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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 10 Circles Ex 10.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Question 1.
Prove that the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:

Circles with centres O and P intersect each other at points A and B.
In ∆ OAP and ∆ OBR
OA = OB (Radii of circle with centre O)
PA = PB (Radii of circle with centre P)
OP = OP (Common)
∴ By SSS rule, ∆ OAP = ∆ OBP
∴ ∠OAP = ∠OBP (CPCT)
Thus, OP subtends equal angles at A and B. Hence, the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Answer:

Draw the perpendicular bisector of AB to intersect AB at M and draw the perpendicular bisector of CD to intersect CD at N.
Both these perpendicular bisectors pass through centre O and since AB || CD; M, O and N are collinear points.
Now, MB = \(\frac{1}{2}\)AB = \(\frac{5}{2}\) = 2.5 cm,
CN = \(\frac{1}{2}\)CD = \(\frac{11}{2}\) = 5.5 cm and MN = 6 cm.
Let ON = x cm s
∴ OM = MN – ON = (6 – x) cm
Suppose the radius of the circle is r cm.
∴ OB = OC = r cm
In ∆ OMB, ∠M = 90°
∴OB² = OM² + MB²
∴ r² = (6 – x)² + (2.5)²
∴ r² = 36 – 12x + x² + 6.25 ………….. (1)
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + CN²
∴ r² = (x)² + (5.5)²
∴ r² = x² + 30.25 ………………. (2)
From (1) and (2),
36 – 12x + x² + 6.25 = x² + 30.25
∴ – 12x = 30.25 – 6.25 – 36
∴- 12x = – 12
∴x = 1
Now, r² = x² + 30.25
∴ r² = (1)2 + 30.25
∴ r² = 31.25
∴ r = √31.25 (Approximately 5.6)
Thus, the radius of the circle is √31.25 (approximately 5.6) cm.
Note: If the calculations are carried out in simple fractions, then MB = \(\frac{5}{2}\) cm, CN = \(\frac{11}{2}\) cm and radius is \(\frac{5 \sqrt{5}}{2}\) (approximately 5.6) cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chords is at distance 4 cm from the cehtre, what is the distance of the other chord from the centre?
Answer:

In a circle with centre O, chord AB is parallel to chord CD, AB = 8 cm and CD = 6 cm.
Draw OM ⊥ AB, ON ⊥ CD, radius OB and radius OC.
Then, MB = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 8 = 4 cm,
NC = \(\frac{1}{2}\)CD = \(\frac{1}{2}\) × 6 = 3cm and ON = 4cm.
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + NC² = 4² + 3² = 16 + 9 = 25
∴ OC = 5 cm
∴ OB = 5 cm (OB = OC = Radius)
In ∆ OMB, ∠M = 90°
∴ OB² = OM² + MB²
∴ 5² = OM² + 4²
∴ 25 = OM² + 16
∴ OM² = 9
∴ OM = 3 cm
Thus, the distance of the other chord from the centre is 3 cm.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:

In ∆ ABE, ∠AEC is an exterior angle.
∴ ∠AEC = ∠ABE + ∠BAE
∴ ∠ABE = ∠AEC – ∠BAE
∴ ∠ABC = ∠AEC – ∠DAE ……………. (1)
Now, ∠AEC = \(\frac{1}{2}\) ∠AOC (Theorem 10.8)
and ∠ DAE = \(\frac{1}{2}\) ∠DOE (Theorem 10.8)
Substituting above values in (1),
∠ABC = \(\frac{1}{2}\) ∠AOC – \(\frac{1}{2}\)∠DOE
∴ ∠ABC = \(\frac{1}{2}\) (∠AOC – ∠DOE)
Here, ∠AOC is the angle subtended by chord AC at the centre and ∠DOE is the angle subtended by chord DE at the centre.
Thus, ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Note: There is no need for chords AD and CE to be equal.

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:

ABCD is a rhombus and its diagonals intersect at M.
∴ ∠BMC is a right angle.
A circle is drawn with diameter BC.
There are three possibilities for point M:
(1) M lies in the interior of the circle,
(2) M lies in the exterior of the circle.
(3) M lies on the circle.
According to (1), if M lies in the interior of the circle, then BM produced will intersect the circle at E. Then, ∠BEC is an angle in a semicircle and hence a right angle, i.e.,
∠MEC = 90°.
In ∆ MEC, ∠ BMC is an exterior angle.
∴ ∠ BMC > ∠ MEC, i.e., ∠ BMC > 90°. In this situation, ∠ BMC is an obtuse angle which contradicts that ∠ BMC = 90°.
Similarly, according to (2), if M lies in the exterior of the circle, then ∠BMC is an acute angle which contradicts that ∠BMC 90°. Thus, possibilities (1) and (2) cannot be true.
Hence, only possibility (3) is true, i.e., M lies on the circle.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary at) E. Prove that AE = AD.
Answer:

Here, the circle through A, B and C intersects CD at E.
∴ Quadrilateral ABCE is cyclic.
ABCD is a parallelogram.
∴ ∠ABC = ∠ADC
∴ ∠ABC = ∠ADE
In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180°
∴ ∠ADE + ∠AEC = 180° ……………… (1)
Moreover, ∠AEC and ∠AED form a linear pair.
∴ ∠AED + ∠AEC = 180° ………………. (2)
From (1) and (2),
∠ADE + ∠AEC = ∠AED + ∠AEC
∴ ∠ ADE = ∠ AED
Thus, in ∆ AED, ∠ADE = ∠AED.
∴ AE = AD (Sides opposite to equal angles)
Note: If the circle intersect CD produced, l then also the result can be proved in similar way.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer:

Chords AC and BD of a circle bisect each other at point O.
Hence, the diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD Is a parallelogram.
∴ ∠BAC = ∠ACD (Alternate angles formed by transversal AC of AB || CD)
Moreover, ∠ACD = ∠ABD (Angles in same segment)
∴ ∠BAC = ∠ABD
∴ ∠BAO = ∠ABO
∴ In A OAB, OA = OB.
But, OA = OC and OB = OD
∴ OA = OB = OC = OD
∴ OA + OC = OB + OD
∴ AC = BD
Thus, the diagonals of parallelogram ABCD are equal.
∴ ABCD is a rectangle.
∴ ∠ABC = 90°
Hence, ∠ABC is an angle in a semicircle and AC is a diameter.
Similarly, ∠BAD = 90°.
Hence, ∠BAD is an angle in a semicircle and BD is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\)A, 90° – \(\frac{1}{2}\)B and 90° – \(\frac{1}{2}\)C.
Answer:

The bisectors of ∠A, ∠B and ∠ C of ∆ ABC intersect the circumcircle of ∆ ABC at D, E and F respectively. .
∠FDE = ∠FDA + ∠EDA (Adjacent angles)
= ∠ FCA + ∠ EBA (Angles in same segment)
= \(\frac{1}{2}\)∠C + \(\frac{1}{2}\)∠B (Bisector of angles in ∆ ABC)
= \(\frac{1}{2}\)(∠ B + ∠ C)
= \(\frac{1}{2}\)(180° – ∠A) [∠A + ∠B + ∠C = 180°)
= 90° – \(\frac{1}{2}\) ∠A
Thus, ∠FDE = 90° – \(\frac{1}{2}\) ∠A.
Similarly, ∠ DEF = 90° – \(\frac{1}{2}\) ∠B and
∠ EFD = 90° – \(\frac{1}{2}\) ∠C.

Question 9.
Two congruent circles intersect each Other at points A and B. Through A any line segment PAQ is drawn so that P 9 lie-on. , the two circles. Prove that BP = BQ.
Answer:

Two congruent circles with centres X and Y intersect at A and B.
Hence, AB is their common chord.
In congruent circles, equal chords subtend equal angles at the centres.
∴ ∠AXB = ∠AYB
In the circle with centre X, ∠AXB = 2∠APB and in the circle with centre Y, ∠AYB = 2∠AQB.
∴ 2∠ APB = 2∠ AQB
∴ ∠APB = ∠AQB
∴ ∠QPB = ∠PQB
Thus, in ∆ BPQ, ∠QPB = ∠PQB
∴ QB = PB (Sides opposite to equal angles)
Hence, BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:

In ∆ ABC, the bisector of ∠A intersects the circumcircle of ∆ ABC at D.
∴∠BAD = ∠CAD
Aso, ∠BAD = ∠BCD and ∠CAD = ∠CBD (Angles in same segment)
∴ ∠BCD = ∠CBD
Thus, in ∆ BCD, ∠BCD = ∠CBD
∴BD = CD (Sides opposite to equal angles)
Thus, point D is equidistant from B and C.
Hence, D is a point on the perpendicular bisector of BC.
Thus, the bisector of ∠ A and the perpendicular bisector of side BC intersect at D and D is a point on the circumcircle of ∆ ABC.
Thus, in ∆ ABC, if the angle bisector of ∠A and the perpendicular bisector of side BC intersect, they intersect on the circumcircle of ∆ ABC.
Note: In ∆ ABC, if AB = AC, then the bisector of ∠A and the perpendicular bisector of side BC will coincide , and would not intersect in a single point.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.4

1. Write the following statements as algebraic equations:

Question (i)
The sum of x and 3 gives 10.
Solution:
The sum of x and 3 = x + 3
It gives 10.
∴ The algebraic equation is x + 3 = 10

Question (ii)
5 less than a number ‘a’ is 12.
Solution:
5 less than a number ‘a’ = a – 5
It is 12
∴ The algebraic equation is a – 5 = 12

Question (iii)
2 more than 5 times of p gives 32.
Solution:
2 more than 5 times of p = 5p + 2
It gives 32
∴ The algebraic equation is 5p + 2 = 32

Question (iv)
Half of a number is 10.
Solution:
Let Half of a number x = \(\frac {x}{2}\)
It is 10
∴ The algebraic equations is
\(\frac {x}{2}\) = 10

Question (v)
Twice of a number added to 3 gives 17.
Solution:
Let the number be x
Twice of a number added to 3 = 2x + 3
It gives 17
∴ The algebraic equation is 2x + 3 = 17

2. Write the L.H.S. and R.H.S. for the following equations:

Question (i)
l + 5 = 8
Solution:
L.H.S. = l + 5, R.H.S. = 8

Question (ii)
13 = 2m + 3
Solution:
L.H.S. = 13, R.H.S. = 2m + 3

Question (iii)
\(\frac {t}{4}\) = 6
Solution:
L.H.S. = \(\frac {t}{4}\), R.H.S. = 6

Question (iv)
2h – 5 = 13
Solution:
L.H.S. = 2h – 5, R.H.S. = 13

Question (v)
\(\frac {5x}{7}\) = 15.
Solution:
L.H.S. = \(\frac {5x}{7}\), R.H.S. = 15.

3. Solve the following equations by trial and error method:

Question (i)
x + 2 = 7
Solution:
x + 2= l
We try different values of x to make L.H.S. = R.H.S.

Value of JC	L.H.S. = x + 2	R.H.S. = 7	L.H.S. = R.H.S.
1	1+ 2 = 3	7	No
2	2 + 2 = 4	7	No
3	3 + 2 = 5	7	No
4	4 + 2 = 6	7	No
5	5 + 2 = 7	7	Yes

From the above table we find that L.H.S. = R.H.S. When x = 5

Question (ii)
5p = 20
Solution:
5p = 20
We try different values of p to make L.H.S. = R.H.S.

Value of p	L.H.S. = 5p	R.H.S. = 20	L.H.S. = R.H.S.
1	5 × 1 = 5	20	No
2	5 × 2 = 10	20	No
3	5 × 3 = 15	20	No
4	5 × 4 = 20	20	Yes

From the above table we find that L.H.S. = R.H.S. When p = 4

Question (iii)
\(\frac {a}{5}\) = 2
Solution:
We try different values of a to make L.H.S. = R.H.S.
image
From the above table we find that L.H.S. = R.H.S. When a = 10

Question (iv)
2l – 4 = 8
Solution:
21-4 = 8
We try different values of l to make L.H.S. = R.H.S.

Value of a	L.H.S.	R.H.S. = 8	L.H.S. ff R.H.S.
1	2 × 1 – 4 = 2 – 4 = – 2	8	No
2	2 × 2 – 4 = 4 – 4 = 0	8	No
3	2 × 3 – 4 = 6 – 4 = 2	8	No
4	2 × 4 – 4 = 8 – 4 = 4	8	No
5	2 × 5 – 4 = 10 – 4 = 6	8	No
6	2 × 6 – 4  = 12 – 4 = 8	8	Yes

From the above table we find that L.H.S. = R.H.S. When l = 6

Question (v)
3x + 2 = 11.
Solution:
3x + 2 = 11
We try different values of x to make L.H.S. = R.H.S.

Value of p	L.H.S. = 3x + 2	R.H.S. = 11	L.H.S. = R.H.S.
1	3 × 1 + 2 = 3 + 2 = 5	11	No
2	3 × 2 + 2 = 6 + 2 = 8	11	No
3	3 × 3 + 2 = 9 + 2 = 11	11	Yes

From the above table we find thatL.H.S. = R.H.S. When x = 3

4. Solve the following equations by systematic method.

Question (i)
z – 4 = 10
Solution:
Given Equation is z – 4 = 10 Adding 4 on both sides, we get
2 – 4 + 4 = 10 + 4
⇒ z = 14 is the required solution.

Question (ii)
a + 3 = 15
Solution:
Given equation is a + 3 = 15
Subtracting 3 from both sides, we get
a + 3- 3 = 15 – 3
⇒ a = 12 is the required solution.

Question (iii)
4m = 20
Solution:
Given equation is 4m = 20
Dividing both sides by 4, we get
\(\frac{4 m}{4}=\frac{20}{4}\)
⇒ m = 5 is the required solution.

Question (iv)
3x – 3 = 15
Solution:
Given equation is 3x – 3 = 15
Adding 3 on both sides, we get
3x – 3 + 3 = 15 + 3
⇒ 3x = 18
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{18}{3}\)
⇒ x = 6 is the required solution.

Question (v)
4x + 5 = 13.
Solution:
Given equation is 4x + 5 = 13
Subtracting 5 from both sides, we get
4x + 5 – 5 = 13 – 5
⇒ 4x = 8
Dividing both sides by 4, we get
\(\frac{4 x}{4}=\frac{8}{4}\)
⇒ x = 2 is the required solution.

5. Solve the following equation by transposition:

Question (i)
x – 5 = 6
Solution:
Given equation : x – 5 = 6
∴ x = 6 + 5
(Transposing – 5 to other side, it becomes + 5)
⇒ x = 11 is the required solution.

Question (ii)
y + 2 = 3
Solution:
Given equation : y + 2 = 3
⇒ y = 3 – 2
(Transposing + 2 to other side, it becomes – 2)
∴ y = 1 is the required solution.

Question (iii)
5x = 10
Solution:
Given equation : 5x = 10
⇒ x = \(\frac {10}{5}\)
(Transposing ‘multiplication’, it becomes ‘division’)
∴ x = 2 is the required solution.

Question (iv)
\(\frac {a}{6}\) = 4
Solution:
Given equation : \(\frac {a}{6}\) = 4
⇒ a = 4 × 6
(Transposing ‘division’, it becomes ‘multiplication’)
∴ a = 24 is the required solution.

Question (v)
4y – 2 = 30.
Solution:
Given equation : 4y – 2 = 30
⇒ 4y = 30 + 2
(Transposing – 2, it becomes + 2)
⇒ 4y = 32
⇒ y = \(\frac {32}{4}\)
(Transposing ‘multiplication’, it becomes division)
∴ y = 8 is the required solution.

6. Solve the following equations:

Question (i)
x + 7 = 11
Solution:
Given equation :
x + 7 = 11
⇒ x = 11 – 7
(Transposing 7 to R.H.S.)
⇒ x = 4 is the required solution.

Question (ii)
x – 3 = 15
Solution:
Given equation : x – 3 = 15
⇒ x = 15 + 3
(Transposing – 3 to L.H.S. it becomes + 3)
∴ x = 18 is the required solution

Question (iii)
x – 2 = 13
Solution:
Given equation : x – 2 = 13
⇒ x = 13 + 2
(Transposing – 2 to L.H.S.)
∴ x = 15 is the required solution.

Question (iv)
6x = 18
Solution:
Given equation is 6x = 18
Dividing both sides by 6 we get
\(\frac{6x}{6}=\frac{18}{6}\)
∴ x = 3 is the required solution.

Question (v)
3x = 24
Solution:
Given equation 3x = 24
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{24}{3}\)
∴ x = 8 is the required solution.

Question (vi)
\(\frac {x}{4}\) = 7
Solution:
Given equation :
\(\frac {x}{4}\) = 7
Multiplying both sides by 4, we get
4 × \(\frac {x}{4}\) = 4 × 7
∴ x = 28 is the required solution.

Question (vii)
\(\frac {x}{8}\) = 5
Solution:
Given equation : \(\frac {x}{8}\) = 5
Multiplying both sides by 8, we get
8 × \(\frac {x}{8}\) = 8 × 5
∴ x = 40 is the required solution.

Question (viii)
2x – 5 = 17
Solution:
Given equation : 2x – 5 = 17
⇒ 2x = 17 – 5
(Transposing – 5 to R.H.S.)
⇒ 2x = 22
⇒ x = \(\frac {22}{2}\)
(Dividing both sides by 2)
∴ x = 11 is the required solution.

Question (ix)
4x + 5 = 21
Solution:
Given equation : 4x + 5 = 21
⇒ 4x = 21 + 5
(Transposing 5 to R.H.S.)
⇒ 4x = 16
⇒ x = \(\frac {16}{4}\)
(Dividing both sides by 4)
∴ x = 4 is the required solution.

Question (x)
5x – 2 = 13.
Solution:
Given equation : 5x – 2 = 13
⇒ 5x = 13 + 2
(Transposing – 2 to R.H.S.)
⇒ 5x = 15
⇒ x = \(\frac {15}{5}\)
(Dividing both sides by 5)
∴ x = 3 is the required solution.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 13 Exponents and Powers Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

1. Fill in the blanks :

(i) In the expression 3⁷, base = …………….. and exponent = ……………..
(ii) In the expression \(\left(\frac{2}{5}\right)^{11}\), base = …………….. and exponent = ……………..
Solution:
(i) 3, 7
(ii) \(\frac {2}{5}\), 11

2. Find the value of the following :
(i) 2⁶
(ii) 9³
(iii) 5⁵
(iv) (-6)⁴
(v) \(\left(-\frac{2}{3}\right)^{5}\)
Solution:
(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2
= 64

(ii) 9³ = 9 × 9 × 9
= 729

(iii) 5⁵ = 5 × 5 ×5 × 5 × 5
= 3125

(iv) (-6)⁴ = -6 × -6 × -6 × -6
= 1296

(v) \(\left(-\frac{2}{3}\right)^{5}\) = \(\frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3} \times \frac{-2}{3}\)
= \(-\frac{32}{243}\)

3. Express the following in the exponential form :
(i) 6 × 6 × 6 × 6
(ii) b × b × b × b
(iii) 5 × 5 × 7 × 7 × 7
Solution:
(i) 6 × 6 × 6 × 6 = 6⁴
(ii) b × b × b × b = b⁴
(iii) 5 × 5 × 7 × 7 × 7 = 5² × 7³

4. Simplify the following :

(i) 2 × 10³
Solution:
2 × 10³ = 2 × 10 × 10 × 10
= 2000

(ii) 5² × 3²
Solution:
5² × 3² = 5 × 5 × 3 × 3
= 25 × 9
= 225

(iii) 3² × 10⁴
Solution:
3² × 10⁴ = 3 × 3 × 10000
= 90000

5. Simplify :
(i) (-3) × (-2)³
Solution:
(-3) × (-2)³ = -3 × -2 × -2 × -2
= -3 × -8
= 24

(ii) (-4)³ × 5²
Solution:
(-4)3 × 52= -4 × -4 × -4 × 5 × 5
= 64 × 25
= -1600

(iii) (-1)⁹⁹
Solution:
(-1)⁹⁹ = -1
[(-1)^{odd number} = -1]

(iv) (-3)² × (-5)²
Solution:
(-3)² × (-5)² = -3 × -3 × – 5 × -5
= 9 × 25
= 225

(v) (-1)¹³²
Solution:
(-1)¹³² = 1
[(-1)^{even number} = +1]

6. Identify the greater number in each of the following :

(i) 4³ or 3⁴
Solution:
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
81 > 64
∴ 3⁴ > 4³.

(ii) 5³ or 3²
Solution:
5³ = 5 × 5 × 5 = 125
3² = 3 × 3 = 9
125 > 9
∴ 5³ > 3².

(iii) 2³ or 8²
Solution:
2³ = 2 × 2 × 2 = 8
8² = 8 × 8 = 64
64 > 8
∴ 8² > 2³.

(iv) 4⁵ or 5⁴
Solution:
4⁵ = 4 × 4 × 4 × 4 × 4 = 1024
5⁴ = 5 × 5 × 5 × 5 = 625
1024 > 625
∴ 4⁵ > 5⁴.

(v) 2¹⁰ or 10²
Solution:
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024
10² = 10 × 10 = 100
1024 > 100
∴ 2¹⁰ > 10²

7. Write the following numbers as power of 2 :

(i) 8
Solution:
8 = 2 × 2 × 2
\(\begin{array}{c|c}
2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
= 2³

(ii) 128
Solution:
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁷
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)

(iii) 1024
Solution:
1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
=2¹⁰
\(\begin{array}{l|l}
2 & 1024 \\
\hline 2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)

8. Write the following numbers as power of 3 :

(i) 27
Solution:
27 = 3 × 3 × 3
= 3³
\(\begin{array}{l|l}
3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(ii) 2187
Solution:
2187 = 3 × 3 × 3 × 3 × 3 × 3 × 3
= 3⁷
\(\begin{array}{l|l}
3 & 2187 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

9. Find the value of x in each of the following:

(i) 7x = 343
Solution:
343 =7 × 7 × 7 = 7³
7^x = 343
7^x = 73
∴ x = 3

(ii) 9x = 729
Solution:
729 =9 × 9 × 9
= 9³
9^x = 729
9^x = 9³
∴ x = 3.

(iii) (-8)^x = -512
Solution:
512 = 8 × 8 × 8
= 8³
(-8)^x = -512
(-8)^x = (-8)³
∴ x = 3.

10. To what power (-2) should be raised to get 16 ?
Solution:
Let power raised be x
16 = 2 × 2 × 2 × 2
= 2⁴
(-2)^x = 2⁴
(-2)^x = (-2)⁴
[(-1)^{even number} = +1]
∴ x = 4.

11. Write the prime factorization of the following numbers in the exponential form :

(i) 72
Solution:
72 = 2 × 2 × 2 × 3 × 3
= 2³ × 3²
\(\begin{array}{l|l}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(ii) 360
Solution:
360 = 2 × 2 × 2 × 3 × 3 × 5
= 2³ × 3² × 5¹
\(\begin{array}{c|c}
2 & 360 \\
\hline 2 & 180 \\
\hline 2 & 90 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

(iii) 405
Solution:
405 = 3 × 3 × 3 × 3 × 5
= 3⁴ × 5¹
\(\begin{array}{l|l}
3 & 405 \\
\hline 3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

(iv) 648
Solution:
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2³ × 3⁴
\(\begin{array}{c|c}
2 & 648 \\
\hline 2 & 324 \\
\hline 2 & 162 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)

(v) 3600
Solution:
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²
\(\begin{array}{c|c}
2 & 3600 \\
\hline 2 & 1800 \\
\hline 2 & 900 \\
\hline 2 & 450 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.3

1. Pick the algebraic expressions and the arithmetic expressions from the following:

Question (i)
(i) 2l – 3
(ii) 5 × 3 + 8
(iii) 6 – 3x
(iv) 51
(v) 2 × (21 – 18) + 9
(vi) \(\frac {6a}{5}\) + 2
(vii) 7 × 20 + 5 + 3
(viii) 8.
Solution:
Algebraic Expressions :
2l – 3, 6 – 3x, 51, \(\frac {6a}{5}\) + 2
Arithmetic Expressions :
5 × 3 + 8, 2 × (21 – 18) + 9, 7 × 20 + 5 + 3, 8.

2. Write the terms for the following expressions:

Question (i)
2y + 5z
Solution:
Terms of 2y + 5z = 2y, 5z

Question (ii)
6x – 3y + 8
Solution:
Terms of 6x – 3y + 8 = 6x, -3y, 8

Question (iii)
7a
Solution:
Terms of 7a = 7a

Question (iv)
3l – 5m + 2n
Solution:
Terms of 31 – 5m + 2n = 31, -5m, 2n

Question (v)
\(\frac {2l}{3}\) + x.
Solution:
Terms of = \(\frac {2l}{3}\) + x = \(\frac {2l}{3}\), x

3. Tell how the following expressions are formed.

Question (i)
a + 11
Solution:
a + 11 = a is increased by 11

Question (ii)
12 – x
Solution:
12 – x = x is subtracted from 12

Question (iii)
3z + 8
Solution:
3z + 8 = Three time of z is increased by 8

Question (iv)
6 – 5l
Solution:
6 – 5l = 5 times of l is subtracted from 6

Question (v)
\(\frac {5a}{4}\)
Solution:
\(\frac {5a}{4}\) = 5 times a is divided by 4.

4. Give expressions for the following:

Question (i)
10 is added to p
Solution:
10 is added to p = p + 10

Question (ii)
5 is subtracted from y
Solution:
S is subtracted from y = y – 5

Question (iii)
d is divided by 3
Solution:
d is divided by 3 = \(\frac {d}{3}\)

Question (iv)
l is multiplied by – 6
Solution:
l is multiplied by – 6 = – 6l

Question (v)
m is subtracted from l
Solution:
m is subtracted from 1 = 1 – m

Question (vi)
11 is added to 3x
Solution:
11 is added to 6x = 6x + 11

Question (vii)
y is divided by -2 and then 2 is added to the result
Solution:
y is multiplied by – 2 and then 2 is added to the result = – 2y + 2

Question (viii)
c is divided by 5 and then 7 is multiplied to the result
Solution:
c is divided by 5 and thus 7 is multiplied to the result = \(\frac {7c}{5}\)

Question (ix)
x is multiplied by 3 then subtracted this result from y
Solution:
x is multiplied by 3 then subtracted this result from y = y – 3x

Question (x)
a is added to b then c is multiplied with this result.
Solution:
a is added to b then c is multiplied by this result = (a + b) c

5. Write the number which is 15 less than y.
Solution:
The number which is 15 less than y = y – 15

6. Write the number which is 3 more than a.
Solution:
The number which is 3 more than a = a + 3

7. Find the number which is 1 more than twice of x.
Solution:
The number which is 1 more than twice of x = 2x + 1

8. Find the number which is 7 less than 5 times of y.
Solution:
The number which is 7 less than 5 times of y = 5y – 7

9. Somi’s present age is ‘a’ years. Express the following in algebraic form:

Question (i)
Her age after 15 years.
Solution:
Somi’s present age = ‘a’ years
Her age after 15 years = (a + 15) years

Question (ii)
Her age 2 years ago.
Solution:
Her age 2 years ago = (a – 2) years

Question (iii)
If Somi’s father’s age is 5 more than twice of her present age, express her father’ age.
Solution:
Somi’s father’s age is 5 more than twice of her present age
∴ Her father’s age = (2a + 5) years.

Question (iv)
If Somi’s sister is 4 years younger to her. Express her sister’s age.
Solution:
Somi’s sister is 4 years younger to her
∴ Her sister’s age = (a – 4) years

Question (v)
If Somi’s mother is 3 less than 3 times her present age. Express her mother’s age.
Solution:
Somi’s mother is 3 less than 3 times her present age
∴ Her mother’s age = (3a – 3) years.

10. The length of a floor is 10 more than two times of breadth what is the length if breadth is l meters?
Solution:
The breadth of floor = l metres
The length of the floor is 10 more than two times of its breadth
∴ Length of floor = (2l + 10) metres.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.2

1. Each side of equilateral triangle is denoted by ‘a’ then express the perimeter of the triangle using ‘a’.

Solution:
Each triangle of equilateral triangle = a
∴ Perimeter of equilateral triangle
= a + a + a = 3a

2. An isosceles triangle is shown. Express its perimeter in terms of ‘l’ and ‘b’.

Solution:
Perimeter of isosceles triangle = l + l + b
= 21 + b

3. Each side of regular hexagon is denoted by ‘S’ then express the perimeter of the regular hexagon using ‘S’.

Solution:
Each side of regular hexagon = S
Perimeter of regular hexagon
=S + S + S + S + S + S
= 6S

4. The cube has 6 faces and all of them are identify squares. If l is the length of an edge of a cube, find the total length of all edges of the cube in terms of ‘l’?

Solution:
Length of each edge of a cube = l
There are 12 edges of a cube
Total length of all edges of the cube
= 12 × l = 12l

5. Write commutative property of addition using variables x and y.
image
Solution:
According to commutative property of addition.
If the order of numbers, in addition, is changed it does not change their sum.
∴ x + y = y + x

6. Write associative property of multiplication using variables l, m and n.
Solution:
According to associative property of multiplication.
If three numbers can be multiplied in any order, it does not change their product.
∴ l × (m × n) = (l × m) × n

7. Write distributive property of multiplication over addition in terms of variables p, q and r respectively.
Solution:
According to Distributive property of multiplication over addition
p × (q + r) = p × q + p × r

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.1

1. Find the rule which gives the number of matchsticks required to make the following ‘it’ matchstick patterns. Use a variables to write the rule:

Question (i)
A pattern of letter T as

Solution:
Number of matchsticks required in a pattern of letter T = 2
Number of matchsticks required in ‘n’ patterns = 2n

Question (ii)
A pattern of letter E as

Solution:
Number of matchsticks required in a pattern of letter E = 4

Number of matchsticks required in V patterns of letter E = 4n

Question (iii)
A pattern of letter F as

Solution
Number of matchsticks required in a pattern of letter F = 3

Number of matchsticks required in ‘n’ patterns of letter F = 3 n

Question (iv)
A pattern of letter C as

Solution:
Number of matchsticks required in a pattern of letter C = 3

Number of matchsticks required in ‘n’ patterns of letter C = 3n

Question (v)
A pattern of letter S as

Solution:
Number of matchsticks required in a pattern of letter S = 5

Number of matchsticks required in V patterns of letter S = 5 n

2. Students are sitting in rows. There are 12 students in row. What is the rule which gives the number of students in ‘n’ rows? (Represent by table)
Solution:
Let us make a table for the number of students in ‘n’ rows.

Number of Rows	1	2	3	4	…..	10	……	n
Number of Students	12	24	36	48	……	120	……	12 n

It is observed from the table that
Total number of students in ‘n’ number of rows
= (Number of Students) × (Number of rows)
= 12 × n = 12n

3. The teacher distributes 3 pencils to a student What is the rule which gives the number of pencils, if there are ‘a’ number of students?
Solution:
We know
Total number of pencils
= Number of pencils × Number of students
= 3 × a = 3a

4. There are 8 pens in a pen stand. What is the rule that gives the total cost of the pens if the cost of each pen is represented by a variable ‘c’?
Solution:
We know
Total cost of the pens in ₹
= Number of pens × cost of 1 pen
= 8 × c = 8c

5. Gurleen is drawing pictures by joining dots. To make one picture,’she has to join 5 dots. Find the rule that gives the number of dots, if the number of pictures is represented by the symbol ‘p’.
Solution:
We know
Total number of dots = Number of dots × Number of pictures
= 5p

6. The cost of a dozen bananas is ₹ 50. Find the rule of total cost of bananas if there are ‘d’ dozens bananas.
Solution:
We know
Total cost of bananas in ₹
= Cost of one dozen × Number of bananas
= 50 × d
= 50d

7. Look at the following matchsticks patterns of squares given below. The squares are not separate as there are two adjoined adjacent squares have a common match stick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares.

(Hint: If you remove the vertical stick at the end you will get a patterns of C)
Solution:

Fig. No.	No. of Squares	Number of matchsticks	Pattern
(i)	1	4	3 x 1+ 1
(ii)	2	7	3 × 2 + 1
(iii)	3	10	3 × 3 + 1

Thus, we get the rule the number of matchsticks = 3x + 1 or 1 + 3x where x is the number of squares.

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

PSEB 6th Class Maths Chapter 6 Decimals MCQ Questions

Multiple Choice Questions

Question 1.
3 + \(\frac {2}{10}\) = ………….
(a) 302
(b) 3.2
(c) 3.02
(d) 30.2.
Answer:
(b) 3.2

Question 2.
200 + 4 + \(\frac {5}{10}\) = …………
(a) 24.5
(b) 204.05
(c) 204.5
(d) 24.05.
Answer:
(c) 204.5

Question 3.
\(\frac {7}{100}\) = …………..
(a) .07
(b) 700
(c) .007
(d) 7.
Answer:
(a) .07

Question 4.
50 + \(\frac {3}{1000}\) = ………….
(a) 50.3
(b) 503000
(c) 50.0003
(d) 50.003.
Answer:
(d) 50.003.

Question 5.
Seventy and four thousandths = …………….
(a) 74000
(b) 70.004
(c) .00074
(d) .074.
Answer:
(b) 70.004

Question 6.
2.03 in expanded form = ……….
(a) 2 + \(\frac {3}{10}\)
(b) 20 + \(\frac {3}{10}\)
(c) 2 + \(\frac {3}{100}\)
(d) 20 + \(\frac {3}{100}\)
Answer:
(c) 2 + \(\frac {3}{100}\)

Question 7.
2.5 = ……….. .
(a) \(\frac {5}{2}\)
(b) \(\frac {25}{2}\)
(c) \(\frac {5}{10}\)
(d) \(\frac {1}{4}\)
Answer:
(a) \(\frac {5}{2}\)

Question 8.
\(\frac {13}{2}\) = …………….
(a) 6
(b) 6.1
(c) 1.3
(d) 6.5.
Answer:
(d) 6.5.

Question 9.
Which of the following decimals is largest?
(a) 0.5
(b) 0.05
(c) 0.51
(d) 0.005.
Answer:
(c) 0.51

Question 10.
Which of the following decimals is smallest?
(a) 2.13
(b) .213
(c) 21.3
(d) 213.
Answer:
(b) .213

Question 11.
75 g = ……. kg.
(a) .075 kg
(b) .75 kg
(c) 7.5 kg
(d) 75 kg.
Answer:
(a) .075 kg

Question 12.
27 mm = ………….. cm.
(a) .27 cm
(b) 27 cm
(c) 2.7 cm
(d) .027 cm.
Answer:
(c) 2.7 cm

Question 13.
2.5 + 4.23 = ……………
(a) 4.48
(b) 6.73
(c) 4.73
(d) 6.48.
Answer:
(b) 6.73

Question 14.
15 + 3.84 = ………… .
(a) 3.99
(b) 18.99
(c) 3.84
(d) 18.84
Answer:
(d) 18.84

Question 15.
13.5 – 4.23 = …………….
(a) 2.87
(b) 7.29
(c) 9.27
(d) 9.37.
Answer:
(c) 9.27

Question 16.
20 – 12.56 = …………..
(a) 7.44
(b) 8.44
(c) 9.44
(d) 6.44.
Answer:
(a) 7.44

Question 17.
14.8 + 2.62 – 8.4 = …………….. .
(a) 8.02
(b) 9.12
(c) 9.02
(d) 6.44.
Answer:
(c) 9.02

Question 18.
517 ml = …………… l.
(a) 5.07 l
(b) 5.7 l
(c) 5.70 l
(d) 5.007 l.
Answer:
(d) 5.007 l

Question 19.
12 kg 85 g = ……………. kg.
(a) 12.085 kg
(b) 12.85 kg
(c) 128.5 kg
(d) 12.0085 kg.
Answer:
(a) 12.085 kg

Question 20.
235 paise = …………..
(a) ₹ 235
(b) ₹ 23.5
(c) ₹ 2.35
(d) ₹ .235.
Answer:
(c) ₹ 2.35

Question 21.
Express 88 m as km using decimals:
(a) 0.88 km
(b) 8.8 km
(c) 0.088 km
(d) 0.0088 km.
Answer:
(c) 0.088 km

Question 22.
In the following lists which numbers are in the descending order?
(a) 0.355, 0.4, 0.43, 0.355
(b) 0.4, 0.43, 0.444, 0.355
(c) 0.43, 0.355, 0.444, 0.4
(d) 0.444, 0.43, 0.4, 0.355.
Answer:
0.444, 0.43, 0.4, 0.355.

Question 23.
In the following lists which numbers are in the descending order?
(a) 19.4, 0.3, 10.6, 205.9
(b) 205.9, 10.6, 0.3, ,19.4
(c) 205.9, 19.4, 10.6, 0.3
(d) 0.3, 10.6, 19.4, 205.9.
Answer:
205.9, 19.4, 10.6, 0.3

Question (iv)
In the following lists which numbers are in the ascending order?
(a) 0.7, 20.9, 14.6, 600.8
(b) 0.7, 14.6, 20.9, 600.8
(c) 600.8, 14.6, 20.9, 0.7
(d) 14.6,20.9,0.7,600.8.
Answer:
0.7, 14.6, 20.9, 600.8

Question (v)
Express 30 mm as cm using decimals:
(a) 3,0 cm
(b) 0.30 cm
(c) 0.03 cm
(d) 0.003 cm.
Answer:
3,0 cm

Fill in the blanks:

Question (i)
15 cm as m using decimals is …………… m.
Answer:
0.15

Question (ii)
75 paise as ₹ using decimals is ₹ ………….. .
Answer:
₹ 0.75

Question (iii)
9 cm 8 mm as cm using decimals is …………… m.
Answer:
0.98

Question (iv)
27 m = ………….. cm.
Answer:
2.7

Question (v)
15 + 3.84 = ……………… .
Answer:
18.84

Write True/False:

Question (i)
The word decimal comes from Latin word “Decem.” (True/False)
Answer:
True

Question (ii)
\(\frac {1}{10}\) is read as one tenth. (True/False)
Answer:
True

Question (iii)
10 + 3 + \(\frac{2}{10}=\frac{15}{10}\) (True/False)
Answer:
False

Question (iv)
Seven and three-tenths is written as 7.3. (True/False)
Answer:
True

Question (v)
Twenty-four point five is written as 24.5. (True/False)
Answer:
True

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 1.
State which pairs of triangles in Fig. are similar. Write the similarity criterion used by you for answering the queStion and also write the pairs of similar triangles in the symbolic form:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution:
(i) In ∆ABC and ∆PQR,
∠A = ∠P (each 60°)
∠B = ∠Q (each 80°)
∠C = ∠R (each 40°)
∴ ∆ABC ~ PQR [AAA Similarity criterion]

(ii) In ∆ABC and ∆PQR,
\(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{2}{4}=\frac{1}{2}\) …………….(1)

\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\) ……………..(2)

\(\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{2.5}{5}=\frac{1}{2}\) ……………(3)
From (1), (2) and (3),
\(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{1}{2}\)

∴ ΔABC ~ ΔQRP [By SSS similarity criterion]

(iii) In ΔLMP and ΔDEF,
\(\frac{\mathrm{MP}}{\mathrm{DE}}=\frac{2}{4}=\frac{1}{2}\)

\(\frac{\mathrm{PL}}{\mathrm{DF}}=\frac{3}{6}=\frac{1}{2}\) \(\frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5}=\frac{27}{50}\)

\(\)
∴ Two Triangles are not similar.

(iv) In ΔMNL and ΔPQR,
\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}\)

∠M = ∠Q (each 70°)

\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2.5}{5}=\frac{1}{2}\)

∴ ΔMNL ~ ΔPQR [By SAS similarity cirterion]

(v) In ΔABC and ΔDEF,
\(\frac{\mathrm{AB}}{\mathrm{DF}}=\frac{2.5}{5}=\frac{1}{2}\)

\(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{3}{6}=\frac{1}{2}\)

But ∠B ≠ ∠F
∴ ΔABC and ΔDEF are not similar.

(vi) In ΔDEF, ∠D = 70°, ∠E = 80°
∠D + ∠E + ∠F = 180°
70° + 80° + ∠F = 180° [Angle Sum Propertyl
∠F= 180° – 70° – 80°
∠F = 30°
In ΔPQR,
∠Q = 80°, ∠R = 30°
∠P + ∠Q + ∠R = 180°
(Sum of angles of triangle)
∠P + 80° + 30° = 180°
∠P = 180° – 80° – 30°
∠P = 70°
In ΔDEF and ΔPQR,
∠D = ∠P (70° each)
∠E = ∠Q (80° each)
∠F = ∠R (30° each)
∴ ΔDEF ~ ΔPQR (AAA similarity criterion).

Question 2.
In Fig., ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. FInd ∠DOC, ∠DCO and ∠OAB.

Solution:
Given that: ∠BOC = 125°
∠CDO = 70°
DOB is a straight line
∴ ∠DOC + ∠COB = 180°
[Linear pair Axiom]
∠DOC + 125° = 180°
∠DOC = 180°- 125°
∠DOC = 55°
∠DOC = ∠AOB = 55°
[Vertically opposite angle]
But ΔODC ~ ΔOBA
∠D = ∠B = 70°
In ΔDOC, ∠D + ∠O + ∠C = 180°
70° + 55° + ∠C = 180°
∠C= 180° – 70° – 55°
∠C = 55°
∠C = ∠A = 55°
Hence ∠DOC = 55°
∠DCO = 55°
∴ ∠OAB = 55°.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathbf{O A}}{\mathbf{O C}}=\frac{\mathbf{O B}}{\mathbf{O D}}\).
Solution:

Given: In Trapezium ABCD, AB || CD, and diagonal AC and BD intersects each other at O.
To Prove = \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\) (Given)
Proof: AB || DC (Given)
In ΔDOC and ΔBOA,
∠1 = ∠2 (alternate angle)
∠5 = ∠6 (vertical opposite angle)
∠3 = ∠4 (alternate angle)
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]
∴ \(\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}}\)
[If two triangle are similar corresponding sides are Proportional }
⇒ \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{DO}}\)
Hence Proved.

Question 4.
In Fig., \(\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Given that,
\(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and
∠1 = ∠2

To Prove. PQS – ITQR
Proof: In ΔPQR,
∠1 = ∠2 (given)
∴ PR = PQ
[Equal angle have equal side opposite to it]
and = \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) (given)
or \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PQ}}\) [PR = PQ]
⇒ \(\frac{\mathrm{QS}}{\mathrm{QR}}=\frac{\mathrm{PQ}}{\mathrm{QT}}\)
In ΔPQS and ΔTQR,
\(\frac{\mathrm{QS}}{\mathrm{QR}}=\frac{\mathrm{PQ}}{\mathrm{QT}}\)
∠1 = ∠1 (common)
∴ ∆PQS ~ ∆TQR [SAS similarity criterion]
Hence proved.

Question 5.
S and T are points on skies PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
S and T are the points on side PR and QR such that ∠P = ∠RTS.

To Prove. ∆RPQ ~ ∆RTS
Proof: In ∆RPQ and ∆RTS
∠RPQ = ∠RTS (given)
∠R = ∠R [common angle]
∴ RPQ ~ ARTS
[By AA similarity critierion which is the required result.]

Question 6.
In figure ∆ABE ≅ ∆ACD show that ∆ADE ~ ∆ABC.

Solution:
Given. ∆ABC in which ∆ABE ≅ ∆ACD
To Prove. ∆ADE ~ ∆ABC
Proof. ∆ABE ≅ ∆ACD (given)
AB = AC (cpct) and AE = AD (cpct)
\(\frac{A B}{A C}=1\) ……………..(1)
\(\frac{A E}{A D}=1\) …………….(2)
From (1) and (2).
\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
In ∆ADE and ∆ABC,
\(\frac{\mathrm{AD}}{\mathrm{AE}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
∠A = ∠A (common)
∴ ∆ADE ~ ∆ABC [By SAS similarity criterion].

Question 7.
In Fig., altitudes AD and CE of ∆ABC intersect each other at the point P.

Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Solution:
Given. ∆ABC, AD ⊥ BC CE⊥AB,

To Prove. (i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Proof:
(i) In ∆AEP and ∆CDP,
∠E = ∠D (each 90°)
∠APE = ∠CPD (vertically opposite angle)
∴ ∆AEP ~ ∆CDP [By AA similarity criterion].

(ii) In ∆ABD and ∆CBE,
∠D = ∠E (each 90°)
∠B = ∠B (common)
∴ ∆ABD ~ ∆CBE [AA Similarity criterion]

(iii) In ∆AEP and ∆ADB.
∠E = ∠D (each 90°)
∠A = ∠A (common)
∴ ∆AEP ~ ∆ADB [AA similarity criterion].

(vi) In ∆PDC and ∆BEC,
∠C = ∠C
∠D = ∠E
∴ ∆SPDC ~ ∆BEC [AA similarity criterion].

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that AABE – &CFB.
Solution:
Given. Parallelogram ABCD. Side AD is produced to E, BE intersects DC at F.

To Prove. ∆ABE ~ ∆CFB
Proof. In ∆ABE and ∆CFB.
∠A = ∠C (opposite angle of || gm)
∠ABE = ∠CFB (alternate angle)
∴ ∆ABE ~ ∆CFB (AA similarity criterion)

Question 9.
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) ∆ABC ~ ∆AMP
(ii) \(\frac{\mathbf{C A}}{\mathbf{P A}}=\frac{\mathbf{B C}}{\mathbf{M P}}\)

Solution:
Given. ∆ABC and ∆AMP are two right triangles right angled at B and M.
To Prove. (i) ∆ABC ~ ∆AMP
(ii) \(\frac{\mathbf{C A}}{\mathbf{P A}}=\frac{\mathbf{B C}}{\mathbf{M P}}\)
Proof. In ∆ABC and ∆AMP,
∠A = ∠A (common)
∠B = ∠M (each 90°)
(i) ∴ ∆ABC ~ ∆AMP (AA similarity criterion)

(ii) ∴ \(\frac{\mathrm{AC}}{\mathrm{AP}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
[If two triangles are similar corresponding sides]
\(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
Hence Proved.

Q. 10.
CD and GH are respectively the vectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and
∆EFG respectively. If ∆ABC ~ ∆FEG, show
(i) \(\frac{\mathbf{C D}}{\mathbf{G H}}=\frac{\mathbf{A C}}{\mathbf{F G}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF

Given. In ∆ABC and ∆EFG, CD and OH are bisector of ∠ACB and ∠EGF
i.e. ∠1 = ∠2
and ∠3 = ∠4
and ∆ABC ~ ∆FEG
To Prove. (i) = \(\frac{\mathbf{C D}}{\mathbf{G H}}=\frac{\mathbf{A C}}{\mathbf{F G}}\)
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
Proof.
(i) Given that, ∆ABC ~ ∆FEG
∴ ∠A = ∠F; ∠B = ∠E
and ∠C = ∠C
[∵ The corresponding angles of similar triangles are equal]
Consider, ∠C = ∠C [Proved above]
\(\frac{1}{2}\) ∠C = \(\frac{1}{2}\) ∠G
∠2 = ∠4 or ∠1 = ∠3
Now, in ∆ACD and ∆FGH
∠A = ∠F [Proved above]
∠2 = ∠4 [Proved above]
∴ ∠ACD ~ ∠FGH [∵ AA similarity creterion]
Also, \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)
[∵ Corresponding sides are in proportion].

(ii) In ∆DCB and ∆HGE,
∠B = ∠E [Proved above]
∠1 = ∠3 [Proved above]
∴ ∆DCB ~ ∆HGE [∵ AA similarity criterion]

(iii) In ∆DCA and ∆HGF
∠A = ∠F [Proved above]
∠2 = ∠4 [Proved above]
∴ ∆DCA ~ ∆HGF [∵ AA similarity criterion].

Question 11.
In Fig., E is a point on side CB produced of an Isosceles triangle ABC with AB = AC. IfAD ⊥BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution:
Given. ∆ABC, isosceles triangle with AB = AC AD ⊥ BC, side BC is produced to E. EF ⊥ AC
To Prove. ∆ABD ~ ∆ECF
Proof. ∆ABC is isosceles (given)
AB = AC
∴ ∠B = ∠C [Equal sides have equal angles opposite to it)
In ∆ABD and ∆ECF,
∠ABD = ∠ECF (Proved above)
∠ADB = ∠EFC (each 90°)
∴ ∠ABD – ∠ECF [AA similarity).

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see Fig.). Show that ∆ABC ~ ∆PQR.

Solution:
Given. ∆ABC and ∆PQR, AB, BC, and median AD of ∆ABC are proportional to side PQ; QR and median PM of ∆PQR,
i.e., \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

To prove: ∆ABC ~ ∆PQR
Construction: Produce AD to E such that AD = DE and Produce PM to N such that PM = MN join BE, CE, QN and RN
Proof: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (given) …………..(1)
BD = DC (given)
AD = DE (construction)
Diagonal bisects each other ¡n quadrilateral ABEC
∴ Quadrilateral ABEC is parallelogram
Similarly PQNR is a parallelogram
∴ BE = AC (opposite sides of parallelogram) and QN = PR
\(\frac{\mathrm{BE}}{\mathrm{AC}}=1\) ……………(i)
\(\frac{\mathrm{QN}}{\mathrm{PR}}=1\) …………..(ii)
From (i) and (ii),
\(\frac{\mathrm{BE}}{\mathrm{AC}}=\frac{\mathrm{QN}}{\mathrm{PR}}\)
⇒ \(\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)
But \(\frac{A B}{P Q}=\frac{A C}{P R}\) (Given)
∴ \(\frac{B E}{Q N}=\frac{A B}{P Q}\) …………..(2)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) From (1)
= \(\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AE}}{\mathrm{PN}}\) …………..(3)
From (2) and (3),
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∆ABE ~ ∆PQN [Sides are Proportional]
∴ ∠1 = ∠2 …………….(4) [Corresponding angle of similar triangle]
|| ly ∆ACE ~ ∆PRN ……….(5) [Corresponding angle of similar triangle]
Adding (4) and (5).
∠1 + ∠3 = ∠2 + ∠4
∠A = ∠P
Now in ∆ABC and ∆PQR,
∠A = ∠P (Proved)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) (given)
∴ ∆ABC ~ ∆PQR [By using SA similarity criterion]
Hence Proved.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB. CD.
Solution:
Given. ∆ABC, D is a point on side BC such that ∠ADC = ∠BAC
To Prove. CA² = BC × CD
Proof. In ABC and ADC,

∠C = ∠C (common)
∠BAC = ∠ADC (given)
∴ ∆ABC ~ ∆DAC [by AA similarity criterion]
∴ \(\frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
[If two triangles are similar corresponding sides are proportional]
AC² = BC. DC Hence Proved.

Question 14.
Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another
triangle PQR. Prove that ∆ABC ~ ∆PQR.
Solution:

Given: Two ∆s ABC and PQR. D is the mid-point of BC and M is the mid-point of QR. and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) ………..(1)
To Prove: ∆ABC ~ ∆PQR
Construction:
Produce AD to E such that AD = DE
Join BE and CE.
Proof. In quad. ABEC, diagonals AE and
BC bisect each other at D.
∴ Quad. ABEC is a parallelogram.
Similarly it can be shown that quad PQNR is a parallelogram.
Since ABEC is a parallelogram
∴. BE = AC ………….(2)
Similarly since PQNR is a || gm
∴ QN = PR ………….(3)
Dividing (2) by (3), we get:
\(\frac{B E}{Q N}=\frac{A C}{P R}\) …………….(4)
Now \(\frac{\mathrm{AD}}{\mathrm{PM}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∠BAE = ∠QPN ………….(5)
From (1), (4) and (5), we get:
\(\frac{A D}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
Thus in ∆s ABE and PQN, we get:
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ ∆ABC ~ ∆PQN
∴ ∠BAE = ∠QPN ………..(6)
Similarly it can be proved that
∆AEC ~ ∆PNR
∴ ∠EAC = ∠NPR …………..(7)
Adding (6) and (7), we get:
∠BAE + ∠EAC = ∠QPN + ∠NPR
i.e., ∠BAC = ∠QPR
Now in ∆ABC and ∆PQR.
\(\frac{A B}{P Q}=\frac{A C}{P R}\)
and included ∠A = ∠P
∴ ∆ABC ~ ∆QPR (By SAS criterion of similarity).

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:

Length of vertical stick = 6 m
Shadow of stick = 4 m
Let height of tower be H m
Length of shadow of tower = 28 m
In ∆ABC and ∆PMN,
∠C = ∠N (angle of altitude of sun)
∠B = ∠M (each 90°)
∴ ∆ABC ~ ∆PMN [AA similarity criterion]
∴ \(\frac{\mathrm{AB}}{\mathrm{PM}}=\frac{\mathrm{BC}}{\mathrm{MN}}\)
[If two triangles are similar corresponding sides are proportional]
∴ \(\frac{6}{\mathrm{H}}=\frac{4}{28}\)
H = \(\frac{6 \times 28}{4}\)
H = 6 × 7
H = 42 m.
Hence, Height of Tower = 42 m.

Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that \(\frac{\mathbf{A B}}{\mathbf{P Q}}=\frac{\mathbf{A D}}{\mathbf{P M}}\).

Solution:
Given: ∆ABC and ∆PQR, AD and PM are median and ∆ABC ~ ∆PQR
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Proof. ∆ABC ~ ∆PQR (given)
∴ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
(If two triangles are similar corrosponding sides are Proportional)
∠A = ∠P
(If two triangles are similar corrosponding angles are equal)
∠B = ∠Q
∠C = ∠R
D is mid Point of BC
∴ BD = DC = \(\frac{1}{2}\) BC ……………..(2)
M is mid point of OR
∴ QM = MR = \(\frac{1}{2}\) QR …………….(3)
\(\frac{A B}{P Q}=\frac{B C}{Q R}\)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\) (from(2)and(3))
\(\frac{A B}{P Q}=\frac{B D}{Q M}\)
∠ABD = ∠PQM (given)
∆ABC ~ ∆PQM (By SAS similarity criterion)
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
[If two triangles are similar corresponding sides are proportional].

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 12 Algebraic Expressions MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 12 Algebraic Expressions MCQ Questions

Multiple Choice Questions :

Question 1.
On subtracting 9 from -q, we get:
(a) 9 – q
(b) q – 9
(c) 9 + q
(d) 9 – q
Answer:
(b) q – 9

Question 2.
The numerical coefficient of variable in expression 5 – 3t² is :
(a) 3
(b) -3
(c) – 3²
(d) 2
Answer:
(b) -3

Question 3.
In the expression 5y² + 7x, the coefficient of y² is :
(a) 5
(b) 7
(c) -5
(d) 2
Answer:
(a) 5

Question 4.
The sum of 3mn, -5mn, 8mn, -4mn is :
(a) 10 mn
(b) – 8 mn
(c) 12 mn
(d) 2 mn.
Answer:
(d) 2 mn.

Question 5.
If m = 2, the value of 3m – 5 is :
(a) 6
(b) 1
(c) 11
(d) -1.
Answer:
(b) 1

Question 6.
If m = 2, the value of 9 – 5m is :
(a) -1
(b) 1
(c) 19
(d) 13
Answer:
(a) -1

Question 7.
If p = – 2, the value of 4p + 7 is :
(a) 15
(b) 18
(c) 20
(d) -1.
Answer:
(d) -1.

Question 8.
If a = 2, b = – 2, the value of a² + b² is :
(a) 0
(b) 4
(c) 8
(d) 10
Answer:
(c) 8

Fill in the blanks :

Question 1.
On subtracting 5 from x we get ……………
Answer:
x – 5

Question 2.
The vable of 4x + 7 for x = 2 is ……………
Answer:
15

Question 3.
The sum of -4xy, 2xy, 3xy is ……………
Answer:
xy

Question 4.
A symbol having a fixed numerical value is called ……………
Answer:
constant

Question 5.
Binomial has …………… terms.
Answer:
two

Write True or False :

Question 1.
Every number is a constant. (True/False)
Answer:
True

Question 2.
A symbol which takes on various numerical value is called a variable (True/False)
Answer:
True

Question 3.
Expressions are formed by addition of terms. (True/False)
Answer:
False

Question 4.
7 and 12xy are like terms. (True/False)
Answer:
False

Question 5.
The coefficient of x in 2x + 3y = 6 is 3. (True/False)
Answer:
False