RD Sharma Solutions For Class 7 Maths Chapter 16 Congruence Exercise 16.4

Our expert team have designed the solutions for RD Sharma Solutions for Class 7 Maths Chapter 16 to help students prepare for their exams effortlessly.

RD Sharma Solutions for Class 7 is one of the best reference materials for CBSE students. Learners can download the pdf from the links provided below. The experts suggest students to practise the solutions many numbers of times and yield good results in their exams. In Exercise 16.4 of Chapter 16 Congruence, we shall discuss about the ASA congruence condition for two congruent triangles and their properties.

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Exercise 16.4 Page No: 16.19

1. Which of the following pairs of triangle are congruent by ASA condition?

Solution:

(i) We have,

Since âˆ ABO = âˆ CDO = 45o and both are alternate angles, AB parallel to DC, âˆ BAO = âˆ DCO (alternate angle, AB parallel to CD and AC is a transversal line)

âˆ ABO = âˆ CDO = 45o (given in the figure) Also,

AB = DC (Given in the figure)

Therefore, by ASA Î”AOB â‰… Î”DOC

(ii) In ABC,

Now AB =AC (Given)

âˆ ABD = âˆ ACD = 40o (Angles opposite to equal sides)

âˆ ABD + âˆ ACD + âˆ BAC = 180o (Angle sum property)

40o + 40o + âˆ BAC = 180o

âˆ BAC =180o – 80o =100o

âˆ BAD + âˆ DAC = âˆ BAC

âˆ BAD = âˆ BAC â€“ âˆ DAC = 100o â€“ 50o = 50o

Therefore, by ASA, Î”ABD â‰… Î”ACD

(iii) In Î” ABC,

âˆ A + âˆ B + âˆ C = 180o (Angle sum property)

âˆ C = 180o– âˆ A â€“ âˆ B

âˆ C = 180o â€“ 30o â€“ 90o = 60o

In PQR,

âˆ P + âˆ Q + âˆ R = 180o (Angle sum property)

âˆ P = 180o â€“ âˆ R â€“ âˆ Q

âˆ P = 180o â€“ 60o â€“ 90o = 30o

âˆ BAC = âˆ QPR = 30o

âˆ BCA = âˆ PRQ = 60o and AC = PR (Given)

Therefore, by ASA, Î”ABC â‰… Î”PQR

(iv) We have only

BC = QR but none of the angles of Î”ABC and Î”PQR are equal.

Therefore, Î”ABC is not congruent to Î”PQR

(ii) State the three pairs of matching parts you have used in (i)

(iii) Is it true to say that BD = DC?

Solution:

3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.

Solution:

We have drawn

Î” ABC with âˆ ABC = 65o and âˆ ACB = 70o

We now construct Î”PQR â‰… Î”ABC where âˆ PQR = 65o and âˆ PRQ = 70o

Also we construct Î”PQR such that BC = QR

Therefore by ASA the two triangles are congruent

4. In Î” ABC, it is known that âˆ B = C. Imagine you have another copy of Î” ABC

(i) Is Î”ABC â‰… Î”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 Î”ABC â‰… Î”ACB

(ii) We have used âˆ ABC = âˆ ACB and âˆ ACB = âˆ ABC again.

Also BC = CB

(iii) Yes it is true to say that AB = AC since âˆ ABC = âˆ ACB.

5. In Fig. 38, AX bisects âˆ BAC as well as âˆ BDC. State the three facts needed to ensure that Î”ACD â‰… Î”ABD

Solution:

As per the given conditions,

âˆ CAD = âˆ BAD and âˆ CDA = âˆ BDA (because AX bisects âˆ BAC)

Therefore, by ASA, Î”ACD â‰… Î”ABD

6. In Fig. 39, AO = OB and âˆ A = âˆ B.

(i) Is Î”AOC â‰… Î”BOD

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that âˆ ACO = âˆ BDO?

Solution:

We have

âˆ OAC = âˆ OBD,

AO = OB

Also, âˆ AOC = âˆ BOD (Opposite angles on same vertex)

Therefore, by ASA Î”AOC â‰… Î”BOD