Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal Solutions: Quadrilaterals- 2
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Page 1
?
?
Question:11
In the adjoining figure, ABCD is a parallelogram in which ?A = 72°. Calculate ?B, ?C and ?D.
Solution:
ABCD is parallelogram and ?A = 72° ?.
We know that opposite angles of a parallelogram are equal.
? ?A ?= ?C and ?B ?= ?D ? ? ? ?C = 72
o
?A and ?B are adajcent angles.
i.e., ?A ?+ ?B ? = 180
o
? ?B = 180
o
- ?A
? ?B ? = 180
o
- 72
o
= 108
o
? ? ?B ? = ? ?D = 108
o
Hence, ?B ? = ? ?D = 108
o
? and ?C ? = 72
o
? Question:12
In the adjoining figure, ABCD is a parallelogram in which ?DAB = 80° and ?DBC = 60°. Calculate ?CDB and
?ADB.
Solution:
Given: ABCD is parallelogram and ?DAB = 80° ? and ?DBC = 60°
To find: Measure of ?CDB and ?ADB
In parallelogram ABCD, AD || ? BC
? ?DBC = ? ADB = 60
o
Alternateinteriorangles ...
i
As ?DAB and ?ADC are adajcent angles, ?DAB ?+ ?ADC ? = 180
o
? ?ADC = 180
o
- ?DAB
? ? ?ADC ? = 180
o
- 80
o
= 100
o
Also, ?ADC ? = ?ADB + ?C ??DB
Page 2
?
?
Question:11
In the adjoining figure, ABCD is a parallelogram in which ?A = 72°. Calculate ?B, ?C and ?D.
Solution:
ABCD is parallelogram and ?A = 72° ?.
We know that opposite angles of a parallelogram are equal.
? ?A ?= ?C and ?B ?= ?D ? ? ? ?C = 72
o
?A and ?B are adajcent angles.
i.e., ?A ?+ ?B ? = 180
o
? ?B = 180
o
- ?A
? ?B ? = 180
o
- 72
o
= 108
o
? ? ?B ? = ? ?D = 108
o
Hence, ?B ? = ? ?D = 108
o
? and ?C ? = 72
o
? Question:12
In the adjoining figure, ABCD is a parallelogram in which ?DAB = 80° and ?DBC = 60°. Calculate ?CDB and
?ADB.
Solution:
Given: ABCD is parallelogram and ?DAB = 80° ? and ?DBC = 60°
To find: Measure of ?CDB and ?ADB
In parallelogram ABCD, AD || ? BC
? ?DBC = ? ADB = 60
o
Alternateinteriorangles ...
i
As ?DAB and ?ADC are adajcent angles, ?DAB ?+ ?ADC ? = 180
o
? ?ADC = 180
o
- ?DAB
? ? ?ADC ? = 180
o
- 80
o
= 100
o
Also, ?ADC ? = ?ADB + ?C ??DB
? ? ?ADC ? = ? 100
o
? ?ADB + ?C ??DB = 100
o
...ii
From i
and ii
, we get:
60
o
+ ?C ??DB = 100
o
? ?C ??DB = 100
o
- 60
o
= 40
o
Hence, ?CDB ? = ? 40
o
and ?ADB ? = 60
o
? Question:13
In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ?BAM = ?DAM. Prove that
AD = 2CD.
Solution:
Given: parallelogram ABCD, M is the midpoint of side BC and ?BAM = ?DAM.
To prove: AD = 2CD
Proof:
Since, AD ? BC
and AM is the transversal.
So, ?DAM = ?AMB
Alternateinteriorangles
But, ?DAM = ?BAM
Given
Therefore, ?AMB = ?BAM
? AB = BM
Anglesoppositetoequalsidesareequal.
...1
Now, AB = CD Oppositesidesofaparallelogramareequal.
Page 3
?
?
Question:11
In the adjoining figure, ABCD is a parallelogram in which ?A = 72°. Calculate ?B, ?C and ?D.
Solution:
ABCD is parallelogram and ?A = 72° ?.
We know that opposite angles of a parallelogram are equal.
? ?A ?= ?C and ?B ?= ?D ? ? ? ?C = 72
o
?A and ?B are adajcent angles.
i.e., ?A ?+ ?B ? = 180
o
? ?B = 180
o
- ?A
? ?B ? = 180
o
- 72
o
= 108
o
? ? ?B ? = ? ?D = 108
o
Hence, ?B ? = ? ?D = 108
o
? and ?C ? = 72
o
? Question:12
In the adjoining figure, ABCD is a parallelogram in which ?DAB = 80° and ?DBC = 60°. Calculate ?CDB and
?ADB.
Solution:
Given: ABCD is parallelogram and ?DAB = 80° ? and ?DBC = 60°
To find: Measure of ?CDB and ?ADB
In parallelogram ABCD, AD || ? BC
? ?DBC = ? ADB = 60
o
Alternateinteriorangles ...
i
As ?DAB and ?ADC are adajcent angles, ?DAB ?+ ?ADC ? = 180
o
? ?ADC = 180
o
- ?DAB
? ? ?ADC ? = 180
o
- 80
o
= 100
o
Also, ?ADC ? = ?ADB + ?C ??DB
? ? ?ADC ? = ? 100
o
? ?ADB + ?C ??DB = 100
o
...ii
From i
and ii
, we get:
60
o
+ ?C ??DB = 100
o
? ?C ??DB = 100
o
- 60
o
= 40
o
Hence, ?CDB ? = ? 40
o
and ?ADB ? = 60
o
? Question:13
In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ?BAM = ?DAM. Prove that
AD = 2CD.
Solution:
Given: parallelogram ABCD, M is the midpoint of side BC and ?BAM = ?DAM.
To prove: AD = 2CD
Proof:
Since, AD ? BC
and AM is the transversal.
So, ?DAM = ?AMB
Alternateinteriorangles
But, ?DAM = ?BAM
Given
Therefore, ?AMB = ?BAM
? AB = BM
Anglesoppositetoequalsidesareequal.
...1
Now, AB = CD Oppositesidesofaparallelogramareequal.
? 2AB = 2CD ? (AB +AB) = 2CD
? BM +MC = 2CD
(AB = BM and MC = BM)
? BC = 2CD ? AD = 2CD (AD = BC, Opposite sides of a parallelogram are equal. )
Question:14
In the adjoining figure, ABCD is a parallelogram in which ?A = 60°. If the bisectors of ?A and ?B meet DC at P,
prove that
i ?APB = 90°,
ii AD = DP and PB = PC = BC,
iii DC = 2AD.
Solution:
ABCD is a parallelogram.
? ??A = ??C and ?B = ? ?D Oppositeangles
And ??A + ??B = 180
o
Adjacentanglesaresupplementary
? ??B = 180
o
- ?A
? 180
o
- 60
o
= 120
o
( ? ?A = 60
o
)
? ??A = ??C = 60
o
and ?B = ? ?D ? = 120
o
i
In ? APB, ? ?PAB =
60°
2
= 30°
and ?PBA =
120°
2
= 60°
? ? ? ?APB ? = 180
o
- (30
o
+ 60
o
) = 90
o
ii
In ? ADP, ? ?PAD = 30
o
and ?ADP = 120
o
? ?APB = 180
o
- (30
o
+ 120
o
) = 30
o
Thus, ? ?PAD = ??APB = ?30
o
Hence, ?ADP is an isosceles triangle and AD = DP.
In ? PBC, ? ? PBC = 60
o
, ? ? BPC = 180
o
- (90
o
+30
o
) = 60
o
and
? ? BCP = 60
o
(Opposite angle of ?A)
? ? PBC = ? ? BPC = ? ? BCP
Hence, ?PBC is an equilateral triangle and, therefore, PB = PC = BC. ? iii
DC = DP + PC
From ii
, we have:
DC = AD + BC [AD = BC, opposite sides of a parallelogram]
? DC = AD + AD
? DC = 2 AD
Question:15
Page 4
?
?
Question:11
In the adjoining figure, ABCD is a parallelogram in which ?A = 72°. Calculate ?B, ?C and ?D.
Solution:
ABCD is parallelogram and ?A = 72° ?.
We know that opposite angles of a parallelogram are equal.
? ?A ?= ?C and ?B ?= ?D ? ? ? ?C = 72
o
?A and ?B are adajcent angles.
i.e., ?A ?+ ?B ? = 180
o
? ?B = 180
o
- ?A
? ?B ? = 180
o
- 72
o
= 108
o
? ? ?B ? = ? ?D = 108
o
Hence, ?B ? = ? ?D = 108
o
? and ?C ? = 72
o
? Question:12
In the adjoining figure, ABCD is a parallelogram in which ?DAB = 80° and ?DBC = 60°. Calculate ?CDB and
?ADB.
Solution:
Given: ABCD is parallelogram and ?DAB = 80° ? and ?DBC = 60°
To find: Measure of ?CDB and ?ADB
In parallelogram ABCD, AD || ? BC
? ?DBC = ? ADB = 60
o
Alternateinteriorangles ...
i
As ?DAB and ?ADC are adajcent angles, ?DAB ?+ ?ADC ? = 180
o
? ?ADC = 180
o
- ?DAB
? ? ?ADC ? = 180
o
- 80
o
= 100
o
Also, ?ADC ? = ?ADB + ?C ??DB
? ? ?ADC ? = ? 100
o
? ?ADB + ?C ??DB = 100
o
...ii
From i
and ii
, we get:
60
o
+ ?C ??DB = 100
o
? ?C ??DB = 100
o
- 60
o
= 40
o
Hence, ?CDB ? = ? 40
o
and ?ADB ? = 60
o
? Question:13
In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ?BAM = ?DAM. Prove that
AD = 2CD.
Solution:
Given: parallelogram ABCD, M is the midpoint of side BC and ?BAM = ?DAM.
To prove: AD = 2CD
Proof:
Since, AD ? BC
and AM is the transversal.
So, ?DAM = ?AMB
Alternateinteriorangles
But, ?DAM = ?BAM
Given
Therefore, ?AMB = ?BAM
? AB = BM
Anglesoppositetoequalsidesareequal.
...1
Now, AB = CD Oppositesidesofaparallelogramareequal.
? 2AB = 2CD ? (AB +AB) = 2CD
? BM +MC = 2CD
(AB = BM and MC = BM)
? BC = 2CD ? AD = 2CD (AD = BC, Opposite sides of a parallelogram are equal. )
Question:14
In the adjoining figure, ABCD is a parallelogram in which ?A = 60°. If the bisectors of ?A and ?B meet DC at P,
prove that
i ?APB = 90°,
ii AD = DP and PB = PC = BC,
iii DC = 2AD.
Solution:
ABCD is a parallelogram.
? ??A = ??C and ?B = ? ?D Oppositeangles
And ??A + ??B = 180
o
Adjacentanglesaresupplementary
? ??B = 180
o
- ?A
? 180
o
- 60
o
= 120
o
( ? ?A = 60
o
)
? ??A = ??C = 60
o
and ?B = ? ?D ? = 120
o
i
In ? APB, ? ?PAB =
60°
2
= 30°
and ?PBA =
120°
2
= 60°
? ? ? ?APB ? = 180
o
- (30
o
+ 60
o
) = 90
o
ii
In ? ADP, ? ?PAD = 30
o
and ?ADP = 120
o
? ?APB = 180
o
- (30
o
+ 120
o
) = 30
o
Thus, ? ?PAD = ??APB = ?30
o
Hence, ?ADP is an isosceles triangle and AD = DP.
In ? PBC, ? ? PBC = 60
o
, ? ? BPC = 180
o
- (90
o
+30
o
) = 60
o
and
? ? BCP = 60
o
(Opposite angle of ?A)
? ? PBC = ? ? BPC = ? ? BCP
Hence, ?PBC is an equilateral triangle and, therefore, PB = PC = BC. ? iii
DC = DP + PC
From ii
, we have:
DC = AD + BC [AD = BC, opposite sides of a parallelogram]
? DC = AD + AD
? DC = 2 AD
Question:15
In the adjoining figure, ABCD is a parallelogram in which ?BAO = 35°, ?DAO = 40° and ?COD = 105°. Calculate
i ?ABO,
ii ?ODC,
iii ?ACB,
iv ?CBD.
Solution:
ABCD is a parallelogram.
? AB | | ? DC and BC ?| | ? AD
i
In ?AOB, ?BAO = 35°, ??AOB = ?COD = 105° Verticallyoppositeangels
? ??ABO = 180
o
- (35
o
+ 105
o
) = 40
o
ii
?ODC and ?ABO are alternate interior angles.
? ?ODC = ?ABO = 40
o
iii
?ACB = ? ?CAD = 40
o
Alternateinteriorangles
iv
?CBD = ?ABC - ?ABD ...i
?ABC = 180
o
- ?BAD
Adjacentanglesaresupplementary
? ? ?ABC = 180
o
- 75
o
= 105
o
? ?CBD = 105
o
- ?ABD ( ?ABD = ?ABO)
? ?CBD = 105
o
- 40
o
= 65
o
Question:16
In a || gm ABCD, if ?A = (2x + 25)° and ?B = (3x - 5)°, find the value of x and the measure of each angle of the
parallelogram.
Solution:
ABCD is a parallelogram.
i.e., ?A = ?C and ?B = ?D Oppositeangles
Also, ?A + ?B = 180
o
Adjacentanglesaresupplementary
? ? ? (2x + 25)° ? + (3x - 5)° ? = 180
? ?5x +20 = 180
? ? 5x = 160
? ? x = 32
o
? ??A = 2 ? 32 + 25 = 89
o
and ?B = 3 ? 32 - 5 = 91
o
Hence, x = 32
o
, ?A = ?C = 89
o
and ?B = ?D = 91
o
Question:17
If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram.
Page 5
?
?
Question:11
In the adjoining figure, ABCD is a parallelogram in which ?A = 72°. Calculate ?B, ?C and ?D.
Solution:
ABCD is parallelogram and ?A = 72° ?.
We know that opposite angles of a parallelogram are equal.
? ?A ?= ?C and ?B ?= ?D ? ? ? ?C = 72
o
?A and ?B are adajcent angles.
i.e., ?A ?+ ?B ? = 180
o
? ?B = 180
o
- ?A
? ?B ? = 180
o
- 72
o
= 108
o
? ? ?B ? = ? ?D = 108
o
Hence, ?B ? = ? ?D = 108
o
? and ?C ? = 72
o
? Question:12
In the adjoining figure, ABCD is a parallelogram in which ?DAB = 80° and ?DBC = 60°. Calculate ?CDB and
?ADB.
Solution:
Given: ABCD is parallelogram and ?DAB = 80° ? and ?DBC = 60°
To find: Measure of ?CDB and ?ADB
In parallelogram ABCD, AD || ? BC
? ?DBC = ? ADB = 60
o
Alternateinteriorangles ...
i
As ?DAB and ?ADC are adajcent angles, ?DAB ?+ ?ADC ? = 180
o
? ?ADC = 180
o
- ?DAB
? ? ?ADC ? = 180
o
- 80
o
= 100
o
Also, ?ADC ? = ?ADB + ?C ??DB
? ? ?ADC ? = ? 100
o
? ?ADB + ?C ??DB = 100
o
...ii
From i
and ii
, we get:
60
o
+ ?C ??DB = 100
o
? ?C ??DB = 100
o
- 60
o
= 40
o
Hence, ?CDB ? = ? 40
o
and ?ADB ? = 60
o
? Question:13
In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that ?BAM = ?DAM. Prove that
AD = 2CD.
Solution:
Given: parallelogram ABCD, M is the midpoint of side BC and ?BAM = ?DAM.
To prove: AD = 2CD
Proof:
Since, AD ? BC
and AM is the transversal.
So, ?DAM = ?AMB
Alternateinteriorangles
But, ?DAM = ?BAM
Given
Therefore, ?AMB = ?BAM
? AB = BM
Anglesoppositetoequalsidesareequal.
...1
Now, AB = CD Oppositesidesofaparallelogramareequal.
? 2AB = 2CD ? (AB +AB) = 2CD
? BM +MC = 2CD
(AB = BM and MC = BM)
? BC = 2CD ? AD = 2CD (AD = BC, Opposite sides of a parallelogram are equal. )
Question:14
In the adjoining figure, ABCD is a parallelogram in which ?A = 60°. If the bisectors of ?A and ?B meet DC at P,
prove that
i ?APB = 90°,
ii AD = DP and PB = PC = BC,
iii DC = 2AD.
Solution:
ABCD is a parallelogram.
? ??A = ??C and ?B = ? ?D Oppositeangles
And ??A + ??B = 180
o
Adjacentanglesaresupplementary
? ??B = 180
o
- ?A
? 180
o
- 60
o
= 120
o
( ? ?A = 60
o
)
? ??A = ??C = 60
o
and ?B = ? ?D ? = 120
o
i
In ? APB, ? ?PAB =
60°
2
= 30°
and ?PBA =
120°
2
= 60°
? ? ? ?APB ? = 180
o
- (30
o
+ 60
o
) = 90
o
ii
In ? ADP, ? ?PAD = 30
o
and ?ADP = 120
o
? ?APB = 180
o
- (30
o
+ 120
o
) = 30
o
Thus, ? ?PAD = ??APB = ?30
o
Hence, ?ADP is an isosceles triangle and AD = DP.
In ? PBC, ? ? PBC = 60
o
, ? ? BPC = 180
o
- (90
o
+30
o
) = 60
o
and
? ? BCP = 60
o
(Opposite angle of ?A)
? ? PBC = ? ? BPC = ? ? BCP
Hence, ?PBC is an equilateral triangle and, therefore, PB = PC = BC. ? iii
DC = DP + PC
From ii
, we have:
DC = AD + BC [AD = BC, opposite sides of a parallelogram]
? DC = AD + AD
? DC = 2 AD
Question:15
In the adjoining figure, ABCD is a parallelogram in which ?BAO = 35°, ?DAO = 40° and ?COD = 105°. Calculate
i ?ABO,
ii ?ODC,
iii ?ACB,
iv ?CBD.
Solution:
ABCD is a parallelogram.
? AB | | ? DC and BC ?| | ? AD
i
In ?AOB, ?BAO = 35°, ??AOB = ?COD = 105° Verticallyoppositeangels
? ??ABO = 180
o
- (35
o
+ 105
o
) = 40
o
ii
?ODC and ?ABO are alternate interior angles.
? ?ODC = ?ABO = 40
o
iii
?ACB = ? ?CAD = 40
o
Alternateinteriorangles
iv
?CBD = ?ABC - ?ABD ...i
?ABC = 180
o
- ?BAD
Adjacentanglesaresupplementary
? ? ?ABC = 180
o
- 75
o
= 105
o
? ?CBD = 105
o
- ?ABD ( ?ABD = ?ABO)
? ?CBD = 105
o
- 40
o
= 65
o
Question:16
In a || gm ABCD, if ?A = (2x + 25)° and ?B = (3x - 5)°, find the value of x and the measure of each angle of the
parallelogram.
Solution:
ABCD is a parallelogram.
i.e., ?A = ?C and ?B = ?D Oppositeangles
Also, ?A + ?B = 180
o
Adjacentanglesaresupplementary
? ? ? (2x + 25)° ? + (3x - 5)° ? = 180
? ?5x +20 = 180
? ? 5x = 160
? ? x = 32
o
? ??A = 2 ? 32 + 25 = 89
o
and ?B = 3 ? 32 - 5 = 91
o
Hence, x = 32
o
, ?A = ?C = 89
o
and ?B = ?D = 91
o
Question:17
If an angle of a parallelogram is four-fifths of its adjacent angle, find the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram.
? ? ?A = ?C and ?B = ?D Oppositeangles
Let ?A = x
o
and ?B =
4x
5
°
Now, ? ?A + ?B = 180
o
Adjacentanglesaresupplementary
? x +
4x
5
= 180
o
?
9x
5
= 180
o
? x = 100
o
Now, ?A = 100
o
and ?B =
4
5
×100° = 80
o
Hence, ?A = ?C = 100
o
; ?B = ?D = 80
o
Question:18
Find the measure of each angle of a parallelogram, if one of its angles is 30° less than twice the smallest angle.
Solution:
Let ABCD be a parallelogram.
? ? ?A = ? ?C and ?B = ?D Oppositeangles
Let ?A be the smallest angle whose measure is x
o
.
? ? ? ?B = (2x - 30)
o
Now, ? ?A + ?B = 180
o
Adjacentanglesaresupplementary
? x + 2x - 30
o
= 180
o
? 3x = 210
o
? x = 70
o
? ?? ?B = 2 ? 70
o
- 30
o
= 110
o
Hence, ?A = ?C = 70
o
; ?B = ?D = 110
o
Question:19
ABCD is a parallelogram in which AB = 9.5 cm and its perimeter is 30 cm. Find the length of each side of the
parallelogram.
Solution:
ABCD is a parallelogram.
The opposite sides of a parallelogram are parallel and equal.
? AB = DC = 9.5 cm
Let BC = AD = x
? ? Perimeter of ABCD = AB + BC + CD + DA = 30 cm
? 9.5 + x + 9.5 + x = 30
? 19 + 2x = 30
? 2x = 11
? x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm
Question:20
In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.
( )
( )
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FAQs on RS Aggarwal Solutions: Quadrilaterals- 2 - Mathematics (Maths) Class 9
| 1. What are the different types of quadrilaterals? |  |
Ans. There are several types of quadrilaterals, including squares, rectangles, parallelograms, trapeziums, and rhombuses. Each type has its own unique properties and characteristics.
| 2. How do you identify a parallelogram? | |
Ans. A parallelogram is a quadrilateral where both pairs of opposite sides are parallel. Additionally, the opposite sides of a parallelogram are equal in length. To identify a parallelogram, you can check if both pairs of opposite sides are parallel or if the opposite sides are equal in length.
| 3. What is the sum of the interior angles of a quadrilateral? | |
Ans. The sum of the interior angles of any quadrilateral is always equal to 360 degrees. This means that if you measure the angles of a quadrilateral and add them up, the total sum will always be 360 degrees.
| 4. How do you find the area of a rectangle? | |
Ans. To find the area of a rectangle, you multiply the length of one of its sides (usually called the base) by the length of another side (usually called the height). The formula for finding the area of a rectangle is: Area = Length × Width.
| 5. What is the difference between a square and a rhombus? | |
Ans. A square is a type of rhombus where all four sides are equal in length and all four angles are right angles (90 degrees). However, a rhombus is a quadrilateral where all four sides are equal in length, but the angles are not necessarily right angles. In other words, all squares are rhombuses, but not all rhombuses are squares.
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Dec 27, 2024 Last updated |
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,RS Aggarwal Solutions: Quadrilaterals- 2 | Mathematics (Maths) Class 9
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,RS Aggarwal Solutions: Quadrilaterals- 2 | Mathematics (Maths) Class 9
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Additional Information about RS Aggarwal Solutions: Quadrilaterals- 2 for Class 9 Preparation
RS Aggarwal Solutions: Quadrilaterals- 2 Free PDF Download
The RS Aggarwal Solutions: Quadrilaterals- 2 is an invaluable resource that delves deep into the core of the Class 9 exam.
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RS Aggarwal Solutions: Quadrilaterals- 2 Notes
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The content of RS Aggarwal Solutions: Quadrilaterals- 2 is prepared as per the latest Class 9 syllabus.
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