Ex-16.5, Congruence, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 1:

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State each result in symbolic form. (Fig. 46)

Answer 1:

i)
∠ADC= ∠BCA =90°
AD = BC
and hyp AB = hyp AB
Therefore, by RHS,    △ADB≅△ACB.
.

ii)
AD=AD (Common)  
hyp AC  = hyp AB (Given)
∠ADB+∠ADC = 180°(Linear pair)
∠ADB+90°=180°
∠ADB=180°−90°=90°
∠ADB=∠ADC = 90°
Therefore, by RHS, △ADB=△ADC

iii)
hyp AO = hyp DO
BO = CO
∠B =∠C =90°
Therefore, by RHS, △AOB≅△DOC

iv)
Hyp AC = Hyp CA
BC= DC
∠ABC=∠ADC = 90°
Therefore, by RHS, △ABC≅△ADC

v)
BD = DB
Hyp AB =Hyp BC, as per the given figure.
∠BDA +∠BDC = 180°
∠BDA +90° = 180°
∠BDA= 180°−90°= 90°
∠BDA =∠BDC = 90°
Therefore, by RHS, △ABD≅△CBD  

Question 5:

In Fig. 47, BD and CE are altitudes of ∆ ABC and BD = CE.
(i) Is ∆ BCD ≅ ∆ CBE?
(ii) State the three pairs of matching parts you have used to answer (i).

Answer 5:

(i) Yes, △BCD≅△CBE by RHS congruence condition.
(ii) We have used hyp BC = hyp CB
BD = CE (given in question)
and ∠BDC = ∠CEB =90°