## 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 Â of triangle are congruent. State the result in symbolic form **

**Answer:**

(i)

We have OA = OC and OB = OD and

\(\angle\)

(ii)

We have BD = DC

\(\angle\)

AD = AD

Therefore, by SAS condition, \(\Delta ADB \cong \Delta ADC\)

(iii)

We have AB = DC

\(\angle\)

BD = DB

Therefore, by SAS condition, \(\Delta ABD \cong \Delta CBD\)

(iv)

We have BC = QR

ABC = PQR = 90Â°

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.**

**Answer: **

(i)

AB = AD

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)

AB = AD

\(\angle\)

AC = AC

Therefore by SAS condition, \(\Delta BAC \cong \Delta BAC\)

(iv)

AD = BC

\(\angle\)

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.**

**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?**

**Answer:**

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

Also \(\angle\)

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?**

**Answer: **

(i) We have AB = AC (Given)

\(\angle\)

And AD = AD (common)

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

(ii) We have used AB, AC; \(\angle\)

(iii) Now, \(\Delta ABD \cong \Delta ACD\)

**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 ___**

Answer:

i) AB = AD (given)

\(\angle\)

AC = CA (common)

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

ii) \(\angle\)

\(\angle\)

**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 ? **

**(iv) Does it follow from (iii) that AD || BC? Â **

**Answer: **

(i) Yes by SAS condition of congruency, \(\Delta DCA \cong \Delta BAC\)

(ii) We have used AB = DC, AC = CA and \(\angle\)

(iii) \(\angle\)

(iv) Yes this follows from AD // BC as alternate angles are equal. lf alternate angles are equal the lines are parallel