# RD Sharma Solutions Class 7 Congruence Exercise 16.3

## RD Sharma Solutions Class 7 Chapter 16 Exercise 16.3

Exercise 16.3

Q1. By applying SAS congruence condition, state which of the following pairs  of triangle are congruent. State the result in symbolic form

(i)

We have OA = OC and OB = OD and

$\angle$ AOB = $\angle$ COD which are vertically opposite angles. Therefore by SAS condition, $\Delta AOC \cong \Delta BOD$

(ii)

We have BD = DC

$\angle$ ADB = $\angle$ ADC = 90° and

Therefore, by SAS condition, $\Delta ADB \cong \Delta ADC$.

(iii)

We have AB = DC

$\angle$ ABD = $\angle$ CDB and

BD = DB

Therefore, by SAS condition, $\Delta ABD \cong \Delta CBD$

(iv)

We have BC = QR

ABC = PQR = 90°

And AB = PQ

Therefore, by SAS condition, $\Delta ABC \cong \Delta PQR$.

Q2. State the condition by which the following pairs of triangles are congruent.

(i)

BC = CD and AC = CA

Therefore by SSS condition, $\Delta ABC \cong \Delta ADC$

(ii)

AC = BD

AD = BC and AB = BA

Therefore, by SSS condition, $\Delta ABD \cong \Delta ADC$

(iii)

$\angle$ BAC = $\angle$ DAC and

AC = AC

Therefore by SAS condition, $\Delta BAC \cong \Delta BAC$

(iv)

$\angle$ DAC = $\angle$ BCA and

AC = CA

Therefore, by SAS condition, $\Delta ABC \cong \Delta ADC$

Q3. In figure, line segments AB and CD bisect each other at O. Which of the following statements is true?

(i) $\Delta AOC \cong \Delta DOB$

(ii) $\Delta AOC \cong \Delta BOD$

(iii) $\Delta AOC \cong \Delta ODB$

State the three pairs of matching parts, you have used to arrive at the answer.

We have,

And, CO = OD

Also, AOC = BOD

Therefore, by SAS condition, $\Delta AOC \cong \Delta BOD$

Q4. 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?

We have AO = OB and CO = OD since AB and CD bisect each other at 0.

Also $\angle$ AOC = $\angle$ BOD since they are opposite angles on the same vertex.

Therefore by SAS congruence condition, $\Delta AOC \cong \Delta BOD$

Q5. $\Delta$ ABC is isosceles with AB = AC. Line segment AD bisects $\angle$ A and meets the base BC in D.

(i) Is $\Delta ADB \cong \Delta ADC$?

(ii) State the three pairs of matching parts used to answer (i).

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

(i) We have AB = AC (Given)

$\angle$BAD = $\angle$CAD (AD bisects $\angle$ BAC)

Therefore by SAS condition of congruence, $\Delta ABD \cong \Delta ACD$

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

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

Q6. In Figure, AB = AD and $\angle$ BAC = $\angle$ 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) $\angle$ ABC =

(b) $\angle$ ACD =

(c) Line segment AC bisects ___ and ___

$\angle$ BAC = $\angle$ DAC (given)

AC = CA (common)

Therefore by SAS condition of congruency, $\Delta ABC \cong \Delta ADC$

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

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

Q7. In figure, AB || DC and AB = DC.

(i) Is $\Delta ACD \cong \Delta CAB$?

(ii) State the three pairs of matching parts used to answer (i).

(iii) Which angle is equal to $\angle$ CAD ?

(i) Yes by SAS condition of congruency, $\Delta DCA \cong \Delta BAC$
(ii) We have used AB = DC, AC = CA and $\angle$ DCA = $\angle$ BAC.
(iii) $\angle$ CAD = $\angle$ ACB since the two triangles are congruent.