Congruence (Exercise 16.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download

 Page 1


  
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution: 
(i) We have,
Since ?ABO = ?CDO = 45
o
 and both are alternate angles, AB parallel to DC, ?BAO =
?DCO (alternate angle, AB parallel to CD and AC is a transversal line)
?ABO = ?CDO = 45
o
 (given in the figure) Also,
AB = DC (Given in the figure)
Page 2


  
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution: 
(i) We have,
Since ?ABO = ?CDO = 45
o
 and both are alternate angles, AB parallel to DC, ?BAO =
?DCO (alternate angle, AB parallel to CD and AC is a transversal line)
?ABO = ?CDO = 45
o
 (given in the figure) Also,
AB = DC (Given in the figure)
  
Therefore, by ASA ?AOB ? ?DOC 
(ii) In ABC,
Now AB =AC (Given)
?ABD = ?ACD = 40
o
 (Angles opposite to equal sides)
?ABD + ?ACD + ?BAC = 180
o
 (Angle sum property)
40
o
 + 40
o
 + ?BAC = 180
o
?BAC =180
o
 - 80
o
 =100
o
?BAD + ?DAC = ?BAC
?BAD = ?BAC – ?DAC = 100
o
 – 50
o
 = 50
o
?BAD = ?CAD = 50°
Therefore, by ASA, ?ABD ? ?ACD
(iii) In ? ABC,
?A + ?B + ?C = 180
o 
(Angle sum property)
?C = 180
o
- ?A – ?B
?C = 180
o
 – 30
o
 – 90
o
 = 60
o
In PQR,
?P + ?Q + ?R = 180
o
 (Angle sum property)
?P = 180
o
 – ?R – ?Q
?P = 180
o 
– 60
o 
– 90
o
 = 30
o
?BAC = ?QPR = 30
o 
?BCA = ?PRQ = 60
o
 and AC = PR (Given)
Therefore, by ASA, ?ABC ? ?PQR
(iv) We have only
BC = QR but none of the angles of ?ABC and ?PQR are equal.
Therefore, ?ABC is not congruent to ?PQR
2. In fig. 37, AD bisects A and AD ? BC.
(i) Is ?ADB ? ?ADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
Page 3


  
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution: 
(i) We have,
Since ?ABO = ?CDO = 45
o
 and both are alternate angles, AB parallel to DC, ?BAO =
?DCO (alternate angle, AB parallel to CD and AC is a transversal line)
?ABO = ?CDO = 45
o
 (given in the figure) Also,
AB = DC (Given in the figure)
  
Therefore, by ASA ?AOB ? ?DOC 
(ii) In ABC,
Now AB =AC (Given)
?ABD = ?ACD = 40
o
 (Angles opposite to equal sides)
?ABD + ?ACD + ?BAC = 180
o
 (Angle sum property)
40
o
 + 40
o
 + ?BAC = 180
o
?BAC =180
o
 - 80
o
 =100
o
?BAD + ?DAC = ?BAC
?BAD = ?BAC – ?DAC = 100
o
 – 50
o
 = 50
o
?BAD = ?CAD = 50°
Therefore, by ASA, ?ABD ? ?ACD
(iii) In ? ABC,
?A + ?B + ?C = 180
o 
(Angle sum property)
?C = 180
o
- ?A – ?B
?C = 180
o
 – 30
o
 – 90
o
 = 60
o
In PQR,
?P + ?Q + ?R = 180
o
 (Angle sum property)
?P = 180
o
 – ?R – ?Q
?P = 180
o 
– 60
o 
– 90
o
 = 30
o
?BAC = ?QPR = 30
o 
?BCA = ?PRQ = 60
o
 and AC = PR (Given)
Therefore, by ASA, ?ABC ? ?PQR
(iv) We have only
BC = QR but none of the angles of ?ABC and ?PQR are equal.
Therefore, ?ABC is not congruent to ?PQR
2. In fig. 37, AD bisects A and AD ? BC.
(i) Is ?ADB ? ?ADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
  
Solution: 
(i) Yes, ?ADB??ADC, by ASA criterion of congruency.
(ii) We have used ?BAD = ?CAD ?ADB = ?ADC = 90
o
Since, AD ? BC and AD = DA
ADB = ADC
(iii) Yes, BD = DC since, ?ADB ? ?ADC
3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to
it.
Solution: 
We have drawn 
? ABC with ?ABC = 65
o
 and ?ACB = 70
o
We now construct ?PQR ? ?ABC where ?PQR = 65
o
 and ?PRQ = 70
o
Also we construct ?PQR such that BC = QR 
Therefore by ASA the two triangles are congruent 
Page 4


  
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution: 
(i) We have,
Since ?ABO = ?CDO = 45
o
 and both are alternate angles, AB parallel to DC, ?BAO =
?DCO (alternate angle, AB parallel to CD and AC is a transversal line)
?ABO = ?CDO = 45
o
 (given in the figure) Also,
AB = DC (Given in the figure)
  
Therefore, by ASA ?AOB ? ?DOC 
(ii) In ABC,
Now AB =AC (Given)
?ABD = ?ACD = 40
o
 (Angles opposite to equal sides)
?ABD + ?ACD + ?BAC = 180
o
 (Angle sum property)
40
o
 + 40
o
 + ?BAC = 180
o
?BAC =180
o
 - 80
o
 =100
o
?BAD + ?DAC = ?BAC
?BAD = ?BAC – ?DAC = 100
o
 – 50
o
 = 50
o
?BAD = ?CAD = 50°
Therefore, by ASA, ?ABD ? ?ACD
(iii) In ? ABC,
?A + ?B + ?C = 180
o 
(Angle sum property)
?C = 180
o
- ?A – ?B
?C = 180
o
 – 30
o
 – 90
o
 = 60
o
In PQR,
?P + ?Q + ?R = 180
o
 (Angle sum property)
?P = 180
o
 – ?R – ?Q
?P = 180
o 
– 60
o 
– 90
o
 = 30
o
?BAC = ?QPR = 30
o 
?BCA = ?PRQ = 60
o
 and AC = PR (Given)
Therefore, by ASA, ?ABC ? ?PQR
(iv) We have only
BC = QR but none of the angles of ?ABC and ?PQR are equal.
Therefore, ?ABC is not congruent to ?PQR
2. In fig. 37, AD bisects A and AD ? BC.
(i) Is ?ADB ? ?ADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
  
Solution: 
(i) Yes, ?ADB??ADC, by ASA criterion of congruency.
(ii) We have used ?BAD = ?CAD ?ADB = ?ADC = 90
o
Since, AD ? BC and AD = DA
ADB = ADC
(iii) Yes, BD = DC since, ?ADB ? ?ADC
3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to
it.
Solution: 
We have drawn 
? ABC with ?ABC = 65
o
 and ?ACB = 70
o
We now construct ?PQR ? ?ABC where ?PQR = 65
o
 and ?PRQ = 70
o
Also we construct ?PQR such that BC = QR 
Therefore by ASA the two triangles are congruent 
 
 
 
 
  
 
 
 
4. In ? ABC, it is known that ?B = C. Imagine you have another copy of ? ABC 
 
(i) Is ?ABC ? ?ACB 
(ii) State the three pairs of matching parts you have used to answer (i). 
(iii) Is it true to say that AB = AC? 
 
Solution:  
 
(i) Yes ?ABC ? ?ACB 
 
(ii) We have used ?ABC = ?ACB and ?ACB = ?ABC again. 
Also BC = CB 
 
(iii) Yes it is true to say that AB = AC since ?ABC = ?ACB. 
 
5. In Fig. 38, AX bisects ?BAC as well as ?BDC. State the three facts needed to ensure 
that ?ACD ? ?ABD  
 
Solution: 
As per the given conditions,  
?CAD = ?BAD and ?CDA = ?BDA (because AX bisects ?BAC) 
Page 5


  
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution: 
(i) We have,
Since ?ABO = ?CDO = 45
o
 and both are alternate angles, AB parallel to DC, ?BAO =
?DCO (alternate angle, AB parallel to CD and AC is a transversal line)
?ABO = ?CDO = 45
o
 (given in the figure) Also,
AB = DC (Given in the figure)
  
Therefore, by ASA ?AOB ? ?DOC 
(ii) In ABC,
Now AB =AC (Given)
?ABD = ?ACD = 40
o
 (Angles opposite to equal sides)
?ABD + ?ACD + ?BAC = 180
o
 (Angle sum property)
40
o
 + 40
o
 + ?BAC = 180
o
?BAC =180
o
 - 80
o
 =100
o
?BAD + ?DAC = ?BAC
?BAD = ?BAC – ?DAC = 100
o
 – 50
o
 = 50
o
?BAD = ?CAD = 50°
Therefore, by ASA, ?ABD ? ?ACD
(iii) In ? ABC,
?A + ?B + ?C = 180
o 
(Angle sum property)
?C = 180
o
- ?A – ?B
?C = 180
o
 – 30
o
 – 90
o
 = 60
o
In PQR,
?P + ?Q + ?R = 180
o
 (Angle sum property)
?P = 180
o
 – ?R – ?Q
?P = 180
o 
– 60
o 
– 90
o
 = 30
o
?BAC = ?QPR = 30
o 
?BCA = ?PRQ = 60
o
 and AC = PR (Given)
Therefore, by ASA, ?ABC ? ?PQR
(iv) We have only
BC = QR but none of the angles of ?ABC and ?PQR are equal.
Therefore, ?ABC is not congruent to ?PQR
2. In fig. 37, AD bisects A and AD ? BC.
(i) Is ?ADB ? ?ADC?
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
  
Solution: 
(i) Yes, ?ADB??ADC, by ASA criterion of congruency.
(ii) We have used ?BAD = ?CAD ?ADB = ?ADC = 90
o
Since, AD ? BC and AD = DA
ADB = ADC
(iii) Yes, BD = DC since, ?ADB ? ?ADC
3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to
it.
Solution: 
We have drawn 
? ABC with ?ABC = 65
o
 and ?ACB = 70
o
We now construct ?PQR ? ?ABC where ?PQR = 65
o
 and ?PRQ = 70
o
Also we construct ?PQR such that BC = QR 
Therefore by ASA the two triangles are congruent 
 
 
 
 
  
 
 
 
4. In ? ABC, it is known that ?B = C. Imagine you have another copy of ? ABC 
 
(i) Is ?ABC ? ?ACB 
(ii) State the three pairs of matching parts you have used to answer (i). 
(iii) Is it true to say that AB = AC? 
 
Solution:  
 
(i) Yes ?ABC ? ?ACB 
 
(ii) We have used ?ABC = ?ACB and ?ACB = ?ABC again. 
Also BC = CB 
 
(iii) Yes it is true to say that AB = AC since ?ABC = ?ACB. 
 
5. In Fig. 38, AX bisects ?BAC as well as ?BDC. State the three facts needed to ensure 
that ?ACD ? ?ABD  
 
Solution: 
As per the given conditions,  
?CAD = ?BAD and ?CDA = ?BDA (because AX bisects ?BAC) 
  
AD = DA (common) 
Therefore, by ASA, ?ACD ? ?ABD 
6. In Fig. 39, AO = OB and ?A = ?B.
(i) Is ?AOC ? ?BOD
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ?ACO = ?BDO?
Solution : 
We have 
?OAC = ?OBD, 
AO = OB 
Also, ?AOC = ?BOD (Opposite angles on same vertex) 
Therefore, by ASA ?AOC ? ?BOD
(ii) AC || BD   (by C.P.C.T)
(iii)  ?ACO = ?BDO (by C.P.C.T)