RD Sharma solution class 7 chapter 16 Congruence Exercise 16.3

Exercise 16.3

Page-16.14

Question 1:

By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangles are congruent. State the result in symbolic form

Answer 1:

1) We have OA = OC and OB = OD and$$∠AOB =∠COD which are vertically opposite angles.
Therefore by SAS condition, △AOC ≅ △BOD.

2) We have BD = DC
$$∠ADB =∠ADC = 90°
and AD = AD
Therefore by SAS condition, △ADB ≅ △ADC.

3)   We have AB = DC
$\angle$ABD = ∠CDB and BD = DB
Therefore by SAS condition, △ABD ≅ △CBD.

4)  We have BC = QR
$$∠ABC =∠PQR = 90°
and AB = PQ
Therefore by SAS condition, △ABC ≅ △PQR.

Question 2:

State the condition by which the following pairs of triangles are congruent.

Answer 2:

1)  AB = AD
BC = CD
and AC = CA
Therefore by SSS condition, $△ABC\cong △ADC$.

2) AC = BD
AD = BC and AB = BA
Therefore by SSS condition, $△ABD\cong △BAC$.

3)  AB = AD
$\angle BAC=\angle DAC$
and AC = AC
Therefore by SAS condition, $△BAC\cong △DAC$.

4)  AD = BC
$\angle$ DAC = $\angle$BCA
and AC = CA
Therefore by SAS condition, $△ABC\cong △ADC$.

Page-16.15

Question 3:

In Fig. 30, line segments AB and CD bisect each other at O. Which of the following statements is true?
(i) ∆ AOC ≅ ∆ DOB
(ii) ∆ AOC ≅ ∆ BOD
(iii) ∆ AOC ≅ ∆ ODB.
State the three pairs of matching parts, yut have used to arive at the answer.

Answer 3:

We have AO = OB.
And CO = OD

Also ∠AOC = ∠BOD
Therefore by SAS condition, △AOC ≅ △BOD.

Therefore, statement (ii) is true.

Question 4:

Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles, that are given or otherwise known. Are the two triangles congruent? State in symbolic form. Which congruence condition do you use?

Answer 4:

We have AO = OB and CO = OD since AB and CD bisect each other at O.

Also $\angle AOC = \angle BOD$ since they are opposite angles on the same vertex.
Therefore by SAS congruence condition, $△AOC\cong △BOD$.

Question 5:

∆ ABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.
(i) Is ∆ ADB ≅ ∆ ADC?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Is it true to say that BD = DC?

Answer 5:

(i) We have AB = AC (given)
$\angle BAD =\angle CAD$ (AD bisects $\angle BAC$)
and AD = AD (common)
Therefore by SAS condition of congruence, $△ABD\cong △ACD$.

(ii) We have used AB, AC; $\angle BAD =\angle CAD$; AD, DA.

(iii) Now$△ABD\cong △ACD$ therefore by c.p.c.t BD = DC.

Question 6:

In Fig. 31, AB = AD and ∠BAC = ∠DAC.
(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.
(ii) Complete each of the following, so as to make it true:
(a) ∠ABC = ........
(b) ∠ACD = ........
(c) Line segment AC bisects ..... and .....

Answer 6:

i) AB = AD (given)
$\angle BAC =\angle DAC$ (given)
AC = CA (common)
Therefore by SAS conditionof congruency, $△ABC \cong △ADC$.

ii) $\angle ABC =\angle ADC$ (c.p.c.t)
$\angle ACD =\angle ACB$ (c.p.c.t)

Question 7:

In Fig. 32, AB || DC and AB = DC.
(i) Is ∆ ACD ≅ ∆ CAB?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Which angle is equal to ∠CAD?
(iv) Does it follow from (iii) that AD || BC?

Answer 7:

(i) Yes  $△ACD\cong △CAB$ by SAS condition of congruency.
(ii) We have used AB = DC, AC = CA and $\angle DCA=\angle BAC$.
(iii) $\angle CAD =\angle ACB$ since the two triangles are congruent.
(iv) Yes, this follows from AD$\parallel$BC as alternate angles are equal.If alternate angles are equal the lines are parallel.