv) Given terms are 12, 52, 72, 73
T₁ = 12, T₂ = 52, T₃ = 72, T₄ = 73
or T₁ = 1, T₂ = 25, T₃ = 49, T₄ = 73
T₂ – T₁ = 25 – 1 = 24
T₃ – T₂ =49 – 24= 24
T₄ – T₃ = 73 – 49 = 24
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 24
∴ Common difference = d = 24
T₅ = a + 4d = 1 + 4(24) = 1 + 96 = 97
T₆ = a + 5d = 1 + 5(24) = 1 + 120 = 121
T₇ = a + 6d = 1 +6(24) = 1 + 144 = 145

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.3

1. Fill the suitable integer in box:

Question (i)
(a) 2 + _ = 0
(b) _ + 11 =0
(c) -5 + _ = o
(d) _ + (-9) = 0
(e) 3 + _ = 0
(f) _ + 0 = 0.
Solution:
(a) 2 + -2 = 0
(b) -11 + 11 =0
(c) -5 + 5 = o
(d) 9 + (-9) = 0
(e) 3 + (-3) = 0
(f) 0 + 0 = 0.

2. Subtract using number line:

Question (a)
5 from -7
Solution:
(-7) – 5

Hence , -7 – 5 = -12

Question (b)
-3 from -6
Solution:
-6 – (-3)
= -6 + (additive inverse of -3)
= -6 + (3)

Hence, -6 + 3 = -3

Question (c)
-2 from 8
Solution:
8 – (-2)
= 8 + (additive inverse of -2)
= 8 + (2)

Hence, 8 + 2 = 10

Question (d)
3 from 9.
Solution:
9 – 3

Hence, 9 – 3 = 6

3. Subtract without using number line:

Question (a)
-6 from 16
Solution:
16 – (-6)
= 16 + (6) = 16 + 6
= 22

Question (b)
-51 from 55
Solution:
55 – (-51)
= 55 + (51) = 55 + 51
= 106

Question (c)
75 from -10
Solution:
-10 – 75
= -(10 + 75)
= -85

Question (d)
-31 from -47.
Solution:
-47 – (-31)
= -47 + 31 = -(47 – 31)
= -16

4. Find:

Question (a)
35 – (20)
Solution:
= (35 – 20)
= 15

Question (b)
(-20) – (13)
Solution:
= -(20 + 13)
= -33

Question (c)
(-15) – (-18)
Solution:
= (-15) + (18)
= 3

Question (d)
72 – (90)
Solution:
= -(90 – 72)
= -18

Question (e)
23 – (-12)
Solution:
= 23 + (12)
= 23 + 12
= 35

Question (f)
(-32) – (-40).
Solution:
= 40 – 32
= 8

5. Simplify:

Question (a)
2 – 4 + 6 – 8 – 10
Solution:
= 2 + 6 – 4 – 8 – 10
= 2 + 6 -(4 + 8 + 10)
= 8 – 22
= -14

Question (b)
4 – 2 + 2 – 4 – 2 + 2
Solution:
= 4 + 2 + 2 – 2 – 4 – 2
= 8 – 8
= 0

Question (c)
4 – (-9) + 7 – (-3)
Solution:
=4 + 9 + 7 + 3
= 23

Question (d)
(-7) + (-19) + (-7).
Solution:
= -(7 + 19 + 7)
= – 33

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.2

1. Using number line write the integer which is:

Question (a)
5 less than -1
Solution:
5 less than -1 = ?
We need to find the integer which is 5 less than -1.
So, we shall start with ‘-1 ’ and proceed 5 steps to the left of -1 as shown below:

Therefore 5 less than -1 is -6

Question (b)
5 more than -5
Solution:
5 more than -5 = ?
We need to find the integer which is 5 more than -5.
So, we shall start with -5 and proceed 5 steps to the right of -5 as shown below:

Therefore 5 more than -5 is zero.

Question (c)
2 less than 5
Solution:
2 less than 5 = ?
We need to find the integer which is 2 less than 5.
So, we shall start with 5 and proceed
2 steps to the left of 5 as shown below:

Therefore 2 less than 5 is 3.

Question (d)
3 less than -2.
Solution:
3 less than – 2 = ?
We need to find the integer which is 3 less than -2.
So, we shall start with -2 and proceed 3 steps to the left of -2 as shown below:

Therefore 3 less than -2 is -5.

2. Using number line, add the following integers:

Question (a)
9 + (-3)
Solution:
On number line we shall start from 0 and move 9 steps to the right of zero. Then we shall move three steps to the left of ‘+9’. We finally reach at +6. Thus, +6 is the answer.

Hence, 9 + (-3) = +6

Question (b)
5 + (-11)
Solution:
On number line we shall start from 0 and move 5 steps to the right of zero. Then we shall move eleven steps to the left of ‘+5’. We finally reach at ‘-6’. Thus, -6 is the answer.

Hence, 5 + (-11) = -6

Question (c)
(-1) + (-4)
Solution:
On number line we first move 1 step to the left of zero. Then we moved ahead 4 steps to the left of ‘-1\ We finally reaches at -5. Thus -5 is the answer.

Hence, (-1) + (-4) = -5

Question (d)
(-5) + 12
Solution:
On number line we shall move 5 steps to the left of zero. Then we shall move 12 steps to the right of ‘-5’ and finally reach at ‘+7’. Thus +7 is the answer.

Hence, (-5) + 12 = +7

Question (e)
(-1) + (-2) + (-4)
Solution:
Step I: On number line we shall move one step to the left of zero suggested by minus sign of ‘-1’.
Step II: Then we shall move 2 steps to the left of -1 suggested by minus sign of -2 and we reach at ‘-3’.
Step III: Then again we shall move 4 steps to the left of ‘-3’ suggested by minus sign of -4 and finally we reach at-7.

Hence, (-1) + (-2) + (-4) = -7

Question (f)
(-2) + 4 + (-5)
Solution:
Step I: On number line we shall move 2 steps to the left of zero suggested by minus sign of ‘-2’.
Step II: Then we shall move 4 steps to the right of ‘-2’ suggested by plus sign of ‘44’ and we reach at +2.
Step III: Then we shall move 5 steps to the left of ‘+2’ as suggested by minus sign of ‘-5’ and finally we reach at -3. Thus answer is ‘-3’.

Hence, (-2) + 4 + (-5) = -3

Question (g)
(-3) + (5) + (-4).
Solution:
Step I: On number line we shall move 3 steps to the left of zero as suggested by minus sign of ‘-3’.
Step II: Then we shall move 5 steps to the right of ‘-3’ as suggested by plus sign of ‘+5’ and we reach at ‘+2’.
Step III: Then we shall move 4 steps to the left of ‘+2’ as suggested by minus sign of ‘4’ and finally we reach at ‘-2’. Thus ‘-2’ is the answer.

Hence, (-3) + (5) + (-4) = -2

3. Add without using number line:

Question (a)
18 + 13
Solution:
18 + 13 = 31

Question (b)
18 + (- 13)
Solution:
18 + (- 13) = + (18 – 13) = +5

Question (c)
(-18) + 13
Solution:
(-18) + 13 = – (18 – 13) = -5

Question (d)
(-18) + (-13)
Solution:
(-18) + (-13) = -(18 + 13) = -31

Question (e)
180 + (-200)
Solution:
180 + (-200) = -(200 – 180) = -20

Question (f)
111 + (-67)
Solution:
111 + (-67) = (777 – 67) = 710

Question (g)
1262 + (-366) + (-962)
Solution:
1262 + (-366) + (-962)
= 1262 – (366 + 962)
= 1262 – 1328
= -(1328 – 1262) = -66

Question (h)
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
Solution:
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
= 30 + 21 + 11 + (-27) + (-19) + (-3) + (-9)
= 62 + (-58) = 62 – 58 = 4

Question (i)
(-7) + (-9) + 4 + 16
Solution:
(-7) + (-9) + 4+16 = (-16) + 20 = 4

Question (j)
37 + (-2) + (-65) + (-8).
Solution:
37 + (-2) + (-65) + (-8)
= 37 – (2 + 65 + 8)
= 37 – 75 = -38

4. Write the successor and predecessor of the following:

Question (a)
-15
Solution:
Successor of -15 = -15 + 1 = -14
Predecessor of -15 = -15 – 1 = -16

Question (b)
27
Solution:
Successor of 27 = 27 + 1 = 28
Predecessor of 27 = 27 – 1 = 26

Question (c)
-79
Solution:
Successor of -79 = -79 + 1 = -78
Predecessor of -79 = -79- 1 = -80

Question (d)
0
Solution:
Successor of 0 = 0+ 1 = 1
Predecessor of 0 = 0 – 1 = -1

Question (e)
29
Solution:
Successor of 29 = 29 + 1 = 30
Predecessor of 29 = 29 – 1 = 28

Question (f)
-18
Solution:
Successor of -18 = -18 + 1 = -17
Predecessor of -18 = -18 – 1 = -19

Question (g)
-21
Solution:
Successor of -21 = -21 + 1 = -200
Predecessor of -21 = -21 – 1 = -22

Question (h)
99
Solution:
Successor of 99 = 99 + 1 = 100
Predecessor of 99 = 99 – 1 = 98

Question (i)
-1
Solution:
Successor of-1 = -1 + 1 = 0
Predecessor of -1 = -1 – 1 = -2

Question (j)
-13.
Solution:
Successor of -13 = -13 + 1 = -12
Predecessor of -13 = -13 – 1 = -14

5. Complete the following addition table:

+	-3	-4	-2	+1	+2	+3
-2
-3
0
+1
+2

Solution:

+	-3	-4	-2	+ 1	+2	+3
-2	-5	-6	-4	-1	0	+ 1
-3	-6	-7	-5	-2	-1	0
0	-3	-4	-2	+ 1	+2	+3
+ 1	-2	-3	-1	+2	+3	+4
+2	-1	-2	0	+3	+4	+5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at D show that,

(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Answer:
∆ ABC and ∆ DBC are isosceles triangles on the same base BC.
∴ In ∆ ABC, AB = AC and in ∆ DBC, DB = DC.
In ∆ ABD and ∆ ACD,
AB = AC
DB = DC
and AD = AD (Common)
∴ ∆ ABD s ∆ ACD (SSS rule) [Result (i)]
∴ ∠ BAD = ∠ CAD (CPCT)
In ∆ ABP and ∆ ACP
AB = AC
∠ BAP = ∠ CAP (∵ ∠ BAD = ∠ CAD)
and AP = AP (Common)
∴ ∆ ABP ≅ ∆ ACP (SAS rule) [Result (ii)]
∴ BP = CP (CPCT)
In ∆ DBP and ∆ DCR
DB = DC
BP = CP
and DP = DP (Common)
∴ ∆ DBP ≅ ∆ DCP (SSS rule)
From ∆ ABP ≅ ∆ ACR ∠ BAP = ∠ CAP (CPCT)
∴ AP bisects ∠A.
From ∆ DBP ≅ ∆ DCR ∠BDP = ∠ CDP (CPCT)
∴ DP bisects ∠D.
Thus, AP bisects ∠A as well as ∠D. [Result (iii)]
∆ A ABP ≅ ∆ ACP
∴ BP = CP and ∠ APB = ∠ APC (CPCT)
But, ∠ APB + ∠ APC = 180° (Linear pair)
∴ ∠ APB = ∠ APC = \(\) = 90°
Thus, BP = CP and AP ⊥ BC.
∴ AP is the perpendicular bisector of BC. [Result (iv)]

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A

Answer:
AD is an altitude of A ABC.
∴ ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
hypotenuse AB = hypotenuse AC (Given)
∠ ADB = ∠ ADC (Right angles)
AD = AD (Common)
∴ ∆ ADB ≅ ∆ ADC (RHS rule)
∴ BD = CD and ∠ BAD = ∠ CAD (CPCT)
Now, BD = CD means D is the midpoint of BC.
Hence, AD bisects BC. [Result (i)]
Moreover, ∠ BAD = ∠ CAD and
∠ BAD + ∠ CAD = ∠ BAC.
Hence, AD bisects ∠A. [Result (ii)]

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see the given figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR

Answer:
In ∆ ABC, AM is a median.
∴ BM = CM = \(\frac{1}{2}\) BC
In ∆ PQR, PN is a median.
∴ QN = RN = \(\frac{1}{2}\) QR
Now, BC = QR (Given)
∴ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ ABM and ∆ PQN,
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved)
∴ ∆ ABM ≅ ∆ PQN (SSS rule) [Result (i)]
∴ ∠ ABM = ∠ PQN (CPCT)
∴ ∠ ABC = ∠ PQR
Now, in ∆ ABC and ∆ PQR,
AB = PQ
∠ ABC = ∠ PQR
BC = QR .
∴ ∆ ABC ≅ ∆ PQR (SAS rule) [Result (ii)]

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:
In ∆ FBC and ∆ ECB,
CF = BE (Given)
∠ CFB = ∠ BEC = 90° (Given)
BC = CB (Common)
∴ A FBC ≅ A ECB (RHS rule)
∴ ∠ FBC = ∠ ECB (CPCT)
∴ ∠ ABC = ∠ ACB
Now, in ∆ ABC, ∠ ABC = ∠ ACB
∴ AC = AB (Theorem 7.3)
Hence, ∆ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
Answer:

In ∆ ABC, AP is an altitude.
∴ ∠ APB = ∠ APC = 90°
In ∆ APB and ∆ APC,
∠ APB = ∠ APC = 90°
AB = AC (Given)
AP = AP (Common)
∴ ∆ APB ≅ ∆ APC (RHS rule)
∴ ∠ ABP = ∠ AGP (CPCT)
∴ ∠ ABC = ∠ ACB
Thus, in ∆ ABC, ∠ B = ∠ C.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Answer:

In ∆ ABC, AB = AC
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ OBC = ∠ OCB (BO bisects ∠ ABC and CO bisects ∠ ACB)
Now, in ∆ OBC, ∠ OBC = ∠ OCB
∴ OB = OC (Theorem 7.3)
Similarly, ∠ ABC = ∠ ACB gives
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ ABO = ∠ ACO
Now, in ∆ ABO and ∆ ACO,
AB = AC (Given)
∠ ABO = ∠ ACO
and OB = OC
∴ ∆ ABO ≅ ∆ ACO (SAS rule)
∴ ∠ BAO = ∠ CAO (CPCT)
But, ∠ BAO + ∠ CAO = ∠ BAC (Adjacent angles)
∴ ∠ BAO = ∠ CAO = \(\frac{1}{2}\) ∠ BAC
Thus, AO bisects ∠ A.

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Answer:
In ∆ ABC, AD is the perpendicular bisector of BC.
∴ BD = CD and ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
AD = AD (Common)
∠ ADB = ∠ ADC (Right angles)
and BD = CD
∴ ∆ ADB ≅ ∆ ADC (SAS rule)
∴ AB = AC (CPCT)
Now, in ∆ ABC, AB = AC.
Hence, ∆ ABC is an isosceles triangle in which AB = AC.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Answer:
In ∆ ABC, AC = AB
∴ ∠ ABC = ∠ ACB
∴ ∠ FBC = ∠ ECB
Now, in ∆ FBC and ∆ ECB,
∠ FBC = ∠ ECB
∠ BFC = ∠ CEB (Right angles)
BC = CB (Common)
∴ ∆ FBC ≅ ∆ ECB (AAS rule)
∴ CF = BE (CPCT)
Thus, the altitudes CF and BE on equal sides AB and AC respectively of ∆ ABC are equal.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:
In ∆ ABE and ∆ ACF,
∠ AEB = ∠ AFC (Right angles)
∠ A = ∠ A (Common)
BE = CF (Given)
∴ ∆ ABE ≅ ∆ ACF (AAS rule)
∴ AB = AC (CPCT)
Thus, ∆ ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ ABD = ∠ ACD.

Answer:
∠ ABC and ∠ DBC are adjacent angles.
∴ ∠ ABC + ∠ DBC = ∠ ABD ………… (1)
∠ ACB and ∠ DCB are adjacent angles.
∴ ∠ ACB + ∠ DCB = ∠ ACD ………….. (2)
In ∆ ABC, AB = AC.
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
In ∆ DBC, DB = DC.
∴ ∠ DBC = ∠ DCB (Theorem 7.2)
∴ ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∴ ∠ ABD = ∠ ACD [From (1) and (2))

Question 6.
∆ ABC is an isosceles triangle in which AB = AC. side BA is produced to D such that AD = AB (see the given figure). Show that ∠ BCD is a right angle.

Answer:
AB = AC and AD = AB
∴ AC = AD
In ∆ ABC, AB = AC
∴ ∠ ACB = ∠ ABC (Theorem 7.2) ……………… (1)
In A ADC, AC = AD
∴ ∠ ACD = ∠ ADC (Theorem 7.2) ……………… (2)
Adding (1) and (2),
∠ ACB + ∠ ACD = ∠ ABC + ∠ ADC
∴ ∠ BCD = ∠ DBC + ∠ BDC
(Adjacent angles and A lies on BD)
In ∆ BCD,
∠ DBC + ∠ BDC + ∠ BCD = 180°
∴ ∠ BCD + ∠ BCD = 180° (from (3)]
∴ 2 ∠ BCD = 180°
∴ ∠ BCD = 90°
Thus, ∠ BCD is a right angle.

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.
Answer:

In ∆ ABC, AB = AC
∴ Z C = Z B (Theorem 7.2)
In ∆ ABC,
∠ A + ∠ B + ∠ C = 180°
∴ 90° + ∠ B + ∠ B = 180° (Given and ∠ C = ∠ B)
∴ 2 ∠ B = 90°
∴ ∠ B = 45°
∴ ∠ C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:

∆ ABC is an equilateral triangle.
∴ AB = BC = AC
In ∆ ABC, AB = BC
∴ ∠ C = ∠ A (Theorem 7.2)
In ∆ ABC, AB = AC
∴ ∠ C = ∠ B (Theorem 7.2)
Hence, ∠ A = ∠B = ∠ C.
Now, in ∆ ABC, ∠ A + ∠ B + Z C = 180°
∴ ∠ A = ∠ B = ∠ C = \(\frac{180^{\circ}}{3}\) = 60°
Thus, the angles of ah equilateral triangle are 60° each.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

1. Construct ΔABC, given AB = 6 cm, ∠A = 30° and ∠B = 75°.
Solution:
Given. One side of ΔABC as AB = 6 cm, m∠A = 30° and m∠B = 75°.
To construct: A triangle with one side and these two angles.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔABC and indicate the measures of side and two angles.

Step 2. Draw a ray AB of length 6 cm.

Step 3. At A; draw a ray AX making an angle 30° with AB.

Step 4. With the help of compass. At B; draw a ray BY making an angle of 75° with AB.

Step 5. Both rays AX and BY intersect, at a point. So the point of intersection of the two rays is C Then. ΔABC is now obtained.

2. Construct an isosceles ΔABC such that AB = 5.3 cm and each base angle = 45°.
Solution:
Given : Isosceles ΔABC with AB = 5.3 cm each base angle = 45°.
To Construct: A triangle with one side and two base angles.
Steps of construction :
Step 1. Draw a rough sketch of ΔABC with given measures

Step 2. Draw a line segment AB = 5.3 cm.

Step 3. Taking A as centre with the help of compass. Draw a ray AX making an angle 45° with AB.

Step 4. With the help of compass and taking B as a centre. Draw a ray BY making an angle 45° with the line segment AB.

Step 5. Rays AX and BY intersect, at a point say C, then ABC is the required triangle.

3. Construct ΔXYZ if XY = 4 cm, ∠X = 45° and ∠Z = 60°.
[Hint : ∠Y = 180° – 45° – 60° = 75°]
Solution:
One side of ΔXYZ as
XY = 4 cm,
∠X = 45°
and ∠Z = 60°.
As we know that by angle-sum property of a triangle; sum of all three angles of a triangle is equal to 180°.
∴ ∠X + ∠Y + ∠Z = 180°
⇒ 45° + ∠Y + 60° = 180°
⇒ 105° + ∠Y = 180°
⇒ ∠Y = 75°.
Now it will be easy to construct triangle with side
XY = 4 cm,
∠X = 45°
and ∠Y = 75°.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔXYZ and indicate the measure of side and two angles.

Step 2. Draw a ray XY of length 4 cm.

Step 3. At X draw a ray XA making an angle of 45° with XY.

Step 4. At Y; draw a ray YB making an angle of 75° with XY.

Step 5. Z has to lie on both rays XA and YB. So, the point of intersection of two rays is Z.
ΔXYZ is now obtained.

4. Examine whether you can construct ΔPQR such that ∠P = 100°, ∠Q = 90° and PQ = 4.3 cm If not possible given reason.
Solution:
No, we cannot construct given ΔPQR.
Reason :
As we know that by angle sum property of a triangle; sum of all three angles a triangle is equal to 180°. But in given question sum of two angles;
m∠P + m∠Q
= 100° + 90°
= 190°
The sum of these two angles should be less than 180°. So triangle with given measures cannot be constructed as it violates the angle sum property of a triangle.

5. Question (i).
In which of the following cases a unique triangle can be drawn ?
(a) BC = 5 cm, ∠B = 90° and ∠C = 100°
(b) AB = 4 cm, BC = 7 cm and CA = 2 cm
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°
(d) An isosceles triangle with length of each equal side equal to 5 cm.
Answer:
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°

Question (ii).
A triangle can be constructed by taking two of its angles as.
(a) 110°, 40°
(b) 70°, 115°
(c) 135°, 45°
(d) 90°, 90°
Answer:
(a) 110°, 40°

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.1

1. Write two examples from day-to-day life in which we can use positive and negative integers.
Solution:
1. If positive represents above sea level, then negative represents below sea level.
2. If positive represents a deposit, negative represents a withdrawal.

2. Write the opposite of the following:

Question (a)
A profit of ₹ 500
Solution:
A loss of ₹ 500

Question (b)
A withdrawal of ₹ 70 from bank account
Solution:
Deposit of ₹ 70 in bank account

Question (c)
A deposit of ₹ 1000
Solution:
Withdrawal of ₹ 1000

Question (d)
326 B.C
Solution:
326 AD

Question (e)
500 m below sea level
Solution:
500 m above sea level

Question (f)
25° above 0°C.
Solution:
25° below 0°C.

3. Represent the situations mentioned in integers.
Solution:
(a) + 500
(b) – 70
(c) + 1000
(d) – 326
(e) – 500 m
(f) + 25.

4. Represent the following situations in integers.

Question (a)
A deposit of ₹ 500.
Solution:
+ 500

Question (b)
An Aeroplane is flying at a height two thousand metre above the sea level.
Solution:
+ 2000

Question (c)
A withdrawal of ₹ 700 from Bank Account.
Solution:
– 700

Question (d)
A diver dives to a depth of 6 feet below ground level.
Solution:
– 6.

5. Represent the following numbers on number line.

Question (i)
(a) – 5
(b) + 6
(c) o
(d) + 1
(e) – 9
(f) – 4
(g) + 8
(h) + 3.
Solution:

6. Integers are represented on a horizontal number line as shown where A represents – 2. With reference to the number line, answer the following questions:

(a) Which point represent – 3?
(b) Locate the point which represents the opposite of B and name it P.
(c) Write integers for the points C and E.
(d) Which point marked on the number line has the least value?
Solution:

(a) Point B represents – 3.
(b) Point P represents + 3.
(c) Point C represents -7 and Point E represents + 4.
(d) Point C has the least value – 7.

7. In each of the following pairs, which number is to the right of other on the number line?

Question (i)
(a) 2 9
(b) -3, -8
(c) 0, -5
(d) -11, 10
(e) -9, 9
(f) 2, – 200.
Solution:
(a) 9
(b) – 3
(c) 0
(d) 10
(e) 9
(f) 2

8. Write all the integers between the given pairs (write them in increasing order)

Question (a)
0 and -6
Solution:
-5, -4, -3, -2, -1

Question (b)
-6 and +6
Solution:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Question (c)
-9 and -17
Solution:
-16, -15, -14, -13, -12, -11, -10

Question (d)
-19 and -5.
Solution:
-18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6.

9.

Question (a)
Write five negative integers greater than ‘-15’.
Solution:
Five negative integers greater than ‘-15’ are:
-14, -13, -12, -11, -10

Question (b)
Write five integers smaller than ‘-20’.
Solution:
Five integers smaller than ‘-20’ are:
-21, -22, -23, -24, -25

Question (c)
Write five integers greater than 0.
Solution:
Five integers greater than 0 are:
1,2, 3, 4, 5

Question (d)
Write five integers smaller than 0.
Solution:
Five integers smaller than 0 are:
-1, -2, -3, -4, -5.

10. Encircle the greater integer in each given pair.

(a) -5, -7
(b) 0,-3
(e) 5, 7
(d) -9, 0
(e) -9, -11
(f) -4, 4
(g) -10, -100
(h) 10, 100.
Solution:
(a) -5
(b) 0
(c) 7
(d) 0
(e) -9
(f) 4
(g) -10
(h) 100.

11. Arrange the following integers in ascending order:

Question (a)
0, -7, -9, 5, -3, 2, -4
Solution:
Ascending order of given integers is:
-9, -7, -4, -3, 0, 2, 5

Question (b)
8, -3, 7, 0, -9, -6.
Solution:
Ascending order of given integers is:
-9, -6, -3, 0, 7, 8.

12. Arrange the following integers in descending order:

Question (a)
-9, 3, 4, -6, 8, -3
Solution:
8, 4, 3, -3 -6, -9

Question (b)
4, 8,-3,-2, 5, 0.
Solution:
8, 5, 4, 0, -2, -3.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ACBD. AC = AD and AB bisects ∠ A (see the given figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Answer:
In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠ BAC = ∠ BAD (AB bisects ∠ A)
AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (SAS rule)
∴ BC = BD (CPCT)
Thus, BC and BD are equal.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC

Answer:
In ∆ ABD and ∆ BAC,
AD = BC (Given)
∠ DAB = ∠ CBA (Given)
AB = BA (Common)
∴ ∆ ABD ≅ ∆ BAC (SAS rule)
∴ BD = AC (CPCT)
∴ ∠ ABD = ∠ BAC (CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.

Answer:
AD and BC are equal perpendiculars to line segment AB.
∴ AD = BC and ∠ OAD = ∠ OBC = 90°.
Now, in ∆ ADO and ∆ BCO,
AD = BC
∠ OAD = ∠ OBC
∠ AOD = ∠ BOC (Vertically opposite angles)
∴ ∆ ADO ≅ ∆ BCO (AAS rule)
∴ OA = OB (CPCT)
CD intersects AB at O and OA = OB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:
∆ ABC ≅ ∆ CDA.

Answer:
l || m and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
p l| q and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
Now, in ∆ ABC and ∆ CDA,
∠ BCA = ∠ DAC
∠ BAC = ∠ DCA
AC = CA (Common)
∴ ∆ ABC ≅ ∆ CDA (ASA rule)

Question 5.
Ray l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see the given figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Answer:
l is the bisector of ∠ PAQ and B is any point on l.
∴ ∠ PAB = ∠ QAB
BP and BQ are perpendiculars from B to AP and AQ.
∴ ∠ BPA = ∠ BQA = 90°.
Now, in ∆ APB and ∆ AQB,
∠ PAB = ∠ QAB
∠ BPA = ∠ BQA
AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (AAS rule)
∴ BP = BQ (CPCT)
BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.
∴ BP is the distance of B from AP and BQ is the distance of B from AQ.
Thus, B is equidistant from the arms of ∠ A.

Question 6.
In the given figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Answer:
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ DAE (Adjacent angles)
Now, in ∆ BAC and ∆ DAE,
AC = AE (Given)
AB = AD (Given)
∠ BAC = ∠ DAE
∴ ∆ BAC ≅ ∆ DAE (SAS rule)
∴ BC = DE (CPCT)

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

Answer:
∠ BAD = ∠ ABE
∴ ∠ PAD = ∠PBE (∵ P lies on AB.)
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE (Adjacent angles)
P is the midpoint of AB.
∴ AP = BP
Now, in ∆ DAP and ∆ EBP
∠ PAD = ∠ PBE
∠ APD = ∠ BPE
AP = BP
∴ ∆ DAP ≅ ∆ EBP (ASA rule)
∴ AD = BE (CPCT)

Question 8.
In right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = \(\frac{1}{2}\) AB

Answer:
In ∆ AMC and ∆ BMD,
AM = BM (∵ M is the midpoint of AB.)
CM = DM (Given)
∠ AMC = ∠ BMD (Vertically opposite angles)
∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]
∴ ∠ MCA = ∠ MDB (CPCT)
∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.
∴ AC || BD
Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.
∴ ∠ DBC + ∠ ACB = 180°
∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)
∴ ∠ DBC = 90°
Thus, ∠ DBC is a right angle. [Result (ii)]
Now, ∆ AMC ≅ ∆ BMD
∴ AC = BD
In ∆ DBC and ∆ ACB,
BD = CA
∠ DBC = ∠ ACB (Right angles)
BC = CB (Common)
∴ ∆ DBC ≅ ∆ ACB [Result (iii)]
∴ DC = AB (CPCT)
DM = CM and M lies on line’ segment CD.
∴ DC = 2 CM
∴ AB = 2CM
∴ \(\frac{1}{2}\)AB = CM
∴ CM = \(\frac{1}{2}\)AB