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Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4

Question 1.
State whether True or False:
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution :
(a) False
(b) True
(c) True
(d) False
(e) False
(f) False
(g) False
(h) False

Question 2.
Identify all the quadrilaterals that have:
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares as well as rhombuses have four sides of equal length.
(b) Squares as well as rectangles have four right angles.

Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) A square is a four sided closed figure, so it is a quadrilateral.
(ii) The opposite sides of a square are equal and parallel, so it is a parallelogram.
(iii) All the sides of a square are equal, so it is a rhombus.
(iv) Each angle of a square is a right angle, so it is a rectangle.

Question 4.
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) The diagonals of the following quadrilaterals bisect each other :

• Parallelogram
• Rectangle
• Square
• Rhombus

(ii) The diagonals of the following quadrilaterals are perpendicular bisectors of each other :

• Square
• Rhombus

(iii) The diagonals are equal in following quadrilaterals :

• Square
• Rectangle

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:

• All the angles have measure less than 180°.
• Both diagonals lie wholly in the interior of the rectangle.

∴ The rectangle is a convex quadrilateral.

Question 6.
ABC is a right angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Solution:
Produce \(\frac {1}{2}\) to D such that BO = OD.
Joining \(\frac {1}{2}\) and \(\frac {1}{2}\), we get a □ ABCD.
AO = OC (Given)
BO = OD (Constration)
∴ The diagonals AC and BD bisect each other.
∴ □ ABCD is a parallelogram.
∠B is a right angle. (Given)
∴ □ ABCD is a rectangle.
∴ BO = OD = AO = OC
∴ Point O is equidistant from points A, B and C.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers InText Questions

Try These : [Textbook Page No. 250]

1. Write the following numbers in generalised form:

Question (i)

1. 25
2. 73
3. 129
4. 302

Solution:

1. 25 = 10 × 2 + 5 [ ∵ ab = 10a + b]
2. 73 = 10 × 7 + 3 [ ∵ ab = 10a + b]
3. 129 = 100 × 1 + 10 × 2 + 9 [ ∵ abc = 100a + 10b + c]
4. 302 = 100 × 3 + 10 × 0 + 2 [ ∵ abc = 100a + 10b + c]

Question (ii)
Write the following in the usual form:

1. 10 × 5 + 6
2. 100 × 7 + 10 × 1 + 8
3. 100 × a + 10 × c + b

Solution:

1. 10 × 5 + 6 = 50 + 6 = 56
2. 100 × 7 + 10 × 1 + 8
   = 700 + 10 + 8 = 718
3. 100 × a + 10 × c + b
   = 100a + 10c + b = acb

Try These : [Textbook Page No. 251 ]

1. Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 27
Solution:
Let Sundaram choose the number = 27
Then reversed number = 72
∴ Sum of these two numbers
= 27 + 72 = 99
Now, 99 = 11 (9) = 11 (2 + 7)
= 11 (Sum of the digits of the chosen number)

Question (ii)
2. 39
Solution:
Let Sundaram choose the number = 39
Then reversed number = 93
∴ Sum of these two numbers
= 39 + 93 = 132
Now, 132 = 11 (12) = 11 (3 + 9)
= 11 (Sum of the digits of the chosen number)

Question (iii)
3. 64
Solution:
Let Sundaram choose the number = 64
Then reversed number = 46
∴ Sum of these two numbers
= 64 + 46 = 110
Now, 110= 11 (10) = 11 (6+ 4)
= 11 (Sum of the digits of the chosen number)

Question (iv)
4. 17
Solution:
Let Sundaram choose the number =17
Then reversed number = 71
∴ Sum of these two numbers
= 17 + 71 = 88
Now, 88 = 11 (8) = 11 (1 + 7)
= 11 (Sum of the digits of the chosen number)
[Note: From above results, it is clear that sum of 2-digit number and number obtained by interchanging its digit is multiple of 11, i.e., is divisible by 11, leaving remainder 0.]

Try These [Textbook Page No. 251]

Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 17
Solution:
Let Sundaram choose the number =17
Then reversed number = 71
∴ Difference of these numbers
= 71 – 17 = 54
Now, 54 = 9(6) = 9(7- 1)
= 9 (Difference of the digits of the chosen number)

Question (ii)
2. 21
Solution:
Let Sundaram choose the number = 21
Then reversed number =12
∴ Difference of these numbers
= 21 – 12 = 9
Now, 9 = 9(1) = 9(2 – 1)
= 9 (Difference of the digits of the chosen number)

Question (iii)
3. 96
Solution:
Let Sundaram choose the number = 96
Then reversed number = 69
∴ Difference of these numbers = 96 – 69 = 27
Now, 27 = 9 (3) = 9 (9-6)
= 9 (Difference of the digits of the chosen number)

Question (iv)
4. 37
Solution:
Let Sundaram choose the number = 37
Then reversed number = 73
∴ Difference of these numbers = 73 – 37 = 36
Now, 36 = 9(4) = 9(7 – 3)
= 9 (Difference of the digits of the chosen number)
[Note: From above results, it is clear that difference of 2-digit number and number obtained by interchanging its digit is a multiple of 9, i.e., is divisible by 9, leaving remainder 0.]

Try These: [Textbook Page No. 252]

Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end:

Question (i)
1. 132
Solution:
Let Minakshi choose the number =132
Then reversed number = 231
∴ Difference of these numbers = 231 – 132 = 99
Now, 99 ÷ 99 = 1, so remainder = 0

Question (ii)
2. 469
Solution:
Let Minakshi choose the number = 469
Then reversed number = 964
∴ Difference of these numbers = 964 – 469 = 495
Now, 495 ÷ 99 = 5, so remainder = 0

Question (iii)
3. 737
Solution:
Let Minakshi choose the number = 737
Then reversed number = 737
∴ Difference of these numbers = 737 – 737 = 0
Now, 0 ÷ 99 = 0, so remainder = 0

Question (iv)
4. 901
Solution:
Let Minakshi choose the number = 901
Then reversed number =109
∴ Difference of these numbers = 901 – 109 = 792
Now, 792 ÷ 99 = 8, so remainder = 0

[Note: From above results, it is clear s that difference of 3-digit number and number obtained by reversing | its digits (Interchanging unit and ; hundred’s place) is multiple of 99, i.e., is divisible by 99, leaving ’ remainder 0.]

Try These : [Textbook Page No. 253]

Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 417
Solution:
Let Sundaram choose the number = 417
By interchanging the digits, we get two ) numbers, which are 741 and 147.
∴ Sum of these three numbers = 417 + 741 + 174 = 1332
Now, dividing the sum by 37,
1332 ÷ 37 = 36, so remainder = 0

Question (ii)
2. 632
Solution:
Let Sundaram choose the number = 632
By interchanging digits, we get two numbers, which are 263 and 326.
∴ Sum of these three numbers = 632 + 263 + 326 = 1221
Now, dividing the sum by 37,
1221 ÷ 37 = 33, so remainder = 0

Question (iii)
3. 117
Solution:
Let Sundaram choose the number =117 By interchanging digits, we get two numbers, which are 711 and 171.
∴ Sum of these three numbers = 117 + 171 + 711 = 999
Now, dividing the sum by 37,
999 ÷ 37 = 27, so remainder = 0

Question (iv)
4. 937
Solution:
Let Sundaram choose the number = 937
By interchanging digits, we get two numbers, which are 379 and 793.
∴ Sum of these three numbers = 937 + 379 + 793 = 2109
Now, dividing the sum by 37,
2109 ÷ 37 = 57, so remainder = 0
[Note: Sum of 3-digit number and numbers formed by interchanging their digit is divisible by 37, leaving no remainder. ]

Try These : [Textbook Page No. 257]

Question (i)
If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N? (The ones digit, when divided by 5, must leave a remainder of 3. So the ones digit must be either 3 or 8.)
Solution:
The ones digit, when divided by 5 leaves a remainder of 3. So the ones digit must be either 3 or 8.
(Given in textbook, but let us make it simple by solving numerically.)

Question (ii)
If the division N ÷ 5 leaves a remainder of 1, what might be the ones digit of N?
Solution:
The ones digit, when divided by 5 leaves a remainder of 1. So the ones digit must be either 1 or 6.

Question (iii)
If the division N ÷ 5 leaves a remainder of 4, what might be the ones digit of N?
Solution:
The ones digit, when divided by 5 leaves a remainder of 4. So the ones digit must be either 4 or 9.

Try These : [Textbook Page No. 257 – 258]

Question (i)
If the division N ÷ 2 leaves a remainder of 1, what might be the ones digit of N? (N is odd; so its ones digit is odd. Therefore, the ones digit must be 1, 3, 5, 7 or 9.)
Solution:
N is odd, so its ones digit is odd. Therefore, the ones digit must be 1, 3, 5, 7 or 9.
(Given in textbook, but let us make it simple by solving numerically.)

Question (ii)
If the division N ÷ 2 leaves no remainder (i.e., zero remainder), what might be the one’s digit of N?
Solution:
Here, the remainder = 0. So N is an even number, i.e., its ones digit is even. Therefore, the one’s digit must be 0, 2, 4, 6 or 8.

Question (iii)
Suppose that the division N ÷ 5 leaves a remainder of 4 and the division N ÷ 2 leaves a remainder of 1. What must be the ones digit of N?
Solution:
Here, the remainder is 4. So ones digit of N should be 4 or 9. Again, N ÷ 2 leaves a remainder 1.
So ones digit of N is odd, i.e., ones digit is one of these 1, 3, 5, 7 or 9.
∴ 9 is a common ones digit in both the cases.
Therefore, ones digit of N must be 9.

Try These : [Textbook Page No. 259]

Check the divisibility of the following numbers by 9:

Question (i)
1. 108
Solution:
Sum of digits of 108 = 1 + 0 + 8 = 9
Now, 9 ÷ 9 = 1 and remainder = 0
Thus, 108 is divisible by 9.

Question (ii)
2.616
Solution:
Sum of digits of 616 = 6 +1 + 6 = 13
Now, 13 ÷ 9 = 1 and remainder = 4
Thus, 616 is not divisible by 9.

Question (iii)
3. 294
Solution:
Sum of digits of 294 = 2 + 9 + 4 = 15
Now, 15 ÷ 9 = 1 and remainder = 6
Thus, 294 is not divisible by 9.

Question (iv)
4. 432
Solution:
Sum of digits of 432 = 4 + 3 + 2 = 9
Now, 9 ÷ 9 = 1 and remainder = 0
Thus, 432 is divisible by 9.

Question (v)
5. 927
Solution:
Sum of digits of 927 = 9 + 2 + 7 = 18
Now, 18 ÷ 9 = 2 and remainder = 0
Thus, 927 is divisible by 9.

Think, Discuss and Write: [Textbook Page No. 259]

1. You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible by 3 which is a factor of 9. Can you say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m?
Solution:
Yes, if a number is divisible by any number m, then it will also be divisible by each of the factors of m. e.g. 12 is divisible by 6.
Factors of 6 are 2 and 3.
∴ 12 is also divisible by 2 and 3.

2.

Question (i)
Write a 3-digit number abc as 100a + 10b + c
= 99a + 11b + (a – b + c)
= 11 (9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about (a – b + c) ? Is it necessary that (a + c – b) should be divisible by 11?
Solution:
If the number abc is divisible by 11, then (a-b + c) is either 0 or a multiple of 11. Yes, it is necessary that (a + c-b) should be divisible by 11.

Question (ii)
Write a 4-digit number abed as 1000a + 100b + 10c + d = (1001a + 99b + 11c) – (a – b + c – d) = 11 (91a + 9b + c) + [(b + d) – (a + c)] If the number abed is divisible by 11, then what can you say about ((b + d) – (a + c)]?
Solution:
If the number abcd is divisible by 11, then [(b + d) – (a + c)] must be divisible by 11, i.e., [(b + d) – (a + c)] must be 0 or a multiple of 11.

Question (iii)
From (i) and (ii) above, can you say that a number will be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11?
Solution:
Yes, we can say that a number will be divisible by 11, if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11.

Try These : [Textbook Page No. 260]

Check the divisibility of the following numbers by 3:

Question (i)
1. 108
Solution:
Sum of digits of 108 = 1 + 0 + 8 = 9
Now, 9 is divisible by 3.
(∵ 9 ÷ 3 = 3, remainder = 0)
∴ 108 is divisible by 3.

Question (ii)
2. 616
Solution:
Sum of digits of 616 = 6 + 1 + 6 = 13
Now, 13 is not divisible by 3 leaving remainder 0.
(∵ 13 ÷ 3 = 4, remainder = 1)
∴ 616 is not divisible by 3.

Question (iii)
3. 294
Solution:
Sum of digits of 294 = 2 + 9 + 4 = 15
Now, 15 is divisible by 3.
(∵ 15 ÷ 3 = 5, remainder = 0)
∴ 294 is divisible by 3.

Question (iv)
4. 432
Solution:
Sum of digits of 432 = 4 + 3 + 2 = 9
Now, 9 is divisible by 3.
(∵ 9 ÷ 3 = 3, remainder = 0)
∴ 432 is divisible by 3.

Question (v)
5. 927
Solution:
Sum of digits of 927 = 9 + 2 + 7= 18
Now, 18 is divisible by 3.
(∵ 18 ÷ 3 = 6, remainder = 0)
∴ 927 is divisible by 3.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

In Question 1 to 3, choose the corred option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
A circle with centre O from a point, Q the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm.

∴ ∠QPO = 90°
Now, in right angled ∠OPQ,
OQ² = PQ² + OP²
(25)² = (24)² + OP²
Or 625 = 576 + OP²
Or OP² = 625 – 576
Or OP² = 49 = (7)²
Or OP = 7 cm
∴ Option (A) is correct.

Question 2.
In Fig., if TP and TQ and tangents to a circle with centre O so ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
In figure, OP is radius and PT is tangent to circle.
∠OPT = 90°
Similarly, ∠OQT = 90° and ∠POQ = 110° (Given)
Now, POQT is a Quadrilateral.
∴ ∠POQ + ∠OQT + ∠PTQ + ∠TPO = 360°
110° + 90° + ∠PTQ + 90° = 360°
Or ∠PTQ + 290° = 360°
Or ∠PTQ = 360° – 290°
Or ∠PTQ = 70°
∴ Option (B) is correct.

Question 3.
In tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then LPOA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
In given firgure, OA is radius and AP is a tangent to the circle.

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
Now, in right angled ∆PAO and ∆PBO
∠PAO = ∠PBO = 90°
OP = OP (Common side)
OA = OB (radii of same Circle)
∴ ∆PAO ≅ ∆PBO [RHS congruence]
∴ ∠AOP = ∠BOP [CPCT]
Or ∠AOP =∠BOP = \(\frac{1}{2}\) ∠AOB
Also, In Quad. OAPB,
∠OBP + ∠BPA + ∠PAO + ∠AOP = 360°
90° +80° +90° + ∠AOB = 360°
∠AOB = 360° – 260°
∠AOB = 100°
Form (1) and (2), we get
∠AOP = ∠BOP = \(\frac{1}{2}\) × 100° = 50°

∴ Option (A) is correct.

Question 4.
Prove that, the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given: A circle with center O and AB as its diameter l and m are tangents at points A and B.
To Prove: l || m
Proof: OA is the radius and l is the tangent to the circle.

∴ ∠1 = 90°
Similarly, ∠2 = 90°
Or ∠1 = ∠2 = 90°
But these are alternate angles between two lines, when one transversal cuts them.
∴ l || m
Hence, tangents drawn at the ends of a diameter of a circle are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Given. A circle with centre O. AB its tangent meet circle at P.
i.e., P is the point of contact.

To Prove: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
Construction: Join OP.
Proof: The perpendicular to a tangent line AB through the point of contact passes through the centre of the circle because only one perpendicular, OP can be drawn to the line AB through the point P.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at a distance 5 cm. from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
A circle with centre ‘O’ A is any point outside the circle at a distance of 5 cm from the centre.

Length of tangent = PA = 4 cm
Since, OP is the radius and PA is the tangent to the circle.
∠OPA = 90°
Now, in right angled ∠OPA.
Using Pythagoras Theorem.
OA² = OP² + PA²
(5)² = OP² + (4)²
Or OP² = 25 – 16
Or OP² = 9 = (3)²
Or OP = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of Ihe chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles having same centre O, and radii 5 cm and 3 cm respectively. Let PQ be the chord of larger circle
but tangent to the smaller circle.

Since, OM be the radius of smaller circle and PMQ be the tangent.

∴ ∠OMP = ∠OMQ = 90°
Consider, right angled triangles OMP and OMQ,
∠OMP = ∠OMQ = 90°
OP = OQ [radii of same circle]
OM = OM [common side]
∴ ∆OMP ≅ ∆OMQ [RHS congurence]
∴ PM = MQ [CPCT]
Or PQ = 2PM = 2MQ
Now, in right angled ∆OMQ.
Using Pythagoras Theorem,
OQ² = OM² + MQ²
(5)² = (3)² + (MQ)²
Or MQ = 25 – 9
Or MQ² = 16 = (4)²
Or MQ = 4 cm
Length of chord PQ = 2 MQ = 2 (4) cm = 8 cm
Hence, length of required chord is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to the circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

Solution:
Given: A Quadrilateral ABCD is drawn to circumscribe a circle.
To Prove: AB + CD = AD + BC
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside the circle and BP : BQ are tangents to the circle.
∴ BP = BQ ……………(1)
Similarly, AP = AS ………….(2)
and CR = CQ …………..(3)
Also, DR = DS ………….(4)
Adding (1), (2), (3) and (4), we get
(BP + AP) + (CR + DR) = (BQ + CQ) + (AS + DS)
AB + CD = BC + AD
is the required result.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given: A circle with centre O having two parallel tangents XY and X’Y’ and= another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B.
To Prove: ∠AOB = 90°
Contruction: Join OC, OA and OB.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal.
Now, A is any point outside the circle from two tangents PA and AC are drawn to the circle.
∴ PA = AC
Also, in ∆ POA and ∆ AOC,
PA = AC (Proved)
OA = OA (common side)
OP = OC (radii of same circle)
∴ ∆POA ≅ ∆AOC [SSS congruence]
and ∠PAO = ∠CAO [CPCT]
Or ∠PAC = 2 ∠PAO = 2 ∠CAO ……………(1)
Similarly, ∠QBC = 2∠OBC = 2 ∠OBQ ………………(2)
Now, ∠PAC + ∠QBC = 180°
[Sum of the interior angles on the same side of transversal is 180°]
Or 2∠CAO + 2∠OBC = 180° [Using (1) & (2)]
Or ∠CAO + ∠OBC = 180 = 90° …(3)
Now, in ∆OAB,
∠CAO + ∠OBC + ∠AOB = 180°
90°+ ∠AOB = 180° [Using (3)]
Or ∠AOB = 180° – 90° = 90°
Hence, ∠AOB = 90°.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. [Pb. 20191
Solution:
Given. A circle with centre O. P is any point outside the circle PQ and PR are the tangents to the given circle from point P.

To Prove. ∠ROQ + ∠QPR = 180°
Proof. OQ is the radius and PQ is tangent from point P to the given circle.
∴ ∠OQP = 90° ………..(1)
[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact]
Similarly, ∠ORP = 90°
Now, in quadrilateral ROQP,
∠ROQ + ∠PRO + ∠OQP + ∠QPR = 360°
Or ∠ROQ + 90° + 90° + ∠QPR = 360° [Using (1) & (2)]
Or ∠ROQ + ∠QPR + 180 = 360°
Or ∠ROQ + ∠QPR = 360° – 180°
Or ∠ROQ + ∠QPR = 180°
Hence, the angle between the two tangents drawn from and external point to a circle is supplementary to angle subtended by the line segment joining the pnts of contact at the centre.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Given: A parallelogram ABCD circumscribed a circle with centre O.
To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thé circle and BE; BF are tangents to the circle.

To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thè circle and BE; BF are tangents to the circle.
∴ BE = BF
Similarly AE = AH ……………(2)
and CG = CF
Also DG = DH
Adding (1), (2), (3) and (4), we get
(BE + AE) + (CG + DG) = (BF + CF) – (AH + DH)
Or AB + CD = BC +AD ……….(5)
Now, ABCD is a parallelogram, (Given)
∴ AB = CD and BC = AD …………(6)
From (5) and (6), we get
AB + AB = BC + BC
Or 2AB = 2BC or AB = BC
Or AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence, parallelogram circumscribing a circle is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig). Find the sides AB and AC.

Solution:
A triangle ABC is drawn to circumscribe a circle of radius 4 cm and the sides BC, CA, AB of ∆ABC touch the circle at
D, E, F respectively. Since the lengths of tangents drawn from an external point to a circle are equal.
∴ AE = AF = x cm(say)
CE = CD = 6 cm (Given)
and BF = BD = 5 cm
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC; OE ⊥ AC and OF ⊥ AB.
Also, OE = OD = OF = 4 cm.
Consider, ∆ABC
a = AC = (x + 6) cm ;
b = CB = (6 + 8) cm = 14 cm
c = BA = (8 + x) cm
S = \(\frac{a+b+c}{2}\)
∴ S = \(\frac{x+6+14+8+x}{2}\) = \(\frac{2 x+28}{2}\) = (x + 14)

area (∆ABC)= \(\sqrt{\mathrm{S}(\mathrm{S}-a)(\mathrm{S}-b)(\mathrm{S}-c)}\)

= \(\sqrt{\begin{array}{r}
(x+14)(x+14-\overline{x+6}) \\
(x+14-14)(x+14-\overline{8+x})
\end{array}}\)

= \(\sqrt{(x+14)(8)(x)(6)}\)

= \(\sqrt{48 x^{2}+672 x}\) cm² ………………(1)

area (∆OBC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28cm² …………….(2)

area(∆BOA)= \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (8 + x) × 4 = 28cm²…………….(3)

area (∆AOC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (6 + x) × 4 = 28cm² …………….(4)

From the figure, by addition of areas, we have
Or (∆ABC) = ar (∆OBC) + ar (∆BOA) + ar (∆AOC)
\(\sqrt{48 x^{2}+672 x}\) = 28 + 16 + 2x + 12 + 2x
Or \(\sqrt{48 x^{2}+672 x}\) = 4x + 56
Or 48x² + 672x =4[x+ 14]
Squaring both sides, we get
Or 48x² + 672x = 16 (x + 14)²
Or 48x (x + 14) = 16(x + 14)²
Or 3x = x + 14
Or 2x = 14
Or x = \(\frac{14}{2}\) = 7
∴ AC = (x + 6) cm = (7 + 6) cm= 13 cm
and AB = (x + 8) cm = (7 + 8) cm = 15 cm
Hence, AB = 15 cm and AC = 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given:
A quadrilateral PQRS circumscribing a circle having centre O. Sides PQ, QR, RS and SP touches the circles at L, M, N, T
respectively.

To Prove:
∠POQ + ∠SOR = 180°
and ∠SOP + ∠ROQ = 180°
Construction:
Join OP, OL, OQ, 0M. OR, ON, OS, OT
Proof: Since the two tangents drawn from an external point subtend equal angles at the centre.
∴ ∠2 = ∠3; ∠4 = ∠5 ; ∠6 = ∠7; ∠8 = ∠1
But, sum of all angles around a point is 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
Or ∠1 + ∠2 + ∠2 + ∠5 +∠5 + ∠6 + ∠6 + ∠1 = 360°
Or 2(∠1 + ∠2 + ∠5 + ∠6) = 360°
Or (∠1 + ∠2) + (∠5 + ∠6) = \(\frac{360^{\circ}}{2}\) = 180°
Or ∠POQ + ∠SOR = 180°
Similarly, ∠SOP + ∠ROQ = 180°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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

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

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used:

(i) AD = ……………
(ii) ∠DCB = ……………
(iii) OC = ……………
(iv) m∠DAB + m∠CDA = ……………
Solution:
(i) AD = BC
(∵ Opposite sides are equal)

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

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

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

Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
(i)

Solution:
□ ABCD is a parallelogram.
∴ ZB = ZD
∴ y = 100°
(∵ Opposite angles)
Now, y + z = 180° (∵ Adjacent singles are supplementary)
∴ 100° + z = 180°
∴ z = 180° – 100°
∴ z = 80°
Now, x = z (∵ Opposite angles)
∴ x = 80°
Thus, x = 80°, y = 100° and z = 80°.

(ii)

Solution:
It is a parallelogram.

∴ m∠P + m∠S = 180°
(∵ Adjacent angles are supplementary)
∴ x + 50° = 180°
∴ x = 180° – 50°
∴ x = 130°
x = y (∵ Opposite angles)
∴ y = 130°
Now, m∠Q = 50° (∵ ∠S and ∠Q are opposite angles)
m∠Q + z = 180° (∵ Linear pair of angles)
∴ 50° + z = 180°
∴ z = 180° – 50°
∴ z = 130°
Thus, x = 130°, y = 130° and z = 130°.

(iii)

Solution:
Vertically opposite angles are equal.

∴ m∠AMD = m∠BMC = 90°
∴ x = 90°
In ΔBMC
y + 90° + 30° = 180°
(Angle sum property of a triangle)
∴ y + 120° = 180°
∴ y = 180° – 120° = 60°
Here, ABCD is a parallelogram
∴ \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) is a transversal.
∴ y = z (∵ Alternate angles)
∴ z = 60° (∵ y = 60°)
Thus, x = 90°, y = 60° and z = 60°

(iv)

Solution:
□ ABCD is a Dparallelogram.

∠D = ∠B (∵ Opposite angles)
∴ y = 80°
m∠A + m∠D = 180° (∵ Adjacent angles are supplementary.)
∴ x + y = 180°
∴ x + 80° = 180°
∴ x = 180° – 80°
∴ x = 100°
m∠A = m∠BCD (∵ Opposite angles are equal)
∴ 100° = m∠BCD
Now, z + m∠BCD = 180° (∵ Linear pair of angles)
∴ z + 100° = 180°
∴ z = 180° – 100°
∴ z = 80°
Thus, x = 100°, y = 80° and z = 80°

(v)

Solution:
□ ABCD is a parallelogram.

∴ m∠B = m∠D (∵ Opposite angles)
∴ y = 112°
m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ (40° + z) + 112° = 180°
∴ 40° + z + 112° = 180°
∴ z + 152° = 180°
∴ z = 180° – 152°
∴ z = 28°
Now, \(\overline{\mathrm{DC}} \| \overline{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{AC}}\) is their transversal.
∴ z = x
∴ x = 28° (∵ z = 28°)
Thus, x = 28°, y = 112° and z = 28°

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

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

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

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:

Here, ∠B = ∠D (see figure)
Yet, it is not a parallelogram.

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.

Adjacent angles are supplementary.
∴ m∠A + m∠B = 180°
∴ 3x + 2x – 180°
∴ 5x = 180°
∴ \(\frac{180^{\circ}}{5}\)
∴ m∠A = 3x = 3 × 36° = 108°
m∠B = 2x = 2 × 36° = 72°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 108° and m∠D = 72°
Thus, ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B.

∴ m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ m∠A + m∠A = 180° (m∠B = m∠A)
∴ 2m∠A = 180°
∴ m∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ m∠B = 90°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 90° and m∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°.

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:
∠POA is the exterior angle of ΔHOE
∴ y + z = 70° (Exterior)
∴ m∠HOP = 180° – 70° = 110°
∠HEP = ∠HOP = 110° (Opposite angles of a parallelogram)
∴ x = 110°
∵ \(\overline{\mathrm{EH}} \| \overline{\mathrm{PO}}\), \(\overleftrightarrow{\mathrm{PH}}\) is a transversal.
∴ y = 40° (∵ Alternate angles are equal)
Now, y + z = 70°
∴ 40° + z = 70°
∴ z = 70° – 40° = 30°
Thus, x = 110°, y = 40° and z = 30°

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i)

Solution :
□ GUNS is a parallelogram.
∴ GS = NU and SN = GU (Y Opposite sides)
∴ 3x = 18 and 26 = 3y – 1
∴ x = \(\frac {18}{3}\)
∴ x = 6

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

(ii)

Solution:
□ RUNS is a parallelogram.
∴ Its diagonals bisect each other.
x + y = 16 ……(1)
and y + 7 = 20
y = 20 – 7 = 13 …..(2)
Substituting value of y in (1)
x + y = 16
x + 13 = 16
∴ x = 16 – 13
= 3
Thus, x = 3 cm and y = 13 cm

Question 9.

In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
□ RISK is a parallelogram.
∴ m∠R + m∠K = 180°
(∵ Adjacent angles are supplementary)
∴ m∠R + 120° = 180°
∴ m∠R = 180° – 120°
∴ m∠R = 60°
∠R and ∠S are opposite angles of parallelogram.
∴ m∠S = 60°
□ CLUE is a parallelogram.
∴ m∠E = m∠L = 70° (∵ Opposite angles)
In ΔESQ
∴ m∠E + m∠S + x = 180° (∵ Angle sum property of triangle)
∴ 70° + 60° + x = 180°
∴ 130° + x = 180°
∴ x = 180° – 130°
∴ x = 50°
Thus, x = 50°

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel ? (Fig3.32)

Solution:
Since, 100° + 80° = 180°
i. e., ∠M and ∠L are supplementary.
∴ \(\overline{\mathrm{NM}} \| \overline{\mathrm{KL}}\)
(∵ Interior angles along on the same side of the transversal are supplementary)
One pair of opposite side of □ LMNK is parallel.
∴ □ LMNK is a trapezium.

Question 11.
Find m∠C in figure 3.33 if \(\overline{\mathbf{A B}} \| \overline{\mathbf{D C}}\).

Solution:
In □ ABCD, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) and BC is their transversal,
∴ m∠B + m∠C = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ 120° + m∠C = 180°
∴ m∠C = 180° – 120° = 60°
Thus, m∠C = 60°

Question 12.
Find the measure of ∠P and ∠S if \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) in figure 3.34. (If you find m∠R, is there more than one method to find m∠P ?)

Solution:
In □ PQRS \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
∴ □ PQRS is a trapezium.
\(\overleftrightarrow{\mathrm{PQ}}\) is a transversal of \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
m∠P + m∠Q = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ mZP + 130° = 180°
∴ mZP = 180° – 130°
∴ mZP = 50°
In □ PQRS, ∠R = 90°
\(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\), \(\overleftrightarrow{\mathrm{RS}}\) is their transversal.
∴ m∠S + m∠R = 180°
∴ m∠S + 90° = 180°
∴ m∠S= 180°-90°
∴ m∠S = 90°
Yes, the sum of the measures of all angles of quadrilateral is 360°. ∠P and ∠R can be found.
m∠P + m∠Q + m∠R + m∠S = 360°
∴ m∠P + 130° + 90° + 90° = 360°
∴ m∠P + 310° = 360°
∴ m∠P = 360° – 310°
∴ m∠P = 50°

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers Ex 16.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution:
21y5 is a multiple of 9 (given)
∴ Sum of digits of 21y5 = 2 + 1 + y + 5
= 8 + y
Therefore, (8 + y) should be 0, 9, 18, …, etc.
8 + y = 0 is not possible ∴ 8 + y = 9
∴ y = 9 – 8 = 1
Thus, the value of y is 1.

Verification:
21y5 = 2115 (∵ y = 1)
∴ Sum of digits of 2115 = 2 + 1 + 1 + 5 = 9 (9 ÷ 9 = 1, remainder = 0)
∴ 2115 is divisible by 9.
(Note: Here, verification is given to explain you.]

2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution:
31z5 is a multiple of 9. (given)
∴ Sum of digits of 31z5 = 3 + 1 + z + 5
= z + 9
Therefore, (z + 9) should be 0, 9, 18, …, etc.
Since, z is a digit, it should be either 0 or 9.
Hence, z = 0 or 9.

Verification:
31z5 = 3105 (∵ z = 0)
∴ Sum of digits of 3105
= 3 + 1 + 0 + 5 = 9
(9 ÷ 9 = 1, remainder = 0)
∴ 3105 is divisible by 9.
31z5 = 3195 (∵ z = 9)
∴ Sum of digits of 3195
= 3 + 1 + 9 + 5 = 18
(18 ÷ 9 = 2, remainder = 0)
∴ 3195 is divisible by 9.

3. If 24x is a multiple of 3, where x is a digit, what is the value of x? (Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 …………. But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution:
24x is a multiple of 3. (given)
∴ Sum of digits of 24x = 2 + 4 + x = 6 + x
Therefore, 6 + x should be 0, 3, 6, 9, 12, …, etc.
Since, 6 + x is a multiple of 3.
∴ 6 + x = 0, 6 + x = 3, 6 + x = 6, 6 + x = 9, 6 + x = 12, 6 + x = 15, 6 + x = 18, ……….
∴ x = -6, x = -3, x = 0, x = 3, x = 6, x = 9, x = 12, ……………..
Here, x = 0, 3, 6, 9 are possible.
Thus, the value of x can be 0 or 3 or 6 or 9.

4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
31z5 is a multiple of 3. (given)
∴ 31z5 is divisible by 3.
Sum of digits of 31z5 = 3 + 1 + z + 5
= 9 + z
∴ 9 + z is divisible by 3.
∴ Value of 9 + z should be 0, 3, 6, 9, 12, 15 or 18.
Since, z is a multiple of 3.
If 9 + z = 0,
∴ z = – 9 which is impossible.
9 + z = 3,
∴ z = – 6 which is impossible.
9 + z = 6,
∴ z = – 3 which is impossible.
9 + z = 9,
∴ z = 0 which is possible.
9 + z = 12,
∴ z = 3 which is possible.
9 + z = 15,
∴ z = 6 which is possible.
9 + z = 18,
∴ z = 9 which is possible.
9 + z = 21,
∴ z = 12 which is impossible.
Thus, the value of z can be 0 or 3 or 6 or 9.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 1.
(a)

Solution:
Sum of all the exterior angles of a polygon = 360°.
∴ x + 125° + 125° = 360°
∴ x + 250° = 360°
∴ x = 360° – 250° (Transposing 250° to RHS)
∴ x = 110°

(b)

Solution:
In this figure, two exterior angles are of 90° each. (one interior angle is 90°)
Sum of all exterior angles of a polygon = 360°.
∴ x + 90° + 60° + 90° + 70° = 360°
∴ x + 310° = 360°
∴ x = 360° – 310° (Transposing 310° to RHS)
∴ x = 50°

Question 2.
Find the measure of each exterior angle of a regular polygon of (i) 9 sides (ii) 15 sides
Solution:
(i) Number of sides (n) = 9
∴ Number of exterior angles = 9
The sum of all exterior angles = 360°.
The given polygon is a regular polygon.
∴ All the exterior angles are equal.
∴ Measure of an exterior angle = \(\frac{360^{\circ}}{9}\).
= 40°

(ii) Number of sides of regular polygon = 15
∴ Number of exterior angles = 15
The sum of all the exterior angles = 360°
The given polygon is a regular polygon.
∴ All the exterior angles are equal.
∴ The measure of each exterior angle = \(\frac{360^{\circ}}{15}\) = 24°

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24° ?
Solution:
Regular polygon is equiangular.
Sum of all the exterior angles = 360°
Measure of an exterior angle = 24°
∴ Number of sides = \(\frac{360^{\circ}}{24^{\circ}}\)
The polygon has 15 sides.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165° ?
Solution:
The given polygon is regular polygon.
Each interior angle = 165°
∴ Each exterior angle = 180° – 165° = 15°
∴ Number of sides = \(\frac{360^{\circ}}{15^{\circ}}\) = 24
The polygon has 24 sides.

Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle as 22° ?
Solution:
Each exterior angle = 22°
∴ Number of sides = \(\frac{360^{\circ}}{22^{\circ}}=\frac{180^{\circ}}{11^{\circ}}\)
The number of sides of a regular polygon must be a whole number.
But, \(\frac {180}{11}\) is not a whole number.
∴ No, exterior angle of a regular polygon cannot be of measure 22°.

(b) Can it be an interior angle of a regular polygon ? Why ?
Solution:
If the measure of an interior angle of a polygon is 22°, then the measure of its exterior angle = 180° – 22° = 158°.
∴ Number of sides = \(\frac{360^{\circ}}{158^{\circ}}=\frac{180^{\circ}}{79^{\circ}}\)
\(\frac {180}{79}\)is not a whole number.
∴ No, 22° cannot be an interior angle of a regular polygon.

Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why ?
Solution:
The minimum number of sides of a polygon = 3
The regular polygon of 3-sides is an equilateral triangle.
Each interior angle of an equilateral triangle = 60°.
Hence, the minimum possible interior angle of a polygon = 60°.

(b) What is the maximum exterior angle possible for a regular polygon ?
Solution:
The sum of an exterior angle and its corresponding interior angle is 180°. (Linear pair)
And minimum interior angle of a regular polygon = 60°.
∴ The maximum exterior angle of a regular polygon = 180° – 60° = 120°.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers Ex 16.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

Find the values of the letters in each of the following and give reasons for the steps involved:

Question 1.

Solution:
Here, we have two letters A and B, whose values are to be found.

By observing the ones column, we have A + 5 and we get 2 from this.
1. e., a number whose ones digit is 2, for this A has to be 7.
∴ A + 5 = 7 + 5 = 12

Now, for the sum in tens column, we have 1 + 3 + 2 = B
∴ B = 6
Thus,

Thus, A = 7 and B = 6

Question 2.

Solution:
Here, we have three letters A, B and C, whose values are to be found.

By observing the ones column, we have A + 8 and we get 3 from this.
i. e., a number whose ones digit is 3, for this A has to be 5.
∴ A + 8 = 5 + 8 = 13
Now, for the sum in tens column, we have 1 + 4 + 9 = CB
∴ CB = 14
Here, B = 4 and C = 1

Thus, A = 5, B = 4 and C = 1

Question 3.

Solution:
Here, we have A, whose value is to be found.

Since, product of ones digit
A × A = A, so it must be 1, 5 or 6.
When A = 1, then 1 1 But, the product is 9 A, so A = 1 is not possible.

When A = 5, then
But, the product is 9 A, so A = 5 is not possible.

When A = 6, then

Thus, A = 6

Question 4.

Solution:
Here, we have two letters A and B, whose values are to be found.

By observing the ones column, we have B + 7 and we get A from this, i. e., a number whose ones digit is A.
Now, for the sum in tens column, we have A + 3 and we get 6 from this. Therefore, the value of A must be 2. (Keeping in mind that carry over 1 is to be considered.)
If A = 2, then

Then B + 7 gives 2, so B + 37 must be 5 and sum in tens 6 2 column is 1 + 2 + 3 = 6, so it is correct.

Question 5.

Solution:
Here, we have three letters A, B and C, whose values are to be found.

Units digit of 3 × B is B, so B must be either 0 or 5.

When B = 0, then

When B = 5, then

Now, units digit of 3 × A is A. So A must be either 0 or 5, but A cannot be 0, because if A = 0, then AB becomes one-digit number. So A must be 5 and multiplication is either 55 × 3 or 50 × 3.
55 × 3 = 165, here A = 6, so this is not possible.
∴ 50 × 3 = 150

Thus, A = 5, B = 0 and C = 6

Question 6.

Solution:
Here, we have three letters A, B and C, whose values are to be found.

Units digit B × 5 = B, so B must be either 0 or 5.
When B = 0, then

When B = 5, then

Now, units digit of 5 × A = A, so A must be either 0 or 5.
There are three letters as a product. So A ≠ 0, but A = 5.
So multiplication is either 50 × 5 or 55 × 5.
50 × 5 = 250 and 55 × 5 = 275, so 55 × 5 is not correct.
So 50 × 5 = 250

Thus, A = 5, B = 0 and C = 2.

Question 7.

Solution:
Here, we have two letters A and B, whose values are to be found.

Units digit B × 6 = B, so B must be 2, 4, 6 or 8.
∴ Possible values of product BBB are 222, 444, 666 or 888.
If we divide these numbers by 6, then quotient should be A2, A4, A6 or A8.
Now, 222 ÷ 6 = 37, remainder = 0
But, the quotient is not as A2, so B = 2 is not possible.

Thus, A = 7 and B = 4

Question 8.

Solution:
Here, we have two letters A and B, whose values are to be found.

By observing the sum in unit column, we have 1 + B = 0. So here is a number whose unit digit is 0, so B must be 9.
Now, for sum in tens column, we have 1 + A + 1 = 9.
So A must be 7.

Thus, A = 7 and B = 9

Question 9.

Solution:
Here, we have two letters A and B, whose values are to be found.

By observing the sum in units column, we have B + 1 = 8.
∴ B must be 7.
Now,

By observing the sum in tens digit column, we have A + 7 = 1, i.e., whose unit’s digit is 1, so A must be 4.

Thus, A = 4 and B = 7

Question 10.

Solution:
Here, we have two letters A and B, whose values are to be found.

By observing the sum in tens column, we have 2 + A = 0, so A must be 8.
Then,

Now, by observing the sum in units column, we have 8 + B = 9, so B must be 1.

Thus, A = 8 and B = 1.

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

PSEB 8th Class English Letter/Application Writing

पत्र लिखना एक कला है। एक अच्छा पत्र लिख कर आप शत्रु का दिल भी जीत सकते हैं। पत्र मुख्यतः तीन प्रकार के होते हैं-व्यक्तिगत, व्यावसायिक अथवा अधिकारियों को लिखे गए पत्र।

व्यक्तिगत पत्र (Personal Letters)-व्यक्तिगत पत्र वे पत्र होते हैं जो हम अपने मित्रों या सगे-सम्बन्धियों को लिखते हैं। ऐसे पत्र प्रायः समाचारों से भरे होते हैं। इन पत्रों में हम अपने घर, परिवार, स्कूल या किसी अनुभव का वर्णन करते हैं। ये पत्र एक विशेष विधि द्वारा लिखे जाते हैं। इन्हें ऐसे लिखा जाता है जैसे कि आप किसी व्यक्ति से बातचीत कर रहे हों। पते के अतिरिक्त इन पत्रों के छ: भाग होते हैं।

1. लिखने वाले का पता (The address of the sender)
2. तिथि (Date)
3. सम्बोधन या अभिवादन (The salutation or greeting)
4. विषय-वस्तु (The body of letter)
5. विधिवत् अन्त (The subscription or complimentary close)
6. हस्ताक्षर (Signature)

प्रत्येक भाग को लिखने की अपनी अलग विधि होती है। आओ हम एक-एक करके इनका अध्ययन करें-
1. The Address of the Sender. पहले पत्र भेजने वाले का पता पृष्ठ पर सबसे ऊपर दायें कोने में लिखा जाता था। प्रत्येक लाइन में अन्त में Comma और आखिरी लाइन के अन्त में Full Stop लगाया जाता था।

परन्तु अब पता लिखते समय विराम चिन्ह लगाने की विधि में अन्तर आ गया है। अब पंक्तियों के अन्त में Commas और Full Stop का प्रयोग नहीं होता। इसके अतिरिक्त अब Sender’s Address दाहिने कोने की बजाये बाएं कोने में लिखा जाता है।

15 Model Town
Rajpura
2. Date. तिथि लिखने वाले के पते के बिल्कुल नीचे लिखी जाती है। तिथि लिखने की कई विधियां हैं, जैसे-
15 Model Town
Rajpura
March 12, 20……
या
12th March 20…..
या
12 March 20……
3. The Salutation or Greeting. (i) मित्र को पत्र लिखते समय उसे उसके नाम से सम्बोधित कीजिए; जैसे-Dear Mohan, जहां तक हो सके मित्र को उसी नाम (संक्षिप्त) से संबोधित कीजिए जिस नाम से आप बातचीत करते समय उसे सम्बोधित करते हैं; जैसे,
Dear Monu
Dear Sonu

(ii) माता-पिता, भाइयों, बहनों अथवा घनिष्ठ सगे सम्बन्धियों को आप यूं सम्बोधित कर सकते हैं-
My dear Father or Daddy/Dad
My dear Mother or Mamma/Mom
My dear Brother My dear Gopal
My dear Kamla
नोट-यदि आप Address में विराम चिन्ह नहीं लगाते तो आप सम्बोधन में भी इनका प्रयोग न करें।

4. The Body of the Letter-यह पत्र का सबसे महत्त्वपूर्ण भाग है। एक अच्छा पत्र वह माना जाता है जो सरल हो और जिसमें पाठक की रुचि बनी रहे। पत्र की प्रत्येक पंक्ति अर्थपूर्ण एवं रोचक हो। पत्र ऐसे लिखना चाहिए मानो आपका सम्बन्धी आपके सामने बैठा हो और आप से बातें कर रहा हो। पत्र का यह भाग लिखते समय विराम चिन्हों और Grammar के अन्य सभी नियमों का पूरा पालन करें।

5. The Subscription or Complimentary Close-(i) मित्र को लिखे गए पत्र का अन्त इस प्रकार करें
Yours sincerely
या
Your sincere friend

(ii) सगे-सम्बन्धियों को लिखे गये पत्रों का अन्त इस प्रकार करें-
Yours affectionately

(iii) निम्नलिखित phrases के साथ भी व्यक्तिगत पत्रों का अन्त किया जा सकता है- . Ever sincerely yours
या
Your loving son/Love
या
Lovingly yours
नोट-यदि आपने सम्बोधन के समय comma नहीं लगाया तो आप पत्र का अन्त करने वाले phrase के साथ भी comma न लगायें।

6. Signature-व्यक्तिगत पत्र में आप को अपने पूरे हस्ताक्षर नहीं करने चाहिएं। आपको या तो नाम का पहला हिस्सा लिखना चाहिए या फिर वह नाम लिखें जिस नाम से आप अपने सगे-सम्बन्धियों या मित्रों द्वारा पुकारे जाते हैं। जैसे-
Sohan
या
Sonu आप Sohan Lal Gupta अर्थात् पूरा नाम न लिखें।

Salutation and Subscription at a Glance

Official Letters या Business Letters.
(i) ऐसे पत्रों में कोई विशेष अन्तर तो नहीं होता। अन्तर केवल इतना है कि इन पत्रों के प्राय: छ: की बजाए सात भाग होते हैं।

(ii) ऐसे पत्रों में पत्र लिखने वाले का पता और तिथि व्यक्तिगत पत्रों की तरह ही लिखे जाते हैं।’

(iii) ऐसे पत्रों में उस फर्म (Firm) या व्यक्ति का पूरा पता भी पत्र में लिखा जाता है जिसे पत्र लिखा जा रहा हो, जैसे-
Messrs Malhotra Book Depot
(Producers of Quality Books)
Railway Road
Jalandhar City

(iv) ऐसे पत्रों में सम्बोधन की विधि इस प्रकार होती है
Dear Sir
या
Dear Sirs
या
Madam
या
Sir
नोट-यदि व्यक्ति आप से परिचित है तो आप उसे यूं भी सम्बोधित कर सकते हैं-
Dear Mr. Sharma

(v) ऐसे पत्रों की विषय-वस्तु संक्षिप्त, स्पष्ट और विनम्र-भाषी होनी चाहिए।

(vi) ऐसे पत्रों का अन्त यूं करना चाहिए
Yours truly
या
Yours faithfully

(vii) आप माननीय व्यक्तियों अथवा उच्च पद पर आसीन व्यक्तियों को पत्र लिखते हुए Yours respectfully या Yours obediently का प्रयोग भी कर सकते हैं। नाम से सम्बोधित पत्रों का अन्त Yours sincerely से ही करना चाहिए।

(viii) ऐसे पत्रों में हस्ताक्षर पूरे होने चाहिएं और हस्ताक्षर के पश्चात् लिखने वाले का नाम या पद लिखा जाना चाहिए, जैसे-
Yours faithfully
Mohan Lal Sharma
Manager

Important Applications and Letters

1. Application for Marriage Leave
Write an application to the Headmistress of your school for marriage leave.

The Headmistress
Khalsa Public School
Nawanshahr
Madam

I beg to say that the marriage of my elder brother/sister takes place next week. I am to help my parents in making marriage arrangements. So I cannot come to school. Kindly grant me leave for five days. I shall be very thankful to you for this.

Yours obediently
Jaspinder Kaur
Roll No. 25
23. VIII A
March 15, 20 ………….

Word-Meanings-Marriage = विवाह, Takes place = होगी, Arrangements = व्यवस्थाएं, Much = अधिक.

2. Leave for Urgent Work
Write an application to the Principal of your school for leave for a day.

The Principal
A.B. Sen. Sec. School
Patiala Sir

I beg to say that I have an urgent piece of work at home. So I cannot come to school. Kindly grant me leave for today. I shall be thankful to you for this.

Yours obediently
Raman
Roll No. 25
VIII C
February 5, 20 ……

Word-Meanings-Urgent piece of work = आवश्यक कार्य, Grant = प्रदान करो

3. Application for a Testimonial
Imagine you are Kavita, a student of Govt. Sen. Sec School, Ludhiana. Write an application to the Principal of your school requesting him to send a testimonial as you are applying for the post of a clerk.

11 Happy House
Model Town
Jalandhar
The Principal
Govt. Sen. Sec. School
Ludhiana
Sir

I beg to say that I am an old student of your school. I have to apply for the post of a clerk in a bank. I need a testimonial from you for that. I am writing down the following details for reference.

I passed my Sen. Sec. Examination in 2001 with 600 marks. I stood first in the district. I was a member of the school hockey team. I took part in the debates also and won many trophies and cups for the school. I was in the good books of my teachers.

Kindly send me the testimonial as soon as possible and oblige.

Yours obediently
Kavita
March 14, 20……

Word-Meanings-Testimonial = अचरण पत्र, In the good books of = नजरों में अच्छा, Oblige = कृतार्थ करें।

4. Application for Change of Section
Write an application to the Principal of your school requesting him to change your section.

The Principal
Govt. Sen. Sec. School
Sirhind
Sir

I am a student of VIII B of your school. I live in Main Bazaar. All my friends are in VIII A Section of the school. So I feel very lonely in VIII B.

Besides, I have to do my home task alone at home. Whenever I am unable to attend the school, I cannot do my homework. There is none to tell me about the homework given by the class teacher. The boys of my street can help me in other ways also. I hope you will appreciate my problem and change my section.
Thanking you

Yours obediently
Kirpal Singh
Roll No. 46
VIII B
March 15, 20……

Word-Meanings- Feel = अनुभव करना, Ways = ढंग, Lonely = अकेला, Appreciate = समझना।

5. For Admission to the Next Class
Write an application to the Headmaster of your school requesting him for admission to the next class.

The Headmaster
Govt. High School
Phagwara
Sir

I beg to say that I am a student of 7th class of your school. I fell ill in March. Still I appeared in the annual examination. I failed but my class-fellows went to the next class.

I am a good student. I got first division in the half-yearly examination. Kindly give me a test on any day and admit me to the 8th class. I assure you that I shall get first division in Middle Standard Examination.
Thanking you

Yours obediently
Mohan Lal
VIII C
March 5, 20…….

Word-Meanings- Annual = वार्षिक, Division = श्रेणी.

6. Application for Late Fee
Write an application to the Headmaster of your school requesting him to permit you to pay your fee for the month late by ten days.

The Headmaster
Govt. High School
Hoshiarpur
Sir

I beg to say that I am a student of VIII A of your school. Tomorrow is fee day and I am unable to pay it. My father has gone to Delhi. He will come back in ten days. Kindly allow me to pay my fee late by ten days.

Thanking you

Yours obediently
Sohan Lal
VIII A
March 9, 20……..

Word-Meanings- Unable = असमर्थ, Allow = अनुमति देना

7. Permission to take part in Games
Write an application to the Headmaster of your school requesting him to permit you to take part in the evening games.

The Headmaster
A.B.C. High School
Moga
Sir

I am a student of VIII D of your school. Our school is preparing for the Distt. Sports Meet. There are regular games in the school in the evening. I am also fond of games. I am a good player of football. Last year I was a member of the school football team. But this year my name is not in the players’ list because I could not perform well in the final trial. I request you to give me another chance and permit me to take part in the evening games. I assure you that I will try my best to improve my performance.
Thanking you

Yours obediently
Vijay Kumar
Roll No. 25
VIII D
Dated : March 9, 20…………

Word Meanings-Take part = भाग लेना, Permit = अनुमति देना, Assure = विश्वास दिलाना।

8. Permission to go on a Historical Tour
Write an application to your Principal requesting him/her to permit you to go on a historical tour.

The Principal
Govt. Sen. Sec. School
Ladowal
Madam

We, the students of class VIII, beg to seek your permission for a historical tour. The school will close for the summer vacation next week. We want to see the Taj Mahal. We also want to go to Fatehpur Sikri. On our way back, we want to visit Delhi. The trip will be very useful for us. It will give us first hand knowledge of History. The trip will cost two hundred rupees per head. Our teacher of History has agreed to take us to these places. We are 30 girls in all. I hope you will arrange this trip.
Thanking you

Yours obediently
Manjit Kaur
VIII A
April 30, 20……

Word-Meanings- Summer = गर्मी, Trip = भ्रमण, Useful = उपयोगी, Knowledge = ज्ञान

11. On Recovery From Illness
Write a letter to your friend congratulating him on his recovery from long illness.

135 New Road
Nawanshahr
March 18, 20…..
Dear Mohan

I am glad to know that you have recovered from long illness. I congratulate you on your recovery. You worked very hard. So you fell ill. Now please take care of your health.

Take long walks in the morning. Drink milk and eat fruit. All this will make you healthy soon.
With best wishes.

Yours sincerely
Hardeep

Word-Meanings-Glad = प्रसन्न, Recovered = ठीक हो गये हो, Take care of = ध्यान रखो, Congratulate = बधाई देना

12. Invitation on Brother’s Marriage
Write a letter to your friend inviting him to your brother’s marriage.

Govt. High School
Amritsar
March 4, 20………
My dear Kamlesh

You will be glad to know that the marriage of my elder brother comes off on March 9, 20……. The marriage party will leave Amritsar for Delhi the same day. We invite you to join us in our joys.

You know that Delhi is a historical city. There are many buildings worth-seeing. You will see the Red Fort, the Qutab Minar and Jantar Mantar. Rina and Tina have also been invited. They will reach here on Sunday. We will have good time together.
I hope you will reach in time.

Yours sincerely
Mitlesh
Address:
Kumari Kamlesh
45 Mall Road
Shimla

Word-Meanings-Comes off = पड़ती है, Marriage party = बारात, Worth-seeing = दर्शनीय

13. Inviting a Friend to the Birthday Party
Suppose you are Harish. You live at 38 Manavta Park, Hoshiarpur. Invite your friend to come to your birthday party.

38 Manavta Park
Hoshiarpur
February 22, 20……
My dear Surinder

You will be glad to know that my birthday falls on next Monday. There will be a tea party in the evening. I have invited all my friends to the party. I cannot forget you on this day. Please reach here on Sunday evening. We will have a good time together.
Thanking you

Yours sincerely
Harish

Word-Meanings-Falls on = पड़ता है, Invited = आमन्त्रित किया है, Forget = भूलना।

14. To Uncle for a Birthday Gift
Suppose you are Poonam. You live at 232, Phase IX, Mohali. Your uncle has sent you a wrist-watch on your birthday. Write a letter of thanks to your uncle.

232 Phase
IX Mohali
March 8, 20……
My dear Uncle

It is very kind of you to remember me on my birthday. You have sent me a beautiful wrist-watch as a gift. It shows your love for me. I received many gifts that day but I liked your gift the most. Everybody praised it.

The watch will help me a lot. It will make my life regular. I shall never be late for school now. I thank you for this lovely gift. I assure you thar I shall keep this watch with great care.

With regards
Yours lovingly
Poonam
Address :
Shri Manohar Singh
Joginder Nagar
Rohtak

Word-Meanings-Gift = उपहार, The most = सबसे बढ़कर, Praised = प्रशंसा की, A lot = बहुत ज्यादा, Regular = नियमित, Care = ध्यान.

15. To Younger Brother to Take Interest in Studies
Write a letter to your younger brother/sister scolding him/her for neglecting studies.

18 Mohan Nagar
Batala
February 18, 20…….
My dear Suman

I received your progress report yesterday. You have failed in all the subjects. You are not working hard. You are neglecting your studies. It is very bad. Final examinations are drawing near. Be careful. Do not waste your time. Work hard. Finish your syllabus in time.
I hope you will act upon my advice.

Yours affectionately
Mohan

Word-Meanings Subjects = विषय, Neglecting = उपेक्षा कार रहे, Waste = नष्ट करना, Avoid = दूर रहना, Act upon = अमल करना।

16. Invitation for Summer Vacation
Write a letter to your friend asking him to spend a part of his summer vacation with you.

1407 Green Avenue
Amritsar
February 18, 20 ……
My dear Gopal

Your school has closed for the summer vacation. You are free now. I invite you to come to Amritsar. Amritsar is a holy city. Here are many places worth-seeing. You will see Sri Harmandar Sahib, the Durgayana Mandir, the Jallianwala Bagh and other important places.

We shall also study together. I hope you will reach here soon.

Yours sincerely
Ramesh

Word-Meanings-Invite = बुलाना, Holy = पठित्र, Worth-seeing = देखने योग्य।

17. To a Friend on his Failure
Write a letter to your friend who has failed in the examination, asking him not to lose heart but try again.

15 New Colony
Kotkapura
March 18, 20 ……
Dear Raman

Your result is out today. It is very sad that you have failed. It is your own fault. You never worked hard. You moved in a bad company. The result is before you.

Please act upon my advice. Don’t lose heart. Give up bad company and work hard. You will pass next time.

Yours sincerely
Kamal
Address:
Mr. Raman
370 Nai Basti
Ambala

Word-Meaning-Moved = घूमते रहे, Act upon = अमल करना, Lose heart = धैर्य छोड़ना।

18. Condolence Letter.
Suppose you are Satish. You live at 6 Soni Street, Khanna. Your friend has lost his mother. Write a letter of condolence to him.

6 Soni Street
Khanna
Feb. 20, 20 ……
My dear X

I got your letter yesterday. I was shocked to read it. The sudden death of your mother is. a great loss. I share your sorrow.

I met your mother last month. She looked healthy. Her death is untimely. It is the will of God. We must bow before His will. Please have courage.

Yours sincerely
Satish
Address:
Mr. X
15-New Chowk
ABC

Word Meanings-Shocked = आघात पहुंचा, Sudden = आकस्मिक, Great loss = बहुत बड़ी क्षति, Share = बांटना, Sorrow – दुःख, Untimely death = अकाल मृत्यु, Will = इच्छा।

19. To Avoid Bad Company
Write a letter to your younger brother, advising him to avoid bad company. Examination Hall

………City
March 15, 20…….
My dear Mukesh

I received a letter from your headmaster. I gather that you move in a bad company. Ramesh and Dinesh are your friends these days. Both of them smoke. They go to pictures everyday. You, too, have started smoking in their company. You have become a film-fan. Your headmaster is worried about you.

Dear brother, we have high hopes on you. You are the light of our home. The examinations are drawing near. Please give up your bad company.
I hope you will act upon my advice.

With love
Yours affectionately
Kamleshwar
Address:
Mr. Mukesh Verma
Boys’ Hostel
A.B.C. Sen. Sec. School
…….. City

Word-Meaning-Film-fan = फिल्म देखने के शौकीन, Draw near = निकट आना, Act upon = अमल करना।

20. To Father about Your Success in the Examination
You have passed the Middle Standard Examination. Write a letter to your father telling him about your good result.

36 Raj Nagar
Khanna
July 3, 20 ……
My dear Father

Our result was out yesterday. You will be glad to know that I have stood second in the state. I have secured 85% marks. My teacher and headmaster came to our house in the morning. They blessed and patted me. They congratulated the mother. All missed you badly on the occasion. When are you coming home?

Your loving son
Amit

Word-Meanings-Secure = प्राप्त करना, Missed = याद आई, Occasion = अवसर।

21. Letter about Hostel Life
You are Anil, a student of class VIII. You are residing in hostel. Write a letter to your mother about your hostel life.

Boys’ Hostel
A.B.C. Sen. Sec. School
Chandigarh
My dear Mother

Our new session has started. I have got a good room in the school hostel. My hostel life is well disciplined. There are fixed hours for study, meals and games. We get up in time and go to bed in time. Our hostel warden is kind as well as strict to us. He does not allow us to go out of the hostel after the main gate is closed.
Dear Mother, I miss you very badly. I wish I got wings to fly home.

With regards
Yours lovingly
Anil

Word-Meanings- Session = सत्र, Disciplined = अनुशासित, Fixed = निरिचत, Strict = सख्त

22. To Father About Your Papers
You are appearing in the Middle Standard Examination. Write a letter to your father telling him about your progress in the examination.

8 The Mall
Jalandhar Cantt
March 18, 20…….
My dear Father

I am taking my Middle Standard Examination these days. The English paper was very easy. I hope to get 80 marks in it. The paper in Mathematics was a little tough. But I did all the sums.

Punjabi and Social Studies are easy subjects. I hope to do well in them. I am sure that I shall get good marks. I may win a scholarship.
With regards to dear mother

Your loving son
Arun Kumar

Word-Meanings- Taking = दे रहा हृं, Over = समाप्त, Tough = कठिन, Do well = अच्छ करना

23. To Father for Change of School
Write a letter to your father asking him to allow you to change school.

512 Naya Nagar
…….. City
March 10, 20……..
My dear Father

I feel sad to tell you about my school. Once it was a great school. It was considered one of the best schools in the town. Now it is not. We have no Principal these days. We have no English teacher even. So I do not feel good at studies. I want to change my school. Jain High School is a good school. Kindly allow me to join it.

Your loving son
Raj Kumar
Address:
Mr. Mohan Lal Sharma
1 New Road
Patiala

Word Meanings-Considered = समझा जाता था, Feel good = मन लगना, Allow = अनुमति देना।

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Sentence and its Types Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Sentence and its Types

A sentence is a well arranged group of words which makes complete sense.
शब्दों का वह सुव्यवस्थित समूह जिससे पूरा अर्थ निकले कहलाता है, जैसे-

1. Lakhi Mal was very rich.
2. Where does the koel sit ?
3. Please bring him milk and fruit now.
4. Hurrah ! We have won the match.

Kinds of Sentence

वाक्य पांच प्रकार के होते हैं-

1. Assertive Sentences (साधारण वाक्य)
2. Interrogative Şentences (प्रश्नवाचक वाक्य)
3. Imperative Sentences (आजसयक वाक्य )
4. Exclamatory Sentences (विस्मया दिसूचक वाक्य)
5. Optative Sentences (इच्छासूचक वाक्य)

Parts of a Sentence

वाक्य के दो भाग होते है-

1. Subject (उद्देश्थ)
2. Predicate (विधेय)

1. वाक्य में जिस व्यक्ति पशु स्थान अथवा वस्तु के बारे में कुछ कहा जाता है उसे Subject कहते हैं।
2. Predicate. Subject के बारे में जो कुछ कहा जाता है उसे Predicate कहते हैं।

Sentence	Subject	Predicate
1. Amar lives in Allahabad.	Amar	lives in Allahabad.
2. How does a fish swim ?	a fish	How does swim.
3. May you live long !	you	May live long.
4. Mr. Verma is my teacher.	Mr. Verma	is my teacher.

विशेष- Imperative Sentences का Subject प्राय लुप्त रहता है जैसे-

1. Thank you. (I thank you.)
2. Come in. (You come in.)

Simple and Complex Sentences

What is a Simple Sentence?
(साधारण वाक्य की परिभाषा)

जिस वाक्य में केवल एक Finite Verb हो, वह Simple Sentence कहलाता है। साधारण वाक्य में एक Subject (उद्देश्य) और एक Predicate (विधेय). होता है; जैसे,-

1. Mohan loves his country.
2. Ram plays hockey.
3. Dogs bark.

Note. 1. जिस क्रिया का अपना subject (उद्देश्य) हो, वह Finite Verb (परिमित क्रिया) कहलाती है।
2. ऊपर के प्रत्येक वाक्य में एक-एक Finite Verb (loves, plays, bark) है।
3. प्रत्येक का अपना Subject (Mohan, Ram, Dogs) है और अपना Predicate (loves his country, plays hockey, bark) है।

What is a Complex Sentence?
(मिश्रित वाक्य की परिभाषा)

Complex Sentencé को जानने से पहले Clauses (उप-वाक्य) को समझ लेना आवश्यक है, क्योंकि Clauses के मेल से ही Complex Sentence बनता है।

Clause. शब्दों के ऐसे समूह को Clause (उप-वाक्य) कहते हैं जिसका अपना Subject तथा अपना Predicate हो और साथ में किसी अन्य वाक्य का उप-भाग भी हो।

Kinds of Clauses. Clauses के प्रकार को समझने के लिए हमें नीचे लिखे वाक्य को ध्यान से पढ़ना चाहिए-
I like him because he is brave.
नोट-1. इस वाक्य के दो भाग हैं।
2. प्रत्येक भाग में Subject और Predicate दोनों हैं। इसलिए दोनों ही clause हैं।

पहली Clause (I like him). पहली clause का पूरा अर्थ निकलता है। इसका अर्थ समझने के लिए दूसरी clause की आवश्यकता नहीं है। ऐसी clause को Principal Clause या Main Clause (प्रधान उप-वाक्य) कहते हैं।

दूसरी Clause (because he is brave). दूसरी clause का पूरा अर्थ नहीं निकलता। पूरा अर्थ देने के लिए यह Principal Clause पर depend (आश्रित) है। इसलिए इसे Dependent या Subordinate Clause (आश्रित उप-वाक्य) कहते हैं।

Complex Sentence की परिभाषा-जिस वाक्य में एक Principal Clause (प्रधान उप-वाक्य) तथा एक या एक से अधिक Subordinate Clauses (आश्रित उप-वाक्य) हों, उसे Complex Sentence कहते हैं।

Exercise (With Hints)

Pick out the complex sentences:
1. Rani is a nice girl.
2. I know that Mohan is a good boy.
3. He is the boy whom I saw in the market.
4. The man (who was) on the platform was my brother.
5. I met the girl with blue eyes.
6. I met the old man who had grey hair.
7. I went wherever he went.
8. I want to know where you found my watch.
9. He is the teacher of my choice.
10. Those whom the gods love, die young.
Hints:
2. I know that Mohan is a good boy.
3. He is the boy whom I saw in the market.
4. The man (who was) on the platform was my brother.
6. I met the old man who had grey hair.
7. I went wherever he went.
8. I want to know where you found my watch.
10. Those whom the gods love, die young.

Change of Sentences

A–Affirmative to Negative
नियम- (1) साधारणतया निर्षधात्मक (Negative) वाक्य बनाते समय not का प्रयोग होता है: जैसे-

Affirmative	Negative
1. I am a girl.	1. I am not a girl.
2. We are farmers.	2. We are not farmers.
3. You are a student.	3. You are not a student.
4. He was tall.	4. He was not tall.
5. They were proud.	5. They were not proud.
6. Radha was a fat girl.	6. Radha was not a fat girl.
7. I shall write a letter.	7. I shall not write a letter.
8. You will help me.	8. You will not help me.
9. I am reading a book.	9. I am not reading a book.
10. You are weeping.	10. You are not weeping.
11. I was buying a pen.	11. I was not buying a pen.
12. We were playing cards.	12. We were not playing cards.
13. He was swimming.	13. He was not swimming.
14. She had gone there before.	14. She hadn’t gone there before. Or She had not gone there before.
15. We should finish this job.	15. We should not finish this job.

(ii) Present Indefinite (पहली फार्म) में do not/does not + पहली फार्म तथा Past Indefinite (दूसरी फार्म) में did not + पहली फार्म का प्रयोग होता है: जैसे-

Affirmative	Negative
1. I like mangoes	1. I do not like mangoes.
2. We play hockey.	2. We do not play hockey.
3. You take exercise.	3. You do not take exercise.
4. She speaks the truth.	4. She does not speak the truth.
5. I do it.	5. I do not do it.
6. I took tea.	6. I did not take tea.
7. He read the book.	7. He did not read the book.
8. He sold milk.	8. He did not sell milk.
9. She lived in Delhi.	9. She did not live in Delhi.

(iii) Has, have, had यदि अधिकार (possession) अर्थात् किसी चीज़ का होना व्यक्त करते हों तो प्रायः negative sentences में ‘no’ का प्रयोग होता हैै परन्तु यदि संख्या दी हो तो not का प्रयोग होता है।

1. He has a coat.
He has no coat.

2. He had a book
He had no book.

3. Ram has three books.
Ram has not three books.

विशेष नोट-1. Negative बनाने के लिए isnt, wasn’t, haven’t, shan’t आदि का भी प्रयोग किया जा सकता है। इन छोटे रूपों का अध्ययन इस प्रकार करें

2. Can का Negative cannot में किया जाता है। cannot एक शब्द के रूप में लिखा जाता है।

3. Not का प्रयोग पहले शब्द के तुरन्त बाद करना चाहिए, जैसे, shall have been = shall not have been तथा shall be = shall not be.

4. Everybody का Negative nobody तथा either का neither होता है। जैसे,
Affirmative. Everybody is coming.
Negative. Nobody is coming.
Affirmative. Either he or his friend is coming.
Negative. Neither he nor his friend is coming.

मिश्चित वाक्य (Mixed Sentences)

Affirmative	Negative
1. He is a player.	He is not a player.
2. They were happy.	They were not happy.
3. He has many books.	He does not have many books.
4. We have four toys.	We do not have four toys.
5. The boys had a garland.	The boys had no garland.
6. I read a book.	I do not read a book.
7. He writes a letter.	He does not write a letter.
8. He does his work.	He does not do his work.
9. He went there.	He did not go there.
10. Mohan will tell a lie.	Mohan will not tell a lie.

B-Assertive to Interrogative

Assertive	Interrogative
1. I do my duty.	Do I do my duty ?
2. He can help you.	Can he help you?
3. They laughed at us.	Did they laugh at us ?
4. She is a girl.	Is she a girl ?
5. You are late.	Are you late ?
6. He was happy.	Was he happy?
7. We shall come soon.	Shall we come soon?
8. We built a house there.	Did we build a house there?
9. It may rain today.	May it rain today?
10. We must go there	Must we go there?
11. She does her work.	Does she do her work?

Exercise for Practice

Change the following sentences into Interrogative:
1. They can swim.
2. She has a gold ring.
3. These birds are pretty.
4. We have seen the Taj.
5. She helped me in my work.
6. The carpenter makes furniture.
7. This pen costs ten rupees.
8. The child fell from a tree.
9. He was brushing his teeth.
10. Our soldiers fought bravely.

C-Exclamatory to Assertive

Exclamatory	Assertive
1. What a lovely flower!	It is a very lovely flower.
2. Long live our king!	We wish that our king may live long.
3. Alas ! I shall never be able to meet her again.	It is very sad that I shall never be able to meet her again.
4. Hurrah ! I have won a scholarship.	I am very happy that I have won a scholarship.
5. What a pity you have lost everything!	It is very sad that you have lost everything.
6. Fie ! A soldier and afraid of death!	It is shameful for a soldier to be afraid of death.
7. How cold it is today!	It is very cold today. I am very glad to see you.
8. How glad I am to see you!	It is a very shameful act on your part.
9. What a shameful act on your part !	The traitors must be punished with death.
10. Death to the traitors!	Assertive

Exercise for Practice

I. Change the following Exclamatory sentences into Assertive sentences:
1. Bravo ! You have killed the enemy.
2. Alas ! I am undone.
3. How sweet the child is!
4. Alas ! How stupid I had been !
5. What a terrible noise !
6. How difficult the paper is !
7. May God bless you with a son!
8. What a beautiful wrist-watch!
9. Splendid ! You have won the day!
10. Alas! You have failed.

II. Change the following Affirmative (Declarative) sentences into Interrogative sentences (Questions):

1. He is clever.
2. He was simple.
3. Ram was feeling tired.
4. Sita was angry.
5. They were good friends.
6. I have two books.
7. She has three pencils.
8. We had a good time there.
9. I have to do it.
10. Sohan had finished his work.
11. I shall go there tomorrow.
12. He will play a match.
13. I can do it.
14. He may help you.
15. You could write.
16. The sun does not shine at night.
17. He beats his donkey with a stick.
18. He sees another dog.
19. The horse runs very fast.
20. They play a match.
21. He waited here for an hour.
22. The train started at ten.
23. He never lost heart.
24. I did not find him there.
25. The rose smelt sweet.
Answer:
1. Is he clever ?
2. Was he simple ?
3. Was Ram feeling tired ?
4. Was Sita angry?
5. Were they good friends ?
6. Have I two books ?
7. Has she three pencils ?
8. Had we a good time there ?
9. Have I to do it?
10. Had Sohan finished his work ?
11. Shall I go there tomorrow?
12. Will he play a match ?
13. Can I do it?
14. May he help you ?
15. Could you write ?
16. Does the sun not shine at night ?
17. Does he beat his donkey with a stick ?
18. Does he see another dog?
19. Does the horse run very fast ?
20. Do they play a match ?
21. Did he wait here for an hour ?
22. Did the train start at ten ?
23. Did he ever lose heart?
24. Did I not find him there ?
25. Did the rose smell sweet?

III. Convert the following Interrogative sentences into Assertive (Declarative) sentences:

1. Are you on leave today?
2. Was the train late ?
3. Were the boys not lazy ?
4. Am I strong?
5. Is your sister ill?
6. Has it been raining since morning ?
7. Have you fulfilled your promise ?
8. Had he a pen?
9. Have you a horse ?
10. Has she a lovely voice ?
11. Should I stick to my promise ?
12. Does your father like you?
13. Did she not sing a sweet song?
14. Did he paint the door blue?
15. Can you help me ?
16. Does she sing well ?
17. Do we love our country?
18. Did you have a nice holiday ?
19. Do birds build nests ?
Answer:
1. You are on leave today.
2. The train was late.
3. The boys were not lazy.
4. I am.. strong.
5. Your sister is ill.
6. It has been raining since morning.
7. You have fulfilled your promise.
8. He had a pen.
9. You have a horse.
10. She has a lovely voice.
11. I should stick to my promise.
12. Your father likes you.
13. She did not sing a sweet song.
14. He painted the door blue.
15. You can help me.
16. She sings well.
17. We love our country.
18. You had a nice holiday.
19. Birds build nests.

IV (a) Change the following Positive sentences into their Negative form:

1. Lotus is a very lovely flower.
2. His neighbour was quite well yesterday.
3. Sham has a garland of flowers in his hand.
4. I have a horse.
5. You had corrected me.
6. The cattle graze in the pasture.
7. I get up early in the morning.
8. We saw a snake in the grass.
9. Sit down.
10. Let him die.
Answer:
1. Lotus is not a very lovely flower.
2. His neighbour was not quite well yesterday.
3. Sham has not a garland of flowers in his hand.
4. I do not have a horse.
5. You had not corrected me.
6. The cattle do not graze in the pasture.
7. I do not get up early in the morning.
8. We did not see a snake in the grass.
9. Do not sit down.
10. Let him not die. (Or) Do not let him die.

(b) Change the following Negative sentences into their Positive form:

1. Sohan is not an idle boy.
2. We had not a book.
3. I do not have an umbrella with me.
4. He may not play well today.
5. She cannot tell a lie.
6. Do not bring me a second book of Hindi.
7. Do not let him go there.
8. Do not touch my chair.
9. I did not take the test.
10. You did not attend the class.
Answer:
1. Sohan is an idle boy.
2. We had a book.
3. I have an umbrella with me.
4. He may play well today.
5. She can tell a lie.
6. Bring me a second book of Hindi.
7. Let him go there.
8. Touch my chair.
9. I took the test.
10. You attended the class.

V. Change the following Exclamatory sentences into Assertive sentences:

1. Bravo! You have done well.
2. Alas! I have failed.
3. How beautiful the scenery is!
4. Alas! How foolish I had been!
5. What a disaster the earthquake is!
6. How stiff the paper is!
7. May God reward this act of yours!
8. What a terrible storm it is!
9. Wonderful! I have never seen the like of it earlier.
10. May God pardon this sinner!
Answer:
1. You have done very well.
2. It is very sad that I have failed
3. The scenery is very beautiful.
4. It is very sad that I had been very foolish.
5. The earthquake is a terrible disaster.
6. The paper is very stiff.
7. It is prayed that God may reward this act of yours.
8. It is a very terrible storm.
9. It is surprising that I have never seen the like of it earlier. (or) I wonder if I have ever seen the like of it.
10. It is prayed that God may pardon this sinner.

VI. Pick out the Subject and the Predicate in the following sentences:

1. The sun shines brightly.
2. You speak very hastily.
3. He plays cricket.
4. The rose smells sweet.
5. Children take after their parents.
6. Send it at once.
7. Tell me a story.
8. The old man was listening to the wireless.
9. Slow and steady wins the race:
10. Time and tide wait for nobody.
11. He did his work efficiently.
12. Barking dogs seldom bite.
Answer:

Subject	Predicate
1. The sun	shines brightly.
2. You	speak very hastily.
3. He	plays cricket.
4. The rose	smells sweet.
5. Children	take after their parents.
6. You	send it at once.
7. You	tell me a story.
8. The old man	was listening to the wireless.
9. Slow and steady	wins the race.
10. Time and tide	wait for nobody.
11. He	did his work efficiently.
12. Barking dogs	seldom bite.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 14 Data Handling MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 14 Data Handling MCQ Questions

Multiple Choice Questions.

Question 1.

(a) 30
(b) 40
(c) 50
(d) 5.
Answer:
(c) 50

Question 2.

(a) 1
(b) 14
(c) 21
(d) 28.
Answer:
(c) 21

Question 3.

(a) 6
(b) 1
(c) 5
(d) 8.
Answer:
(b) 1

Question 4.

(a) 200
(b) 2000
(c) 20
(d) 2.
Answer:
(a) 200

Question 5.
……….. represents data through picture of objects.
(a) Bar Graph
(b) Histogram
(c) Pictograph
(d) None of these.
Answer:
(c) Pictograph

Question 6.
Which tally marks represents 14?

Answer:

Question 7.

(a) 5
(b) 10
(c) 12
(d) 160.
Answer:
(b) 10

Question 8.
……………. is method of representing the data in uniform width size horizontal or vertical box with equal spacing.
(a) Histogram
(b) Bar Graph
(c) Pictograph
(d) Tally Marks.
Answer:
(b) Bar Graph

Question 9.
A …………… is a collection of numbers gathered to give some information:
(a) Frequency
(b) Data
(c) Tally mark
(d) None of these.
Answer:
(b) Data

Question 10.
If on a scale 1 unit = 200 then how much quantity does 5 units will represent?
(a) 100
(b) 1000
(c) 300
(d) 600.
Answer:
(b) 1000

Question 11.
In the adjoining bar graph the growth of population of India from 1951 to 2001 is shown. Observe the graph and find in which two consecutive years the population growth is maximum?

(a) Year 1951 to year 1961
(b) Year 1971 to year 1981.
(c) Year 1981 to year 1991
(d) Year 1991 to year 2001.
Answer:
(d) Year 1991 to year 2001.

Question 12.
The bar graph given alongside shows the amount of wheat purchased by government during the year 1998-2002. Read the bar graph and find in which year was the wheat production maximum?

(a) 1998
(b) 1999
(c) 2001
(d) 2002.
Answer:
(d) 2002

Question 13.

(a) 5
(b) 6
(c) 7
(d) 8.
Answer:
(c) 7

Question 14.

(a) 5
(b) 7
(c) 9
(d) 10.
Answer:
(d) 10

Question 15.
There is a circus in a village. The number of children came to see circus from Monday to Friday are shown by a pictograph. Read the following pictograph carefully and answer the questions below:

How many children came to see circus on Tuesday?
(a) 25
(b) 50
(c) 75
(d) 100.
Answer:
(c) 75

Fill in the blanks:

Question (i)

Answer:
Seven

Question (ii)
A data is a collection of …………… gathered to give some information.
Answer:
Numbers

Question (iii)
A diagram that represents stastieal data in the form of pictures is called …………….. .
Answer:
pictograph

Question (iv)
The initial step of any investigation is the ……………. of data.
Answer:
collection

Question (v)
The data can be arranged in a ……………. form using tally marks.
Answer:
tabular

Write True/False:

Question (i)
Pictograph represents the data in form of pictures. (True/False)
Answer:
True

Question (ii)
There are three types of data. (True/False)
Answer:
False

Question (iii)
The primary data is collected directly from of source. (True/False)
Answer:
True

Question (iv)

Answer:
False

Question (v)

Answer:
True