Quadrilaterals - Exercise 10C | Extra Documents & Tests for Class 9 PDF Download

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Question:38
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
i
PQ || AC and PQ =
1
2
AC
ii
PQ || SR
iii
PQRS is a parallelogram.
Solution:
Page 2


 
 
  
     
   ?   
   ?   ? 
 
  
  ?   
  
  
                                   
 
   ?   
   ?          
Question:38
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
i
PQ || AC and PQ =
1
2
AC
ii
PQ || SR
iii
PQRS is a parallelogram.
Solution:
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
i
 PQ || AC and PQ = 
1
2
AC
ii
 PQ || SR
iii
 PQRS is a parallelogram.
Proof:
i
In ? ABC
,
Since, P and Q are the mid points of sides AB and BC, respectively.      Given
? AC ? PQ and PQ =
1
2
AC
            Usingmid -pointtheorem.
ii
In ? ADC
,
Since, S and R are the mid-points of AD and DC, respectively.        Given
? SR ? AC and SR =
1
2
AC
           Usingmid -pointtheorem.
            ...1
From i
and 1
, we get
PQ || SR
iii
From
i and
ii, we get
 
PQ = SR =
1
2
AC
So, PQ and SR are parallel and equal.
Page 3


 
 
  
     
   ?   
   ?   ? 
 
  
  ?   
  
  
                                   
 
   ?   
   ?          
Question:38
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
i
PQ || AC and PQ =
1
2
AC
ii
PQ || SR
iii
PQRS is a parallelogram.
Solution:
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
i
 PQ || AC and PQ = 
1
2
AC
ii
 PQ || SR
iii
 PQRS is a parallelogram.
Proof:
i
In ? ABC
,
Since, P and Q are the mid points of sides AB and BC, respectively.      Given
? AC ? PQ and PQ =
1
2
AC
            Usingmid -pointtheorem.
ii
In ? ADC
,
Since, S and R are the mid-points of AD and DC, respectively.        Given
? SR ? AC and SR =
1
2
AC
           Usingmid -pointtheorem.
            ...1
From i
and 1
, we get
PQ || SR
iii
From
i and
ii, we get
 
PQ = SR =
1
2
AC
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.
Question:39
A square is inscribed in an isosceles  right triangle so that the square and the triangle have one angle common.
Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Solution:
Given: In an isosceles right ?ABC, CMPN is a square.
 
To prove: P bisects the hypotenuse AB i.e., AP = PB.
 
Proof:
In square CMPN,
 
? CM = MP = PN = CN           
Allsidesareequal.
 
Also, ?ABC is an isosceles with AC = BC.
? AN + NC = CM + MB
? AN = MB          ( ? CN = CM)       ...
i
Now,
In ?ANP and ?PMB,
AN = MB               
From(i)
?ANP = ?PMB = 90°
PN = PM             (Sides of square CMPN)
? By SAS congruence criteria,
?ANP ? ?BMP
Hence, AP = PB         
ByCPCT
Question:40
In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a
line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.
Page 4


 
 
  
     
   ?   
   ?   ? 
 
  
  ?   
  
  
                                   
 
   ?   
   ?          
Question:38
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
i
PQ || AC and PQ =
1
2
AC
ii
PQ || SR
iii
PQRS is a parallelogram.
Solution:
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
i
 PQ || AC and PQ = 
1
2
AC
ii
 PQ || SR
iii
 PQRS is a parallelogram.
Proof:
i
In ? ABC
,
Since, P and Q are the mid points of sides AB and BC, respectively.      Given
? AC ? PQ and PQ =
1
2
AC
            Usingmid -pointtheorem.
ii
In ? ADC
,
Since, S and R are the mid-points of AD and DC, respectively.        Given
? SR ? AC and SR =
1
2
AC
           Usingmid -pointtheorem.
            ...1
From i
and 1
, we get
PQ || SR
iii
From
i and
ii, we get
 
PQ = SR =
1
2
AC
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.
Question:39
A square is inscribed in an isosceles  right triangle so that the square and the triangle have one angle common.
Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Solution:
Given: In an isosceles right ?ABC, CMPN is a square.
 
To prove: P bisects the hypotenuse AB i.e., AP = PB.
 
Proof:
In square CMPN,
 
? CM = MP = PN = CN           
Allsidesareequal.
 
Also, ?ABC is an isosceles with AC = BC.
? AN + NC = CM + MB
? AN = MB          ( ? CN = CM)       ...
i
Now,
In ?ANP and ?PMB,
AN = MB               
From(i)
?ANP = ?PMB = 90°
PN = PM             (Sides of square CMPN)
? By SAS congruence criteria,
?ANP ? ?BMP
Hence, AP = PB         
ByCPCT
Question:40
In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a
line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.
Solution:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AD = BC and AB = DC 
AB = AE + BE and DC = DF + FC
? AE = BE = DF = FC   
Now, DF = AE and DF || AE.
i.e., AEFD is a parallelogram ??.
? ? AD || EF ? Similarly, ?BEFC is also a parallelogram. 
? ? EF || BC
? ? AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such
that DF = FC.
These lines AD, EF and BC ? are also cut by the transversal AB at A, E and B, respectively such that ? AE =  BE. ? Similarly, they are also cut by ? GH.
? GP = PH            Byintercepttheorem
 
Question:41
M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point
of intersection O of its diagonals AC and BD. Show that MN is bisected at O.
Solution:
 
Given: A parallelogram ABCD
 
To prove: MN is bisected at O
Proof:
In 
? OAM and 
? OCN,
OA = OC                 (Diagonals of parallelogram bisect each other)
?AOM = ?CON     
Verticallyoppositeangles
?MAO = ?OCN       
Alternateinteriorangles
?
 By ASA congruence criteria,
? OAM ? ? OCN
Page 5


 
 
  
     
   ?   
   ?   ? 
 
  
  ?   
  
  
                                   
 
   ?   
   ?          
Question:38
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
i
PQ || AC and PQ =
1
2
AC
ii
PQ || SR
iii
PQRS is a parallelogram.
Solution:
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
i
 PQ || AC and PQ = 
1
2
AC
ii
 PQ || SR
iii
 PQRS is a parallelogram.
Proof:
i
In ? ABC
,
Since, P and Q are the mid points of sides AB and BC, respectively.      Given
? AC ? PQ and PQ =
1
2
AC
            Usingmid -pointtheorem.
ii
In ? ADC
,
Since, S and R are the mid-points of AD and DC, respectively.        Given
? SR ? AC and SR =
1
2
AC
           Usingmid -pointtheorem.
            ...1
From i
and 1
, we get
PQ || SR
iii
From
i and
ii, we get
 
PQ = SR =
1
2
AC
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.
Question:39
A square is inscribed in an isosceles  right triangle so that the square and the triangle have one angle common.
Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Solution:
Given: In an isosceles right ?ABC, CMPN is a square.
 
To prove: P bisects the hypotenuse AB i.e., AP = PB.
 
Proof:
In square CMPN,
 
? CM = MP = PN = CN           
Allsidesareequal.
 
Also, ?ABC is an isosceles with AC = BC.
? AN + NC = CM + MB
? AN = MB          ( ? CN = CM)       ...
i
Now,
In ?ANP and ?PMB,
AN = MB               
From(i)
?ANP = ?PMB = 90°
PN = PM             (Sides of square CMPN)
? By SAS congruence criteria,
?ANP ? ?BMP
Hence, AP = PB         
ByCPCT
Question:40
In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a
line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.
Solution:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AD = BC and AB = DC 
AB = AE + BE and DC = DF + FC
? AE = BE = DF = FC   
Now, DF = AE and DF || AE.
i.e., AEFD is a parallelogram ??.
? ? AD || EF ? Similarly, ?BEFC is also a parallelogram. 
? ? EF || BC
? ? AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such
that DF = FC.
These lines AD, EF and BC ? are also cut by the transversal AB at A, E and B, respectively such that ? AE =  BE. ? Similarly, they are also cut by ? GH.
? GP = PH            Byintercepttheorem
 
Question:41
M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point
of intersection O of its diagonals AC and BD. Show that MN is bisected at O.
Solution:
 
Given: A parallelogram ABCD
 
To prove: MN is bisected at O
Proof:
In 
? OAM and 
? OCN,
OA = OC                 (Diagonals of parallelogram bisect each other)
?AOM = ?CON     
Verticallyoppositeangles
?MAO = ?OCN       
Alternateinteriorangles
?
 By ASA congruence criteria,
? OAM ? ? OCN
? OM = ON  
CPCT
Hence, MN is bisected at O.
Question:42
In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN ||
PQ meets QR at N. Show that N is the midpoint of QR.
Solution:
 
Given: In trapezium PQRS, PQ || SR, M is the midpoint of PS and MN || PQ.
To prove: N is the midpoint of QR.
Construction: Join QS.
Proof:
In ?SPQ,
Since, M is the mid-point of SP and MO || PQ.
Therefore, O is the mid-point of SQ.     
ByMid -pointtheorem
Similarly, in ?SRQ,
Since, O is the mid-point of SQ and ON || SR           (SR || PQ and MN || PQ)
Therefore, N is the mid-point of QR.     
ByMid -pointtheorem
Question:43
In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ?P meets SR in M. PM and QR both when
produced meet at T. Find the length of RT.
Solution: