rs aggarwal solutions for class 9 maths chapter 10 … · 2020. 11. 2. · rs aggarwal solutions...
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
Exercise 10(B) 1.
Solution:
We know that in a parallelogram opposite angles are equal
So we get
∠ A = ∠ C = 72o
We that that in a parallelogram the sum of all the angles is 360o
It can be written as
∠ A + ∠ B + ∠ C + ∠ D = 360o
By substituting the values in the above equation
72o + ∠ B + 72o + ∠ D = 360o
We know that ∠ A = ∠ C and ∠ B = ∠ D
So we can write it as
2 ∠ B + 144o = 360o
On further calculation
2 ∠ B = 360o – 144o
By subtraction
2 ∠ B = 216o
By division
∠ B = 108o
Therefore, ∠ B = 108o, ∠ C = 72o and ∠ D = 108o.
2.
Solution:
It is given that ABCD is a parallelogram in which ∠ DAB = 80o and ∠ DBC = 60o
We know that opposite angles are equal in parallelogram
So we get
∠ C = ∠ A = 80o
From the figure we know that AD || BC and BD is a transversal
We know that ∠ ADB and ∠ DBC are alternate angles
So we get
∠ ADB = ∠ DBC = 60o
Consider △ ABD
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
Using the sum property of triangle
∠ A + ∠ ADB + ∠ ABD = 180o
By substituting values in the above equation
80o + 60o + ∠ ABD = 180o
On further calculation
∠ ABD = 180o – 80o – 60o
By subtraction
∠ ABD = 180o – 140o
So we get
∠ ABD = 40o
It can be written as
∠ ABC = ∠ ABD + ∠ DBC
By substituting values we get
∠ ABC = 40o + 60o
By addition
∠ ABC = 100o
We know that the opposite angles are equal in a parallelogram
∠ ADC = ∠ ABC = 100o
We get
∠ ADC = ∠ CDB + ∠ ADB
On further calculation
∠ CDB = ∠ ADC – ∠ ADB
By substituting values
∠ CDB = 100o – 60o
By subtraction
∠ CDB = 40o
Therefore, ∠ ADB = 60o and ∠ CDB = 40o.
3.
Solution:
It is given that ABCD is a parallelogram
So we know that AD || BC
From the figure we know that ∠ DAM and ∠ AMB are alternate angles
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
So we get
∠ DAM = ∠ AMB
We know that ∠ BAM = ∠ DAM
It can be written as
∠ BAM = ∠ AMB
From the figure we know that the sides opposite to equal angles are equal
So we get
BM = AB
We know that the opposite sides of a parallelogram are equal
AB = CD
So we can write it as
BM = AB = CD ……. (1)
We know that M is the midpoint of the line BC
So we get
BM = ½ BC
We know that BC = AD
We get
BM = ½ AD
Based on equation (1)
CD = ½ AD
By cross multiplication
AD = 2CD.
Therefore, it is proved that AD = 2CD.
4.
Solution:
(i) We know that opposite angles are equal in a parallelogram.
So we get
∠ C = ∠ A = 60o
We know that the sum of all the angles in a parallelogram is 360o
It can be written as
∠ A + ∠ B + ∠ C + ∠ D = 360o
So we get
∠ B + ∠ D = 360o – (∠ A + ∠ C)
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
By substituting values in the above equation we get
∠ B + ∠ D = 360o – (60o + 60o)
On further calculation we get
∠ B + ∠ D = 360o – 120o
By subtraction
∠ B + ∠ D = 240o
We know that ∠ B = ∠ B
So the above equation becomes
∠ B + ∠ B = 240o
2 ∠ B = 240o
By division
∠ B = ∠ D = 120o
We know that AB || DP and AP is a transversal
From the figure we know that ∠ APD and ∠ PAD are alternate angles
∠ APD = ∠ PAD = 60o/2
∠ APD = ∠ PAD = 30o ……. (1)
We know that AB || PC and BP is a transversal
So we get
∠ ABP = ∠ CPB = ∠ B/2
i.e. ∠ ABP = ∠ CPB = 120o/2
We get ∠ ABP = ∠ CPB = 60o …….. (2)
We know that DPC is a straight line
It can be written as
∠ APD + ∠ APB + ∠ CPB = 180o
By substituting the values we get
30o + ∠ APB + 60o = 180o
On further calculation
∠ APB = 180o – 30o – 60o
By subtraction
∠ APB = 180o – 90o
∠ APB = 90o
Therefore, it is proved that ∠ APB = 90o
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
(ii) From equation (1) we know that
∠ APD = 30o
We know that
∠ DAP = 60o/2
By division
∠ DAP = 30o
So we get
∠ APD = ∠ DAP ……… (3)
From the figure we know that the sides of an isosceles triangle are equal
So we get
DP = AD
We know that
∠ CPB = 60o and ∠ C = 60o
By sum property of triangle
We get
∠ C + ∠ CPB + ∠ PBC = 180o
By substituting the values in the above equation
60o + 60o + ∠ PBC = 180o
On further calculation we get
∠ PBC = 180o – 60o – 60o
By subtraction
∠ PBC = 180o – 120o
∠ PBC = 60o
We know that all the sides of an equilateral triangle are equal
So we get
PB = PC = BC ………. (4)
Therefore, it is proved that AD = AP and PB = PC = BC.
(iii) We know that ∠ DPA = ∠ PAD based on equation (3)
We also know that all the sides are equal in an isosceles triangle
DP = AD
From the figure we know that the opposite sides are equal
So we get
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
DP = BC
Considering equation (4)
DP = PC
From the figure we know that DP = PC and P is the midpoint of the line DC
So we get
DP = ½ DC
By cross multiplication we get
DC = 2AD
Therefore, it is proved that DC = 2AD.
5.
Solution:
(i) From the figure we know that ∠ AOB and ∠ COD are vertically opposite angles
So we get
∠ AOB = ∠ COD = 105o
Consider △ AOB
By sum property of a triangle
∠ OAB + ∠ AOB + ∠ ABO = 180o
By substituting the values in above equation
35o + 105o + ∠ ABO = 180o
On further calculation
∠ ABO = 180o – 35o – 105o
By subtraction
∠ ABO = 180o – 140o
∠ ABO = 40o
(ii) We know that AB || DC and BD is a transversal
From the figure we know that ∠ ABD and ∠ CDB are alternate angles
It can be written as
∠ CDO = ∠ CDB = ∠ ABD = ∠ ABO = 40o
So we get
∠ ODC = 40o
(iii) We know that AB || CD and AC is a transversal
From the figure we know that ∠ ACB and ∠ DAC are alternate opposite angles
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
So we get
∠ ACB = ∠ DAC = 40o
(iv) We know that ∠ B can be written as
∠ B = ∠ CBD + ∠ ABO
So we get
∠ CBD = ∠ B – ∠ ABO
In a parallelogram we know that the sum of all the angles is 360o
So we get
∠ A + ∠ B + ∠ C + ∠ D = 360o
It can be written as
2 ∠ A + 2 ∠ B = 360o
By substituting values in the above equation
2 (40o + 35o) + 2 ∠ B = 360o
On further calculation
2 (75o) + 2 ∠ B = 360o
So we get
150o + 2 ∠ B = 360o
2 ∠ B = 360o – 150o
By subtraction
2 ∠ B = 210o
By division
∠ B = 105o
So we get
∠ CBD = ∠ B – ∠ ABO
By substituting values
∠ CBD = 105o – 40o
By subtraction
∠ CBD = 65o
6.
Solution:
We know that the opposite angles are equal in a parallelogram
Consider parallelogram ABCD
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
So we get
∠ A = ∠ C = (2x + 25) o
∠ B = ∠ D = (3x – 5) o
We know that the sum of all the angles of a parallelogram is 360o
So it can be written as
∠ A + ∠ B + ∠ C + ∠ D = 360o
By substituting the values in the above equation
(2x + 25) + (3x – 5) + (2x + 25) + (3x – 5) = 360o
By addition we get
10x + 40o = 360o
By subtraction
10x = 360o – 40o
So we get
10x = 320o
By division we get
x = 32o
Now substituting the value of x
∠ A = ∠ C = (2x + 25) o = (2(32) + 25) o
∠ A = ∠ C = (64 + 25) o
By addition
∠ A = ∠ C = 89o
∠ B = ∠ D = (3x – 5) o = (3(32) – 5) o
∠ B = ∠ D = (96 – 5) o
By subtraction
∠ B = ∠ D = 91o
Therefore, x = 32o, ∠ A = ∠ C = 89o and ∠ B = ∠ D = 91o.
7.
Solution:
Consider ABCD as a parallelogram
If ∠ A = xo
We know that ∠ B is adjacent to A which can be written as 4/5 xo
Opposite angles are equal in a parallelogram
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
So we get
∠ A = ∠ C = xo and ∠ B = ∠ D = 4/5 xo
We know that the sum of all the angles of a parallelogram is 360o
It can be written as
∠ A + ∠ B + ∠ C + ∠ D = 360o
By substituting the values in the above equation
x + (4/5) x + x + (4/5) x = 360o
By addition we get
2x + (8/5) x = 360o
By taking the LCM as 5
(18/5) x = 360o
By cross multiplication
x = (360 × 5)/18
On further calculation
x = 100o
By substituting the value of x
So we get
∠ A = ∠ C = x = 100o
∠ B = ∠ D = 4/5 xo = (4/5) (100o) = 80o
Therefore, ∠ A = ∠ C = x = 100o and ∠ B = ∠ D = 80o.
8.
Solution:
Consider ABCD as a parallelogram
Let us take ∠ A as the smallest angle
So we get
∠ B = 2 ∠ A – 30o
We know that the opposite angles are equal in a parallelogram
∠ A = ∠ C and ∠ B = ∠ D = 2 ∠ A – 30o
We know that the sum of all the angles of a parallelogram is 360o
It can be written as
∠ A + ∠ B + ∠ C + ∠ D = 360o
By substituting the values in the above equation
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ A + (2 ∠ A – 30o) + ∠ A + (2 ∠ A – 30o) = 360o
On further calculation
∠ A + 2 ∠ A – 30o + ∠ A + 2 ∠ A – 30o = 360o
So we get
6 ∠ A – 60o = 360o
By addition
6 ∠ A = 360o + 60o
6 ∠ A = 420o
By division
∠ A = 70o
By substituting the value of ∠ A
∠ A = ∠ C = 70o
∠ B = ∠ D = 2 ∠ A – 30o = 2 (70o) – 30o
∠ B = ∠ D = 110o
Therefore, ∠ A = ∠ C = 70o and ∠ B = ∠ D = 110o.
9.
Solution:
We know that the perimeter of parallelogram ABCD can be written as
Perimeter = AB + BC + CD + DA
We know that opposite sides of parallelogram are equal
AB = CD and BC = DA
By substituting the values
Perimeter = 9.5 + BC + 9.5 + BC
It is given that perimeter = 30 cm
So we get
30 = 19 + 2BC
It can be written as
2BC = 30 – 19
By subtraction
2BC = 11
By division we get
BC = 5.5 cm
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
Therefore, AB = 9.5 cm, BC = 5.5 cm, CD = 9.5 cm and DA = 5.5 cm.
10.
Solution:
(i) We know that all the sides are equal in a rhombus
Consider △ ABC
We know that AB = BC
It can be written as
∠ CAB = ∠ ACB = xo
By sum property of a triangle
We get
∠ CAB + ∠ ABC + ∠ ACB = 180o
By substituting the values in above equation
x + 110o + x = 180o
By addition we get
2x + 110o = 180o
On further calculation
2x = 180o – 110o
By subtraction
2x = 70o
By division
x = 35o
Therefore, x = 35o and y = 35o.
(ii) We know that all the sides are equal in a rhombus
Consider △ ABD
We get
AB = AD and
∠ ABD = ∠ ADB
It can be written as
x = y ……. (1)
Consider △ ABC
We get
AB = BC and
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ CAB = ∠ ACB
We know that
∠ ACB = 40o
By using the sum property of a triangle
∠ B + ∠ CAB + ∠ ACB = 180o
By substituting values in the above equation
∠ B + 40o + 40o = 180o
On further calculation
∠ B = 180o – 40o – 40o
By subtraction
∠ B = 180o – 80o
So we get
∠ B = 100o
∠ DBC can be written as
∠ DBC = ∠ B – xo
By substituting the values
∠ DBC = 100o – xo
From the figure we know that ∠ DBC and ∠ ADB are alternate angles
∠ DBC = ∠ ADB = yo
By substituting the value of ∠ DBC
100o – xo = yo
Consider the equation (1) we know that x = y
100o – xo = xo
On further calculation
2xo = 100o
By division
xo = 50o
Therefore, x = y = 50o.
(iii) From the figure we know that
∠ A = ∠ C = 62o
Consider △ BCD
We get
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
BC = DC
It can be written as
∠ CDB = ∠ DBC = yo
Using the sum property of triangle
∠ BDC + ∠ DBC + ∠ BCD = 180o
By substituting the values
y + y + 62o = 180o
On further calculation
2y = 180o – 62o
By subtraction
2y = 118o
By division
y = 59o
We know that the diagonals of a rhombus are perpendicular to each other
Consider △ COD as a right angle triangle
∠ DOC = 90o
∠ ODC = y = 59o
It can be written as
∠ DCO + ∠ ODC = 90o
To find ∠ DCO
∠ DCO = 90o – ∠ ODC
By substituting the values
∠ DCO = 90o – 59o
∠ DCO = x = 31o
Therefore, x = 31o and y = 59o.
11.
Solution:
We know that ABCD is a rhombus
It is given that AC = 24cm and BD = 18cm
In a rhombus we know that the diagonals bisect each other at right angles
Consider △ AOB
We know that
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ AOB = 90o
To find AO and BO
We know that
AO = ½ AC
By substituting AC
AO = ½ (24)
AO = 12 cm
BO = ½ BD
By substituting BD
BO = ½ (18)
BO = 9 cm
Based on the Pythagoras Theorem
AB2 = AO2 + OB2
By substituting values
AB2 = 122 + 92
So we get
AB2 = 144 + 81
By addition
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
AB2 = 225
AB = √ 225
So we get
AB = 15 cm
Therefore, the length of each side of the rhombus is 15 cm.
12.
Solution:
We know that the diagonals of a rhombus bisect at right angles
We get
AO = OC = ½ AC
We know that AC = 16 cm
AO = OC = ½ (16) = 8 cm
Consider △ AOB
Using the Pythagoras Theorem
AB2 = AO2 + OB2
By substituting values
102 = 82 + OB2
On further calculation
OB2 = 100 – 64
By subtraction
OB2 = 36
So we get
OB = √ 36
OB = 6 cm
We know that the length of other diagonal
BD = 2 × OB
By substituting the value
BD = 2 × 6
BD = 12 cm
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
From the figure the area of
△ ABC = ½ × AC × OB
By substituting the values
△ ABC = ½ × 16 × 6
By multiplication
△ ABC = 48 cm2
△ ACD = ½ × AC × OD
By substituting the values
△ ACD = ½ × 16 × 6
By multiplication
△ ACD = 48 cm2
So we can calculate the area of rhombus as
Area of rhombus = area of △ ABC + area of △ ACD
We get
Area of rhombus = 48 + 48 = 96 cm2
Therefore, the length of other diagonal is 12 cm and the area of rhombus is 96 cm2.
13.
Solution:
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
(i) From the figure we know that the diagonals are equal and bisect at point O.
Consider △ AOB
We get AO = OB
We know that the base angles are equal
∠ OAB = ∠ OBA = 35o
By using the sum property of triangle
∠ AOB + ∠ OAB + ∠ OBA = 180o
By substituting the values we get
∠ AOB + 35o + 35o = 180o
On further calculation
∠ AOB = 180o – 35o – 35o
By subtraction
∠ AOB = 180o – 70o
∠ AOB = 110o
From the figure we know that the vertically opposite angles are equal
∠ DOC = ∠ AOB = y = 110o
In △ ABC
We know that ∠ ABC = 90o
Consider △ OBC
We know that
∠ OBC = xo = ∠ ABC – ∠ OBA
By substituting the values
∠ OBC = 90o – 35o
By subtraction
∠ OBC = 55o
Therefore, x = 55o and y = 110o.
(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.
Consider △ AOB
We get
OA = OB
We know that the base angles are equal
∠ OAB = ∠ OBA
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
By using the sum property of triangle
∠ AOB + ∠ OAB + ∠ OBA = 180o
By substituting the values
110o + ∠ OAB + ∠ OBA = 180o
We know that ∠ OAB = ∠ OBA
So we get
2 ∠ OAB = 180o – 110o
By subtraction
2 ∠ OAB = 70o
By division
∠ OAB = 35o
We know that AB || CD and AC is a transversal
From the figure we know that ∠ DCA and ∠ CAB are alternate angles
∠ DCA = ∠ CAB = yo = 35o
Consider △ ABC
We know that
∠ ACB + ∠ CAB = 90o
So we get
∠ ACB = 90o – ∠ CAB
By substituting the values in above equation
∠ ACB = 90o – 35o
By subtraction
∠ ACB = x = 55o
Therefore, x = 55o and y = 35o.
14.
Solution:
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
From the figure consider DP bisects the side AB at point P
Construct a line at BD
Consider △ AMD and △ BMD
We know that M is the mid-point of AB
AM = BM
From the figure we know that
∠ AMD = ∠ BMD = 90o
MD is common i.e. MD = MD
By SAS congruence criterion
△ AMD ≅ △ BMD
AD = BD (c. p. c. t)
We know that the sides of a rhombus are equal
So we get
AD = AB
It can be written as
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
AD = AB = BD
We know that △ ADB is an equilateral triangle
So we get
∠ A = 60o
From the figure we know that the opposite angles are equal
∠ C = ∠ A = 60o
We know that
∠ B + ∠ A = 180o
On further calculation
∠ B = 180o – ∠ A
∠ B = 180o – 60o
By subtraction
∠ B = 120o
We know that ∠ D = ∠ B = 120o
Therefore, the angles are ∠ C = ∠ A = 60o and ∠ D = ∠ B = 120o.
15.
Solution:
In △ ABD
We know that AB = AD
From the figure we know that the base angles are equal
∠ ADB = ∠ ABD
We know that ∠ A = 90o
It can be written as
∠ ADB + ∠ ABD = 90o
We know that ∠ ADB = ∠ ABD
So we get
2 ∠ ADB = 90o
By division
∠ ADB = 45o
Consider △ OXB
From the figure we know that ∠ XOB and ∠ DOC are vertically opposite angles
∠ XOB = ∠ DOC = 80o
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
We also know that
∠ ABD = ∠ XBD = 45o
We can write it as
Exterior ∠ AXO = ∠ XOB + ∠ XBD
By substituting the values
xo = 80o + 45o
By addition
xo = 125o
Therefore, the value of x is 125o.
16.
Solution:
Consider △ ABC and △ ADC
We know that the sides of rhombus are equal
AB = AD and BC = CD
AC is common i.e. AC = AC
By SSS congruence criterion
△ ABC ≅ △ ADC
We know that
∠ BAC = ∠ DAC and ∠ BCA = ∠ DCA (c. p. c. t)
Therefore, AC bisects ∠ A as well as ∠ C.
Consider △ BAD and △ BCD
We know that the sides of rhombus are equal
AB = BC and AD = CD
BD is common i.e. BD = BD
By SSS congruence criterion
△ BAD ≅ △ BCD
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
We know that
∠ ABD = ∠ CBD and ∠ ADB = ∠ CDB (c. p. c. t)
Therefore, BD bisects ∠ B as well as ∠ D.
17.
Solution:
Consider △ AMO and △ CNO
We know that AB || CD
From the figure we know that ∠ MAO and ∠ NCO are alternate angles
It is given that AM = CN
We know that ∠ AOM and ∠ CON are vertically opposite angles
∠ AOM = ∠ CON
By ASA congruence criterion
△ AMO ≅ △ CNO
So we get
AO = CO and MO = NO (c. p. c. t)
Therefore, it is proved that AC and MN bisect each other.
18.
Solution:
Consider △ ABQ and △ CDP
We know that the opposite sides of a parallelogram are equal
AB = CD
So we get ∠ B = ∠ D
We know that
DP = AD – PA
i.e. DP = 2/3 AD
BQ = BC – CQ
i.e. BQ = BC – 1/3 BC
BQ = (3-1)/3 BC
We know that AD = BC
So we get
BQ = 2/3 BC = 2/3 AD
We get BQ = DP
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
By SAS congruence criterion
△ ABQ ≅ △ CDP
AQ = CP (c. p. c. t)
We know that
PA = 1/3 AD
We know that AD = BC
CQ = 1/3 BC = 1/3 AD
So we get
PA = CQ
∠ QAB = ∠ PCD (c. p. c. t)… (1)
We know that
∠ QAP = ∠ A – ∠ QAB
Consider equation (1)
∠ A = ∠ C
∠ QAP = ∠ C – ∠ PCD
From the figure we know that the alternate interior angles are equal
∠ QAP = ∠ PCQ
So we know that AQ and CP are two parallel lines.
Therefore, it is proved that PAQC is a parallelogram.
19.
Solution:
We know that ABCD is a parallelogram whose diagonals intersect each other at O
Consider △ AOE and △ COF
We know that ∠ CAE and ∠ DCA are alternate angles
From the figure we know that the diagonals are equal and bisect each other
AO = CO
We know that ∠ AOE and ∠ COF are vertically opposite angles
∠ AOE =∠ COF
By ASA congruence criterion
△ AOE ≅ △ COF
OE = OF (c. p. c. t)
Therefore, it is proved that OE = OF.
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
20.
Solution:
From the figure
∠ DCM = ∠ DCN + ∠ MCN
By substituting values in the above equation
90o = ∠ DCN + 60o
On further calculation
∠ DCN = 90o – 60o
By subtraction
∠ DCN = 30o
Consider △ DCN
Using the sum property
∠ DNC + ∠ DCN + ∠ D = 180o
By substituting values in the above equation
90o + 30o + ∠ D = 180o
On further calculation
∠ D = 180o – 90o – 30o
By subtraction
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ D = 180o – 120o
∠ D = 60o
We know that the opposite angles of parallelogram are equal
∠ B = ∠ D = 60o
It can be written as
∠ A + ∠ B = 180o
By substituting values
∠ A + 60o = 180o
By subtraction
∠ A = 180o – 60o
∠ A =120o
So we get ∠ A = ∠ C =120o
Therefore, ∠ B = ∠ D = 60o and ∠ A = ∠ C =120o.
21.
Solution:
(i) We know that ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C.
So we get
∠ BAC = ∠ DAC …….. (1)
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ BCA = ∠ DCA …….. (2)
We know that every rectangle is a parallelogram
So we get AB || DC and AC is a transversal
From the figure we know that ∠ BAC and ∠ DCA are alternate angles
∠ BAC = ∠ DCA
By considering equation (1)
We get
∠ DAC = ∠ DCA
Consider △ ADC
We know that the opposite sides of equal angles are equal
AD = CD
Since ABCD is a rectangle
We get AB = BC and CD = AD
So we get AB = BC = CD = AD
Therefore, it is proved that ABCD is a square.
(ii) Consider △ BAD and △ BCD
We know that AB = CD and AD = BC
BD is common i.e. BD = BD
By SSS congruence criterion
△ BAD ≅ △ BCD
We know that
∠ ABD = ∠ CBD and ∠ ADB = ∠ CDB (c. p. c. t)
Therefore, diagonal BD bisects ∠ B as well as ∠ D.
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
22.
Solution:
Consider △ OCD and △ OBE
From the figure we know that ∠ DOC and ∠ EOB are vertically opposite angles
∠ DOC = ∠ EOB
We know that AB || CD and BC is a transversal
∠ OCD and ∠ OBE are alternate angles
∠ OCD = ∠ OBE
From the figure we know that AB = CD and BE = AB
So we can write DC = BE
By AAS congruence criterion
△ OCD ≅ △ OBE
OC = OB (c. p. c. t)
Therefore, it is proved that ED bisects BC.
23.
Solution:
Consider △ DEC and △ FEB
From the figure we know that ∠ DEC and ∠ FEB are vertically opposite angles
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
∠ DEC = ∠ FEB
∠ DCE and ∠ FBE are alternate angles
∠ DCE = ∠ FBE
It is given that CE = EB
By AAS congruence criterion
△ DEC ≅ △ FEB
DC = FB (c. p. c. t)
From the figure
AF = AB + BF
We know that BF = DC and AB = DC
So we get
AF = AB + DC
AF = AB + AB
By addition
AF = 2AB
Therefore, it is proved that AF = 2AB.
24.
Solution:
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
We know that l || m and t is a transversal
From the figure we know that ∠ APR and ∠ PRD are alternate angles
∠ APR = ∠ PRD
We can write it as
½ ∠ APR = ½ ∠ PRD
We know that PS and RQ are the bisectors of ∠ APR and ∠ PRD
So we get
∠ SPR = ∠ PRQ
Hence, PR intersects PS and RQ at points P and R respectively
We get
PS || RQ
In the same way SR || PQ
Therefore, PQRS is a parallelogram
We know that the interior angles are supplementary
∠ BPR + ∠ PRD = 180o
From the figure we know that PQ and RQ are the bisectors of ∠ BPR and ∠ PRD
We can write it as
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
2 ∠ QPR + 2 ∠ QRP = 180o
Dividing the equation by 2
∠ QPR + ∠ QRP = 90o …… (1)
Consider △ PQR
Using the sum property of triangle
∠ PQR + ∠ QPR + ∠ QRP = 180o
By substituting equation (1)
∠ PQR + 90o = 180o
On further calculation
∠ PQR = 180o – 90o
By subtraction
∠ PQR = 90o
We know that PQRS is a parallelogram
It can be written as
∠ PQR = ∠ PSR = 90o
We know that the adjacent angles in a parallelogram are supplementary
∠ SPQ + ∠ PQR = 180o
By substituting the values in above equation
∠ SPQ + 90o = 180o
On further calculation
∠ SPQ = 180o – 90o
By subtraction
∠ SPQ = 90o
We know that all the interior angles of quadrilateral PQRS are right angles
Therefore, it is proved that the quadrilateral formed by the bisectors of interior angles is a rectangle.
25.
Solution:
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
It is given that AK = BL = CM = DN
ABCD is a square
So we get
BK = CL = DM = AN …… (1)
Consider △ AKN and △ BLK
It is given AK = BL
From the figure we know that ∠ A = ∠ B = 90o
Using equation (1)
AN = BK
By SAS congruence criterion
△ AKN ≅ △ BLK
We get
∠ AKN = ∠ BLK and ∠ ANK = ∠ BKL (c. p. c. t)
We know that
∠ AKN + ∠ ANK = 90o
∠ BLK + ∠ BKL = 90o
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
By adding both the equations
∠ AKN + ∠ ANK + ∠ BLK + ∠ BKL = 90o + 90o
On further calculation
2 ∠ ANK + 2 ∠ BLK = 180o
Dividing the equation by 2
∠ ANK + ∠ BLK = 90o
So we get
∠ NKL = 90o
In the same way
∠ KLM = ∠ LMN = ∠ MNK = 90o
Therefore, it is proved that KLMN is a square.
26.
Solution:
We know that AR || BC and AB || RC
From the figure we know that ABCR is a parallelogram
So we get
AR = BC …… (1)
We know that AQ || BC and QB || AC
From the figure we know that AQBC is a parallelogram
So we get
QA = BC ……… (2)
By adding both the equations
AR + QA = BC + BC
We know that AR + QA = QR
So we get
QR = 2BC
Dividing by 2
BC = QR/2
BC = ½ QR
Therefore, it is proved that BC = ½ QR.
27.
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RS Aggarwal Solutions for Class 9 Maths Chapter 10
Quadrilaterals
Solution:
We know that AR || BC and AB || RC
From the figure we know that ABCR is a parallelogram
So we get
AR = BC …….. (1)
We know that AQ || BC and QB || AC
From the figure we know that AQBC is a parallelogram
So we get
QA = BC ……… (2)
By adding both the equations we get
AR + QA = BC + BC
We know that AR + QA = QR
So we get
QR = 2BC
It can be written as
BC = QR/2
BC = ½ QR
In the same way
AB = ½ RP and AC = ½ PQ
Perimeter of △ PQR = PQ + QR + RP
It can be written as
Perimeter of △ PQR = 2AC + 2BC + 2AB
By taking 2 as common
Perimeter of △ PQR = 2 (AC + BC + 2AB)
Perimeter of △ PQR = 2 (Perimeter of △ ABC)
Therefore, it is proved that the perimeter of △ PQR is double the perimeter of △ ABC.
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