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 Page 1


Question:1
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
Solution:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
75°+90°+75°+x = 360° ? 240°+x = 360°
? x = 360°-240° ? x = 120°
So, the measure of  the fourth angle is 120°
.
Question:2
The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.
Solution:
Let ?
A = 2x
°
?.
Then ?
B = (4x)
°
; ?
C = (5x)
°
 and ?
D = (7x)
°
Since the sum of the angles of a quadrilateral is 360
o
, we have:
2x + 4x + 5x + 7x = 360
°
   
? 18 x = 360
°
 ? ? x = 20
°
?  ?
A = 40
°
; ?
B = 80
°
; ?
C = 100
°
; ?
D = 140
°
Question:3
In the adjoining figure, ABCD is a trapezium in which AB || DC. If ?A = 55° and ?B = 70°, find ?C and ?D.
Solution:
  We have AB || DC.
 
? A  and  
? D are the interior angles on the same side of transversal line AD, whereas
Page 2


Question:1
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
Solution:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
75°+90°+75°+x = 360° ? 240°+x = 360°
? x = 360°-240° ? x = 120°
So, the measure of  the fourth angle is 120°
.
Question:2
The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.
Solution:
Let ?
A = 2x
°
?.
Then ?
B = (4x)
°
; ?
C = (5x)
°
 and ?
D = (7x)
°
Since the sum of the angles of a quadrilateral is 360
o
, we have:
2x + 4x + 5x + 7x = 360
°
   
? 18 x = 360
°
 ? ? x = 20
°
?  ?
A = 40
°
; ?
B = 80
°
; ?
C = 100
°
; ?
D = 140
°
Question:3
In the adjoining figure, ABCD is a trapezium in which AB || DC. If ?A = 55° and ?B = 70°, find ?C and ?D.
Solution:
  We have AB || DC.
 
? A  and  
? D are the interior angles on the same side of transversal line AD, whereas
? B and  
? C are the interior angles on the same side of transversal line BC. 
Now, 
?A +
?D = 180
°
?  
?D = 180
°
 -
?A   
?
? D = 180
°
 - 55
°
 = 125
°
Again ,
? B +
?C = 180
°
?  
?C  = 180
° 
-
?B  
?
? C = 180
°
 -  70
°
 = 110
°
 
Question:4
In the adjoining figure, ABCD is a square and ?EDC is an equilateral triangle. Prove that
i AE = BE,
ii ?DAE = 15°.
Solution:
Given:  ABCD is a square in which AB = BC = CD = DA. ?EDC is an equilateral triangle in which ED = EC = DC and
 
? EDC  =   
? DEC = 
? ?DCE =  60
°
.
To prove:  AE = BE and 
?DAE = 15
°
? Proof: In ?ADE and ?BCE, we have:
AD = BC             
Sidesofasquare
DE = EC ?            
Sidesofanequilateraltriangle
?ADE  = 
?BCE = 90
° 
+  60
°
 = 150
°
? 
? ?ADE ?  ?BCE
i.e., AE =  BE  
Now, 
?ADE = 150
°
Page 3


Question:1
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
Solution:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
75°+90°+75°+x = 360° ? 240°+x = 360°
? x = 360°-240° ? x = 120°
So, the measure of  the fourth angle is 120°
.
Question:2
The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.
Solution:
Let ?
A = 2x
°
?.
Then ?
B = (4x)
°
; ?
C = (5x)
°
 and ?
D = (7x)
°
Since the sum of the angles of a quadrilateral is 360
o
, we have:
2x + 4x + 5x + 7x = 360
°
   
? 18 x = 360
°
 ? ? x = 20
°
?  ?
A = 40
°
; ?
B = 80
°
; ?
C = 100
°
; ?
D = 140
°
Question:3
In the adjoining figure, ABCD is a trapezium in which AB || DC. If ?A = 55° and ?B = 70°, find ?C and ?D.
Solution:
  We have AB || DC.
 
? A  and  
? D are the interior angles on the same side of transversal line AD, whereas
? B and  
? C are the interior angles on the same side of transversal line BC. 
Now, 
?A +
?D = 180
°
?  
?D = 180
°
 -
?A   
?
? D = 180
°
 - 55
°
 = 125
°
Again ,
? B +
?C = 180
°
?  
?C  = 180
° 
-
?B  
?
? C = 180
°
 -  70
°
 = 110
°
 
Question:4
In the adjoining figure, ABCD is a square and ?EDC is an equilateral triangle. Prove that
i AE = BE,
ii ?DAE = 15°.
Solution:
Given:  ABCD is a square in which AB = BC = CD = DA. ?EDC is an equilateral triangle in which ED = EC = DC and
 
? EDC  =   
? DEC = 
? ?DCE =  60
°
.
To prove:  AE = BE and 
?DAE = 15
°
? Proof: In ?ADE and ?BCE, we have:
AD = BC             
Sidesofasquare
DE = EC ?            
Sidesofanequilateraltriangle
?ADE  = 
?BCE = 90
° 
+  60
°
 = 150
°
? 
? ?ADE ?  ?BCE
i.e., AE =  BE  
Now, 
?ADE = 150
°
DA = DC    
Sidesofasquare
DC = DE     
Sidesofanequilateraltriangle
So, DA = DE
?ADE and ?BCE are isosceles triangles.
i.e.,
?DAE = 
?DEA = 
1
2
(180° - 150°) = 
30°
2
 = 15°
 
Question:5
In the adjoining figure, BM ? AC and DN ? AC. If BM = DN, prove that AC bisects BD.
Solution:
Given: A quadrilateral ABCD, in which BM ?? AC and DN ? AC and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in ?OND and ?OMB, we have:
?OND = ?OMB                 (90
o
 each)
?DON = ? BOM                 
Verticallyoppositeangles
Also, DN = BM                        
Given
i.e., ?OND ? ?OMB             AAScongurencerule
? OD = OB                          
CPCT
?Hence, AC bisects BD.
Question:6
In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
i
AC bisects ?A and ?C,
ii BE = DE,
iii ?ABC = ?ADC.
Page 4


Question:1
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
Solution:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
75°+90°+75°+x = 360° ? 240°+x = 360°
? x = 360°-240° ? x = 120°
So, the measure of  the fourth angle is 120°
.
Question:2
The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.
Solution:
Let ?
A = 2x
°
?.
Then ?
B = (4x)
°
; ?
C = (5x)
°
 and ?
D = (7x)
°
Since the sum of the angles of a quadrilateral is 360
o
, we have:
2x + 4x + 5x + 7x = 360
°
   
? 18 x = 360
°
 ? ? x = 20
°
?  ?
A = 40
°
; ?
B = 80
°
; ?
C = 100
°
; ?
D = 140
°
Question:3
In the adjoining figure, ABCD is a trapezium in which AB || DC. If ?A = 55° and ?B = 70°, find ?C and ?D.
Solution:
  We have AB || DC.
 
? A  and  
? D are the interior angles on the same side of transversal line AD, whereas
? B and  
? C are the interior angles on the same side of transversal line BC. 
Now, 
?A +
?D = 180
°
?  
?D = 180
°
 -
?A   
?
? D = 180
°
 - 55
°
 = 125
°
Again ,
? B +
?C = 180
°
?  
?C  = 180
° 
-
?B  
?
? C = 180
°
 -  70
°
 = 110
°
 
Question:4
In the adjoining figure, ABCD is a square and ?EDC is an equilateral triangle. Prove that
i AE = BE,
ii ?DAE = 15°.
Solution:
Given:  ABCD is a square in which AB = BC = CD = DA. ?EDC is an equilateral triangle in which ED = EC = DC and
 
? EDC  =   
? DEC = 
? ?DCE =  60
°
.
To prove:  AE = BE and 
?DAE = 15
°
? Proof: In ?ADE and ?BCE, we have:
AD = BC             
Sidesofasquare
DE = EC ?            
Sidesofanequilateraltriangle
?ADE  = 
?BCE = 90
° 
+  60
°
 = 150
°
? 
? ?ADE ?  ?BCE
i.e., AE =  BE  
Now, 
?ADE = 150
°
DA = DC    
Sidesofasquare
DC = DE     
Sidesofanequilateraltriangle
So, DA = DE
?ADE and ?BCE are isosceles triangles.
i.e.,
?DAE = 
?DEA = 
1
2
(180° - 150°) = 
30°
2
 = 15°
 
Question:5
In the adjoining figure, BM ? AC and DN ? AC. If BM = DN, prove that AC bisects BD.
Solution:
Given: A quadrilateral ABCD, in which BM ?? AC and DN ? AC and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in ?OND and ?OMB, we have:
?OND = ?OMB                 (90
o
 each)
?DON = ? BOM                 
Verticallyoppositeangles
Also, DN = BM                        
Given
i.e., ?OND ? ?OMB             AAScongurencerule
? OD = OB                          
CPCT
?Hence, AC bisects BD.
Question:6
In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
i
AC bisects ?A and ?C,
ii BE = DE,
iii ?ABC = ?ADC.
Solution:
Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
i
In ?ABC and ?ADC, we have:
AB = AD                                                  Given
BC = DC                                                 Given
AC is common.
i.e., ?ABC ? ?ADC                                    SSScongruencerule
? ?BAC = ?DAC and ?BCA = ?D ?CA        ByCPCT
Thus, AC bisects ?A and ? C.
ii
Now, in ?ABE and ?ADE, we have:
  AB = AD                                      Given
?BAE = ?DAE ?                               Provenabove
 AE is common.
? ?ABE ?  ?ADE                          SAScongruencerule
? BE = DE                                                
ByCPCT
 
iii   ?ABC ?  ?ADC                  Provenabove
? ?ABC = ?AD ?C                           ByCPCT
 
Question:7
In the given figure, ABCD is a square and ?PQR = 90°. If PB = QC = DR, prove that
i QB = RC,
ii PQ = QR,
iii QPR = 45°.
Solution:
Given: ABCD is a square and ?PQR = 90°.
Also, PB = QC = DR
i
We have:
  BC = CD                Sidesofsquare
  CQ = DR                   Given
  BC = BQ + CQ
? CQ = BC - BQ
? DR = BC - BQ           ...i
   
Page 5


Question:1
Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.
Solution:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
75°+90°+75°+x = 360° ? 240°+x = 360°
? x = 360°-240° ? x = 120°
So, the measure of  the fourth angle is 120°
.
Question:2
The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.
Solution:
Let ?
A = 2x
°
?.
Then ?
B = (4x)
°
; ?
C = (5x)
°
 and ?
D = (7x)
°
Since the sum of the angles of a quadrilateral is 360
o
, we have:
2x + 4x + 5x + 7x = 360
°
   
? 18 x = 360
°
 ? ? x = 20
°
?  ?
A = 40
°
; ?
B = 80
°
; ?
C = 100
°
; ?
D = 140
°
Question:3
In the adjoining figure, ABCD is a trapezium in which AB || DC. If ?A = 55° and ?B = 70°, find ?C and ?D.
Solution:
  We have AB || DC.
 
? A  and  
? D are the interior angles on the same side of transversal line AD, whereas
? B and  
? C are the interior angles on the same side of transversal line BC. 
Now, 
?A +
?D = 180
°
?  
?D = 180
°
 -
?A   
?
? D = 180
°
 - 55
°
 = 125
°
Again ,
? B +
?C = 180
°
?  
?C  = 180
° 
-
?B  
?
? C = 180
°
 -  70
°
 = 110
°
 
Question:4
In the adjoining figure, ABCD is a square and ?EDC is an equilateral triangle. Prove that
i AE = BE,
ii ?DAE = 15°.
Solution:
Given:  ABCD is a square in which AB = BC = CD = DA. ?EDC is an equilateral triangle in which ED = EC = DC and
 
? EDC  =   
? DEC = 
? ?DCE =  60
°
.
To prove:  AE = BE and 
?DAE = 15
°
? Proof: In ?ADE and ?BCE, we have:
AD = BC             
Sidesofasquare
DE = EC ?            
Sidesofanequilateraltriangle
?ADE  = 
?BCE = 90
° 
+  60
°
 = 150
°
? 
? ?ADE ?  ?BCE
i.e., AE =  BE  
Now, 
?ADE = 150
°
DA = DC    
Sidesofasquare
DC = DE     
Sidesofanequilateraltriangle
So, DA = DE
?ADE and ?BCE are isosceles triangles.
i.e.,
?DAE = 
?DEA = 
1
2
(180° - 150°) = 
30°
2
 = 15°
 
Question:5
In the adjoining figure, BM ? AC and DN ? AC. If BM = DN, prove that AC bisects BD.
Solution:
Given: A quadrilateral ABCD, in which BM ?? AC and DN ? AC and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in ?OND and ?OMB, we have:
?OND = ?OMB                 (90
o
 each)
?DON = ? BOM                 
Verticallyoppositeangles
Also, DN = BM                        
Given
i.e., ?OND ? ?OMB             AAScongurencerule
? OD = OB                          
CPCT
?Hence, AC bisects BD.
Question:6
In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
i
AC bisects ?A and ?C,
ii BE = DE,
iii ?ABC = ?ADC.
Solution:
Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
i
In ?ABC and ?ADC, we have:
AB = AD                                                  Given
BC = DC                                                 Given
AC is common.
i.e., ?ABC ? ?ADC                                    SSScongruencerule
? ?BAC = ?DAC and ?BCA = ?D ?CA        ByCPCT
Thus, AC bisects ?A and ? C.
ii
Now, in ?ABE and ?ADE, we have:
  AB = AD                                      Given
?BAE = ?DAE ?                               Provenabove
 AE is common.
? ?ABE ?  ?ADE                          SAScongruencerule
? BE = DE                                                
ByCPCT
 
iii   ?ABC ?  ?ADC                  Provenabove
? ?ABC = ?AD ?C                           ByCPCT
 
Question:7
In the given figure, ABCD is a square and ?PQR = 90°. If PB = QC = DR, prove that
i QB = RC,
ii PQ = QR,
iii QPR = 45°.
Solution:
Given: ABCD is a square and ?PQR = 90°.
Also, PB = QC = DR
i
We have:
  BC = CD                Sidesofsquare
  CQ = DR                   Given
  BC = BQ + CQ
? CQ = BC - BQ
? DR = BC - BQ           ...i
   
Also, CD = RC+ DR
? DR = CD -  RC = BC - RC            ...ii
From i
and ii
, we have:
BC - BQ = ?BC - RC
? BQ = RC
ii
In ?RCQ and ?QBP, we have:
PB = QC   Given
BQ = RC  Provenabove
?RCQ = ?QBP   (90
o
 each)
i.e., ?RCQ ? ?QBP       SAScongruencerule
? QR =  PQ                ByCPCT
iii
?RCQ ? ?QBP and QR = PQ Provenabove
? In ?RPQ, ?QPR = ?QRP = 
1
2
(180° - 90°) = 
90°
2
 = 45
°
Question:8
If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.
Solution:
Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral. 
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ?AOC, OA + OC > AC
Also, in ? BOD, OB + OD > BD
Adding these inequalities, we get:
(OA + OC) + (OB + OD) > (AC + BD)
? OA + OB + OC + OD > AC + BD
 
Question:9
In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
i
AB + BC + CD + DA > 2AC
ii
AB + BC + CD > DA
iii
AB + BC + CD + DA > AC + BD

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FAQs on RS Aggarwal Solutions: Quadrilaterals- 1 - Mathematics (Maths) Class 9

1. What are quadrilaterals?

Ans. Quadrilaterals are polygons with four sides and four angles. They can have different types of angles and side lengths, such as squares, rectangles, parallelograms, trapezoids, and rhombuses.

2. How do you classify quadrilaterals based on their angles?

Ans. Quadrilaterals can be classified into three main categories based on their angles: 1) Right angles: Quadrilaterals with all angles measuring 90 degrees, such as squares and rectangles. 2) Obtuse angles: Quadrilaterals with one angle measuring more than 90 degrees, such as parallelograms. 3) Acute angles: Quadrilaterals with all angles measuring less than 90 degrees, such as trapezoids and rhombuses.

3. How do you classify quadrilaterals based on their side lengths?

Ans. Based on their side lengths, quadrilaterals can be classified into different types: 1) Parallelograms: Opposite sides are parallel and equal in length. 2) Rhombuses: All sides are equal in length. 3) Rectangles: All angles are right angles and opposite sides are equal in length. 4) Squares: All angles are right angles and all sides are equal in length. 5) Trapezoids: Only one pair of opposite sides is parallel.

4. How do you find the area of a quadrilateral?

Ans. The area of a quadrilateral can be calculated differently depending on its type. For rectangles and squares, the area is calculated by multiplying the length and width. For parallelograms, the area is calculated by multiplying the base and height. For rhombuses, the area is calculated by multiplying the diagonals and dividing by 2. For trapezoids, the area is calculated by multiplying the sum of the parallel sides by the height and dividing by 2.

5. How do you solve problems involving quadrilaterals?

Ans. To solve problems involving quadrilaterals, it is important to understand their properties and formulas. Identify the given information, determine the type of quadrilateral, and apply the appropriate formulas to find the required measurements or solve the problem. It is also helpful to draw accurate diagrams and use the given information to set up equations or apply geometric principles. Practice solving different types of problems to improve problem-solving skills related to quadrilaterals.

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