Congruence (Exercise 16.5) RD Sharma Solutions | Mathematics (Maths) Class 7 (Old NCERT) PDF Download

 Page 1


 
 
 
 
  
 
 
 
        
 
1. In each of the following pairs of right triangles, the measures of some parts are 
indicated alongside. State by the application of RHS congruence condition which are 
congruent, and also state each result in symbolic form. (Fig. 46) 
 
 
 
Page 2


 
 
 
 
  
 
 
 
        
 
1. In each of the following pairs of right triangles, the measures of some parts are 
indicated alongside. State by the application of RHS congruence condition which are 
congruent, and also state each result in symbolic form. (Fig. 46) 
 
 
 
 
 
 
 
  
 
 
 
 
Solution: 
(i) ?ADB = ?BCA = 90
o 
AD = BC and hypotenuse AB = hypotenuse AB 
Therefore, by RHS ?ADB ? ?ACB 
 
(ii) AD = AD (Common) 
Hypotenuse AC = hypotenuse AB (Given) 
?ADB + ?ADC = 180
o
 (Linear pair) 
?ADB + 90
o
 = 180
o 
?ADB = 180
o
 – 90
o
 = 90
o 
?ADB = ?ADC = 90
o 
Therefore, by RHS ? ADB ? ? ADC 
 
(iii) Hypotenuse AO = hypotenuse DO 
BO = CO  
?B = ?C = 90
o 
Therefore, by RHS, ?AOB??DOC 
 
(iv) Hypotenuse AC = Hypotenuse CA 
BC = DC  
?ABC = ?ADC = 90
o 
Therefore, by RHS, ?ABC ? ?ADC 
 
(v) BD = DB  
Page 3


 
 
 
 
  
 
 
 
        
 
1. In each of the following pairs of right triangles, the measures of some parts are 
indicated alongside. State by the application of RHS congruence condition which are 
congruent, and also state each result in symbolic form. (Fig. 46) 
 
 
 
 
 
 
 
  
 
 
 
 
Solution: 
(i) ?ADB = ?BCA = 90
o 
AD = BC and hypotenuse AB = hypotenuse AB 
Therefore, by RHS ?ADB ? ?ACB 
 
(ii) AD = AD (Common) 
Hypotenuse AC = hypotenuse AB (Given) 
?ADB + ?ADC = 180
o
 (Linear pair) 
?ADB + 90
o
 = 180
o 
?ADB = 180
o
 – 90
o
 = 90
o 
?ADB = ?ADC = 90
o 
Therefore, by RHS ? ADB ? ? ADC 
 
(iii) Hypotenuse AO = hypotenuse DO 
BO = CO  
?B = ?C = 90
o 
Therefore, by RHS, ?AOB??DOC 
 
(iv) Hypotenuse AC = Hypotenuse CA 
BC = DC  
?ABC = ?ADC = 90
o 
Therefore, by RHS, ?ABC ? ?ADC 
 
(v) BD = DB  
 
 
 
 
  
 
 
 
Hypotenuse AB = Hypotenuse BC, as per the given figure, 
?BDA + ?BDC = 180
o 
?BDA + 90
o
 = 180
o 
?BDA= 180
o 
– 90
o
 = 90
o 
?BDA = ?BDC = 90
o 
Therefore, by RHS, ?ABD ? ?CBD 
 
2. ? ABC is isosceles with AB = AC. AD is the altitude from A on BC. 
(i) Is ?ABD ? ?ACD? 
(ii) State the pairs of matching parts you have used to answer (i). 
(iii) Is it true to say that BD = DC? 
 
Solution: 
(i) Yes, ?ABD ? ?ACD by RHS congruence condition. 
 
(ii) We have used Hypotenuse AB = Hypotenuse AC 
AD = DA 
?ADB = ?ADC = 90
o
 (AD ? BC at point D) 
 
(iii)Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)  
Since we have already proved that the two triangles are congruent. 
  
3. ?ABC is isosceles with AB = AC. Also. AD ? BC meeting BC in D. Are the two triangles 
ABD and ACD congruent? State in symbolic form. Which congruence condition do you 
use? Which side of ADC equals BD? Which angle of ? ADC equals ?B? 
 
Solution: 
We have AB = AC …… (i) 
AD = DA (common) …… (ii) 
And, ?ADC = ?ADB (AD ? BC at point D) …… (iii) 
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ?ABD ? ?ACD, the 
triangles are congruent. 
Therefore, BD = CD. 
And ?ABD = ?ACD (corresponding parts of congruent triangles) 
 
4. Draw a right triangle ABC. Use RHS condition to construct another triangle 
congruent to it. 
Page 4


 
 
 
 
  
 
 
 
        
 
1. In each of the following pairs of right triangles, the measures of some parts are 
indicated alongside. State by the application of RHS congruence condition which are 
congruent, and also state each result in symbolic form. (Fig. 46) 
 
 
 
 
 
 
 
  
 
 
 
 
Solution: 
(i) ?ADB = ?BCA = 90
o 
AD = BC and hypotenuse AB = hypotenuse AB 
Therefore, by RHS ?ADB ? ?ACB 
 
(ii) AD = AD (Common) 
Hypotenuse AC = hypotenuse AB (Given) 
?ADB + ?ADC = 180
o
 (Linear pair) 
?ADB + 90
o
 = 180
o 
?ADB = 180
o
 – 90
o
 = 90
o 
?ADB = ?ADC = 90
o 
Therefore, by RHS ? ADB ? ? ADC 
 
(iii) Hypotenuse AO = hypotenuse DO 
BO = CO  
?B = ?C = 90
o 
Therefore, by RHS, ?AOB??DOC 
 
(iv) Hypotenuse AC = Hypotenuse CA 
BC = DC  
?ABC = ?ADC = 90
o 
Therefore, by RHS, ?ABC ? ?ADC 
 
(v) BD = DB  
 
 
 
 
  
 
 
 
Hypotenuse AB = Hypotenuse BC, as per the given figure, 
?BDA + ?BDC = 180
o 
?BDA + 90
o
 = 180
o 
?BDA= 180
o 
– 90
o
 = 90
o 
?BDA = ?BDC = 90
o 
Therefore, by RHS, ?ABD ? ?CBD 
 
2. ? ABC is isosceles with AB = AC. AD is the altitude from A on BC. 
(i) Is ?ABD ? ?ACD? 
(ii) State the pairs of matching parts you have used to answer (i). 
(iii) Is it true to say that BD = DC? 
 
Solution: 
(i) Yes, ?ABD ? ?ACD by RHS congruence condition. 
 
(ii) We have used Hypotenuse AB = Hypotenuse AC 
AD = DA 
?ADB = ?ADC = 90
o
 (AD ? BC at point D) 
 
(iii)Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)  
Since we have already proved that the two triangles are congruent. 
  
3. ?ABC is isosceles with AB = AC. Also. AD ? BC meeting BC in D. Are the two triangles 
ABD and ACD congruent? State in symbolic form. Which congruence condition do you 
use? Which side of ADC equals BD? Which angle of ? ADC equals ?B? 
 
Solution: 
We have AB = AC …… (i) 
AD = DA (common) …… (ii) 
And, ?ADC = ?ADB (AD ? BC at point D) …… (iii) 
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ?ABD ? ?ACD, the 
triangles are congruent. 
Therefore, BD = CD. 
And ?ABD = ?ACD (corresponding parts of congruent triangles) 
 
4. Draw a right triangle ABC. Use RHS condition to construct another triangle 
congruent to it. 
 
 
 
 
  
 
 
 
 
Solution: 
 
Consider 
? ABC with ?B as right angle. 
We now construct another triangle on base BC, such that ?C is a right angle and AB = DC 
Also, BC = CB 
Therefore by RHS, ?ABC ? ?DCB 
  
5.In fig. 47, BD and CE are altitudes of ? ABC and BD = CE. 
(i) Is ?BCD ? ?CBE? 
(ii) State the three pairs or matching parts you have used to answer (i) 
 
Solution: 
(i) Yes, ?BCD ? ?CBE by RHS congruence condition. 
 
(ii) We have used hypotenuse BC = hypotenuse CB 
Page 5


 
 
 
 
  
 
 
 
        
 
1. In each of the following pairs of right triangles, the measures of some parts are 
indicated alongside. State by the application of RHS congruence condition which are 
congruent, and also state each result in symbolic form. (Fig. 46) 
 
 
 
 
 
 
 
  
 
 
 
 
Solution: 
(i) ?ADB = ?BCA = 90
o 
AD = BC and hypotenuse AB = hypotenuse AB 
Therefore, by RHS ?ADB ? ?ACB 
 
(ii) AD = AD (Common) 
Hypotenuse AC = hypotenuse AB (Given) 
?ADB + ?ADC = 180
o
 (Linear pair) 
?ADB + 90
o
 = 180
o 
?ADB = 180
o
 – 90
o
 = 90
o 
?ADB = ?ADC = 90
o 
Therefore, by RHS ? ADB ? ? ADC 
 
(iii) Hypotenuse AO = hypotenuse DO 
BO = CO  
?B = ?C = 90
o 
Therefore, by RHS, ?AOB??DOC 
 
(iv) Hypotenuse AC = Hypotenuse CA 
BC = DC  
?ABC = ?ADC = 90
o 
Therefore, by RHS, ?ABC ? ?ADC 
 
(v) BD = DB  
 
 
 
 
  
 
 
 
Hypotenuse AB = Hypotenuse BC, as per the given figure, 
?BDA + ?BDC = 180
o 
?BDA + 90
o
 = 180
o 
?BDA= 180
o 
– 90
o
 = 90
o 
?BDA = ?BDC = 90
o 
Therefore, by RHS, ?ABD ? ?CBD 
 
2. ? ABC is isosceles with AB = AC. AD is the altitude from A on BC. 
(i) Is ?ABD ? ?ACD? 
(ii) State the pairs of matching parts you have used to answer (i). 
(iii) Is it true to say that BD = DC? 
 
Solution: 
(i) Yes, ?ABD ? ?ACD by RHS congruence condition. 
 
(ii) We have used Hypotenuse AB = Hypotenuse AC 
AD = DA 
?ADB = ?ADC = 90
o
 (AD ? BC at point D) 
 
(iii)Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)  
Since we have already proved that the two triangles are congruent. 
  
3. ?ABC is isosceles with AB = AC. Also. AD ? BC meeting BC in D. Are the two triangles 
ABD and ACD congruent? State in symbolic form. Which congruence condition do you 
use? Which side of ADC equals BD? Which angle of ? ADC equals ?B? 
 
Solution: 
We have AB = AC …… (i) 
AD = DA (common) …… (ii) 
And, ?ADC = ?ADB (AD ? BC at point D) …… (iii) 
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ?ABD ? ?ACD, the 
triangles are congruent. 
Therefore, BD = CD. 
And ?ABD = ?ACD (corresponding parts of congruent triangles) 
 
4. Draw a right triangle ABC. Use RHS condition to construct another triangle 
congruent to it. 
 
 
 
 
  
 
 
 
 
Solution: 
 
Consider 
? ABC with ?B as right angle. 
We now construct another triangle on base BC, such that ?C is a right angle and AB = DC 
Also, BC = CB 
Therefore by RHS, ?ABC ? ?DCB 
  
5.In fig. 47, BD and CE are altitudes of ? ABC and BD = CE. 
(i) Is ?BCD ? ?CBE? 
(ii) State the three pairs or matching parts you have used to answer (i) 
 
Solution: 
(i) Yes, ?BCD ? ?CBE by RHS congruence condition. 
 
(ii) We have used hypotenuse BC = hypotenuse CB 
 
 
 
 
  
 
 
 
BD = CE (Given in question) 
And ?BDC = ?CEB = 90
o