RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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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)

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev             RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev                   RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

 

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 2:

ABC is isosceles with AB = AC. AD is the altitude from A on BC.
(i) Is ∆ ABDACD?
(ii) State the pairs of matching parts you have used to answer (i).
(ii) Is it true to say that BD = DC?

Answer 2:

(i)Yes, ABDACD△ABD≅△ACD by RHS congruence condition.
(ii) We have used Hyp AB = Hyp AC
AD = DA
and ADB = ADC = 90°∠ADB = ∠ADC = 90° (ADBC at point D)
(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.


Question 3:

ABC is isoseles with AB = AC. Also, ADBC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condtion do you use? Which side of ∆ ADC equls BD? Which angle of ∆ ADC equals ∠B?


Answer 3:

We have AB = AC  ......(1)
AD = DA (common)........(2)
and ADC=ADB∠ADC=∠ADB (ADBC at point D)........(3)

Therefore from 1, 2 and 3, by RHS congruence condition,
 △ABD△ACD

Now, the triangles are congruent .     
Therefore, BD= CD.
And ∠ABD=∠ACD (c.p.c.t).

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev


Question 4:

Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.

Answer 4:

Consider
△ABC with ∠B as right angle.
We now construct another right triangle on base BC, such that    ∠C is a right angle and     AB = DC
Also, BC = CB
Therefore, by RHS, △ABC△DCB

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev


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).

RD Sharma Solutions - Ex-16.5, Congruence, Class 7, Math Class 7 Notes | EduRev

Answer 5:

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

 

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