Congruence of Triangles PDF

## Example 6: In Fig. 9.10, PS = QR and PQ = SR. Find the third pair of corresponding sides which make ∆PSR = ∆RQP by SSS congruence condition. **Solution:** In the two triangles RQP and PSR, the two pairs of corresponding sides are: - QR = PS (given) - PQ = SR (given) For SSS congruence conditi...

## Example 6: In Fig. 9.10, PS = QR and PQ = SR. Find the third pair of corresponding sides which make ∆PSR = ∆RQP by SSS congruence condition. **Solution:** In the two triangles RQP and PSR, the two pairs of corresponding sides are: - QR = PS (given) - PQ = SR (given) For SSS congruence condition, the third pair of corresponding sides should be equal, are the remaining two sides of the two triangles. Here, PR is common to both the triangles. Thus, the third pair of corresponding parts to make ∆PSR = ∆RQP by SSS congruence condition are: - RP = PR (common side) ## Worksheet 1: 1. Fill in the blanks so as to make a true statement. - (i) Two line segments are congruent, if they have **Equal length** - (ii) Two angles are congruent, if they have **Equal measure** - (iii) Two squares are congruent, if they have **Equal Sides** - (iv) Two circles are congruent, if they have **Equal Radius** - (v) Two rectangles are congruent, if they have **Equal length and Breadth** ## Fig. 9.11: State which pairs of triangles are congruent by SSS congruence condition. If congruent, write the congruence of triangles in symbolic form. - **(i)** ∆ABC = ∆PQR ## Fig. 9.12: PS = RS and PQ = RQ. - **(i)** Is ∆PQS = ∆RQS? - **(ii)** State the three pairs of matching parts you have used to answer (i). - PQ = RQ (given) - PS = RS (given) - QS = QS (common side) - **(iii)** ∠P = ∠R ## Fig. 9.13: ∆ABC is isosceles with AB = AC. D is the mid-point of base BC. - **(i)** Is ∆ADB = ∆ADC? If yes, by which congruence condition? - **(ii)** State the three pairs of matching parts that you use to arrive at your answer. - AB = AC (given) - ∠BAD = ∠CAD (because AD is altitude) - AD = AD (common side) ## Fig. 9.14: In ∆PQR and ∆XYZ, PQ = XZ and QR = YZ. What additional information is required to make the two triangles congruent by SSS congruence condition. **Solution:** The third side PR should be equal to the third side XY to make the two triangles congruent by SSS congruence condition. ## Fig. 9.15: If ∆ABC = △DEF, fill in the blanks to make each statement true. - (i) AB = DE - (ii) ∠C = ∠F - (iii) EF = BC - (iv) ∠D = ∠A - (v) CA = FD - (vi) ∠B = ∠E ## Activity 3: Draw ∆ABC with AB = 5cm, AC = 3 cm and ∠A=60°. Also draw APQR with PR = 3 cm and ∠P = 60°. **Solution:** Thus, we have: - AB = PQ, AC = PR and ∠A = ∠P Therefore, we can conclude that ∆ABC = ∆PQR ## Activity 4: Construct ∆ABC with AB = 4 cm, AC = 5 cm and the included angle ∠A = 30°. Draw another triangle APQR with PR = 4 cm, PQ = 5 cm the included angle ∠P = 30°. # Fig. 9.16: Then, we have AB = PR, AC = PQ and the included angles ∠A = ∠P. Measure the remaining parts of the two triangles and record the observations in the form of a table as shown: |Remaining parts of ∆ABC | Corresponding parts of APQR | Difference | |---|---|---| | BC = | QR = | BC – QR = | | ∠B = | ∠Q = | ∠B - ∠Q = | | ∠C = | ∠R = | ∠C - ∠R = | You will find that the difference in each measure is either zero (or very near to zero). Thus, BC = QR, ∠B = ∠Q and ∠C = ∠R. Therefore, ∆ABC = ΔPQR. Remember - The angle in the SAS congruence condition should always be the included angle between the two considered sides. Two triangles are congruent, if two sides and the included angle of one triangle are respectively equal to the two corresponding sides and the included angle of the other. This is the SAS Congruence condition. ## Fig. 9.17: But ∠ADB + ∠ADC = 180° ∠ADB = ∠ADC = 90° (linear pair) Hence, AD bisects BC at right angles. ## Worksheet 2: In ∆ABC and △DEF, AB = DE and BC = EF (Fig. 9.19). What additional information is required to make the two triangles congruent by SAS congruence condition? **Solution**: To prove ∆ABC and ∆DEF congruent by SAS congruence condition, we need either ∠B = ∠E, so that AB, ∠B, BC are corresponding sides and angle of both triangles, or AC = DF so that AB, ∠B, AC are corresponding sides and angle of both triangles. ## Fig 9.20: ∆ABC is isosceles with AB = AC Line segment AD bisects ∠A and meets the base BC at D. Find the third pair of corresponding parts which make ∆ADB = ∆ADC by SAS congruence condition. **Solution:** 1. AD is common to both triangles. 2. AB = AC is given. 3. ∠BAD = ∠CAD because AD is the bisector of ∠A. This satisfies the condition for SAS congruence condition. Is it true to say that BD = CD? Why? **Solution:** Yes, because the two triangles are congruent and BD and CD are corresponding sides of the two congruent traingles. ## Fig 9.21: AB || DC and AB = DC **(i)** Is ∠BAC = ∠DCA? Why? **Solution:** Yes, ∠BAC = ∠DCA because they are alternate angles. **(ii)** Is ∆ABC = ∆CDA by SAS congruence condition? **Solution:** Yes, ∆ABC = ∆CDA by SAS congruence condition because: - AB = DC (given) - ∠BAC = ∠DCA (alternate angles) - AC is common side. **(iii)** State the three facts you have used to answer (ii). - (i) AB = DC. - ( ii) ∠BAC = ∠DCA. - (iii) AC = AC (common side). ## Fig. 9.22: In Fig. 9.22, which pairs of triangles are congruent by SAS congruence condition. If congruent, write the congruence of the two triangles in symbolic form. - **(i)** ∆ABC = ∆DEF - **(ii)** ∆PQR = ∆XYZ ## Fig. 9.23 In Fig. 9.23, AB= AD and ∠BAC = ∠DAC. **(i)** State in symbolic form, the congruence of two triangles ABC and ADC is true. Also state the congruence condition used. **Solution:** ∆ABC = ∆ADC, by SAS congruence condition. **(ii)** Complete each of the following, so as to make it true: - (a) ∠ABC = ∠ADC - (b) ∠ACD = ∠ACB - (c) Line segment AC bisects ∠BCD and ∠BAD. ## Activity 5: Draw ∆ABC with BC = 5 cm, ∠B = 40° and ∠C = 30°. Also draw ∆DEF with EF = 5 cm, ∠E = 40° and ∠F = 30°. **Solution** Thus, we have BC = EF, ∠B = ∠E and ∠C = ∠F. Make a trace copy of ∆ABC and try to make it cover ∆DEF. You will observe that with B on E, C on F and A on D, ∆ABC covers ADEF exactly, or the hing ABC → DEF gives an exact cover of the two triangles by each other. So, we can conclude ∆ABC = ∆DEF. ## Activity 6: Draw ∆ABC with AB = 4 cm, ∠A = 60° and ∠B = 40°. Also draw ∆PQR with PQ = 4 cm, ∠P = 60° and ∠Q = 40°. **Solution:** Thus, we have AB = PQ, ∠A = ∠P and ∠B = ∠Q. Therefore, we can conclude that ∆ABC = ∆PQR. ## Worksheet 3: In Fig. 9.28, ∠P = ∠Y = 40° and ∠Q = ∠Z = 60°. Find the third pair of corresponding parts which make ∆PQR ≈ ∆YZX, by ASA congruence condition. **Solution:** The third pair of corresponding parts are: - PR = YX. ## Fig. 9.29: Which pairs of triangles are congruent by ASA congruence condition in Figure 9.29? If congruent, write the congruence of the two triangles in symbolic form. - (i) ∆ABC = ∆XYZ - (ii) ∆ABO = ∆CDO ## Fig. 9.30: OX bisects ∠POR as well as ∠PSR. State the three facts needed to ensure that ∆ORS = ∆QPS. **Solution:** The three facts are: - ∠ROS = ∠QOS (since OX bisects ∠POR). - ∠SOR = ∠QOR (since OX bisects ∠PSR). - OS = OS (common side). ## Fig. 9.31: PS bisects ∠P and PS ⊥ QR. - **(i)** Find the three pairs of matching parts to check whether ∆PSO = ∆PSR or not. - PS = PS (common side) - ∠PSO = ∠PSR (because PS bisects ∠P) - ∠POS = ∠PRS (because PS ⊥ QR) - **(ii)** Is ∆PSO = ∆PSR? - Yes, it is true by ASA congruent condition. - **(iii)** Is it true to say that QS = SR? Why? - Yes, it is true because corresponding sides of congruent triangles are equal. ## Fig. 9.32: AO = BO and ∠A = ∠B. - **(i)** Is ∠AOC = ∠BOD? Why? - Yes, because ∠AOC and ∠BOD are vertically opposite angles and vertically opposite angles are equal. - **(ii)** Is ∆AOC = ∆BOD by ASA congruence condition? - Yes, ∆AOC = ∆BOD by ASA congruence condition. - ∠A =∠B (given) - ∠AOC = ∠BOD (vertically opposite angles) - AO = BO (given) - **(iii)** Is ∠ACO = ∠BDO? Why? - Yes, ∠ACO = ∠BDO because corresponding angles of congruent triangles are equal. ## Worksheet 4: 1. △ABC is isosceles with AB = AC. AD is the altitude from A on BC (Fig. 9.37). - **(i)** Is △ABD = △ACD? Why? - Yes, △ABD = △ACD because: - AB = AC (given) - ∠ADB = ∠ADC = 90° (because AD is the altitude) - AD = AD (common side) - **(ii)** State the three pairs of matching parts you have used to answer (i). - **(iii)** Is it true to say that BD = CD? Why? - Yes, because corresponding sides of congruent triangles are equal. ## Fig. 9.38: Which pairs of triangles are congruent by RHS congruence condition in Figure 9.38? If congruent, write the result in symbolic form. - (i) ∆ABC = ∆DBC. - (ii) ∆ABD = ∆ACD - (iii) ∆ABD = ∆ACD - (iv) ∆ABO = ∆CDO. ## Fig. 9.39: QS and RT are the altitudes of ∆PQR, and QS = RT. - **(i)** Is ∆QRS = ∆RQT by RHS congruence condition? - Yes, ∆QRS = ∆RQT by RHS congruence condition because: - ∠QSR = ∠QTR = 90° (because QS and RT are altitudes). - QR = QR (common side). - QS = RT (given). - **(ii)** State the three pairs of corresponding parts which make ∆QRS = ∆RQT. - (i) ∠QSR = ∠QTR = 90° (because QS and RT are altitudes). - (ii) QR = QR (common side). - (iii) QS = RT (given). ## Fig. 9.40: In Fig. 9.40, ∠B = ∠P = 90°, and side AB = side PR. What additional information is required to make ∆ABC = ARPQ by RHS congruence condition? **Solution:** To use RHS congruence condition, the hypotenuses need to be equal. So, BC should be equal to PQ to make triangles ∆ABC = ARPQ. ## Fig. 9.41: In Fig. 9.41. AC = DB and AB ⊥ BC. Also DC ⊥ BC. State which of the following is true. - (i) ∆ABC = ADBC **Solution:** - (i) ∆ABC = △DBC because: - ∠ABC = ∠DBC = 90° (given) - AC = DB (given) - BC = BC (common side) - **(ii)** ∆ABC = ADCB - **(iii)** ΔΑΒC = ABCD State the three pairs of matching parts you have used to arrive at the answer. - ∠ABC = ∠DBC = 90° (given) - AC = DB (given) - BC = BC (common side) Which side is equal to side AB? Why? **Solution:** BC is equal to side AB because corresponding sides of congruent triangles are equal.