Congruence - Exercise 21(A)

State the condition (SSS or SAS or ASA or RHS) under which △ABC ≅ △PQR in each of the following cases.

(i)

(ii)

(iii)

(iv)

Answer

(i)

AB = QP = 2.5 cm

BC = QR = 4 cm

AC = PR = 3.5 cm

Condition: SSS

(ii)

∠B = ∠Q = 90° $\quad$[Right angle]

AC = PR $\quad$[Hypotenuse]

AB = PQ $\quad$[Side]

Condition: RHS

(iii)

∠A = ∠P = 35° $\quad$[Angle]

AB = PQ $\quad$[Side]

∠B = ∠Q = 90° $\quad$[Angle]

Condition: ASA

(iv)

AC = QR $\quad$[Side]

∠C = ∠R = 90° $\quad$[Angle]

BC = RP $\quad$[Side]

Condition: SAS

State giving reasons, which of the following are pairs of congruent triangles. Write down the names of triangles with congruent sign in each case, wherever they are congruent.

(i)

(ii)

(iii)

(iv)

(v)

Answer

(i)

Consider △ABC,

∠A = 180° - (60° + 50°) = 70°.

Consider △FDE,

∠F = 180° - (60° + 50°) = 70°.

In △ABC and △FDE:

∠A = ∠F = 70° $\quad$[Angle]

AB = FD $\quad$[Side]

∠B = ∠D = 60° $\quad$[Angle]

Since two angles and the included side match, they are congruent by ASA condition of congruence.

△ABC ≅ △FDE

(ii)

In △ABC and △PQR:

AB = PR $\quad$[Side]

∠B = ∠R = 35° $\quad$[Angle]

BC = QR $\quad$[Side]

Since two sides and the included angle match, they are congruent by SAS condition of congruence.

△ABC ≅ △PRQ

(iii)

In △DEH and △PQR:

DH = RP $\quad$[Side]

∠H = ∠P = 90° $\quad$[Right-angle]

HE = PQ $\quad$[Side]

Since two sides and the included right angle match, they are congruent by SAS condition of congruence.

△DEH ≅ △RQP

(iv)

In △ABC and △PQR:

AB = RQ $\quad$[Side]

∠B = ∠Q = 90° $\quad$[Right-angle]

CB = PQ $\quad$[Side]

Since two sides and the included right angle match, they are congruent by SAS condition of congruence.

△ABC ≅ △RQP

(v)

Consider △XYZ,

∠X = 180° - (30° + 80°) = 70°.

In △LMN and △XYZ:

LM = ZX $\quad$[Side]

∠M = ∠X = 70° $\quad$[Angle]

MN = XY $\quad$[Side]

Since two sides and the included angle match, they are congruent by SAS condition of congruence.

△LMN ≅ △ZXY

State, giving reasons, whether the following pairs of triangles are congruent or not:

△ABC in which ∠A = 50°, ∠B = 60°, BC = 4.5 cm and △DEF in which ∠E = 60°, EF = 4.5 cm, ∠F = 70°.

Answer

△ABC in which ∠A = 50°, ∠B = 60°, BC = 4.5 cm and △DEF in which ∠E = 60°, EF = 4.5 cm, ∠F = 70°.

Consider △ABC,

∠C = 180° - (50° + 60°) = 70°.

In △ABC and △DEF:

∠C = ∠F = 70° $\quad$[Angle]

BC = EF = 4.5 cm $\quad$[Side]

∠B = ∠E = 60° $\quad$[Angle]

Since two angles and the included side match,

They are congruent by ASA condition of congruence.

State, giving reasons, whether the following pairs of triangles are congruent or not:

△DEF in which ∠E = 48°, DE = 6 cm, EF = 8 cm and △MNR in which ∠R = 48°, MN = 6 cm, MR = 8 cm.

Answer

△DEF in which ∠E = 48°, DE = 6 cm, EF = 8 cm and △MNR in which ∠R = 48°, MN = 6 cm, MR = 8 cm.

In △DEF, the angle 48° is between the two given sides (SAS).

In △MNR, the angle 48° is at vertex R, which is not the included angle between side MN and MR (the included angle would be ∠M).

The given angle is not in the corresponding "included" position for both triangles.

∴ They are not congruent.

State, giving reasons, whether the following pairs of triangles are congruent or not:

△KLM in which KM = 4 cm, ∠K = 75°, ∠M = 40° and △PQR in which PR = 4 cm, ∠Q = 65°, ∠R = 40°.

Answer

△KLM in which KM = 4 cm, ∠K = 75°, ∠M = 40° and △PQR in which PR = 4 cm, ∠Q = 65°, ∠R = 40°.

Consider △PQR,

∠P = 180° - (65° + 40°) = 75°.

In △KLM and △PQR:

∠K = ∠P = 75° $\quad$[Angle]

KM = PR = 4 cm $\quad$[Side]

∠M = ∠R = 40° $\quad$[Angle]

Since two angles and the included side match,

△KLM ≅ △PRQ by ASA condition of congruence.

State, giving reasons, whether the following pairs of triangles are congruent or not:

△ABC in which AB = 3 cm, ∠A = 90°, BC = 5 cm and △KLM in which KM = 3 cm, ∠K = 90°, LM = 5 cm.

Answer

△ABC in which AB = 3 cm, ∠A = 90°, BC = 5 cm and △KLM in which KM = 3 cm, ∠K = 90°, LM = 5 cm.

In △ABC and △KML:

AB = KM = 3 cm $\quad$[Side]

∠A = ∠K = 90° $\quad$[Right angle]

BC = LM = 5 cm $\quad$[Hypotenuse]

Since one side, the hypotenuse and the right angle match,

△ABC ≅ △KML by RHS condition of congruence.

In the given figure, AB = CD and AD = CB. Prove that :

(i) △ABD ≅ △CDB

(ii) ∠A = ∠C

Answer

(i) △ABD ≅ △CDB

In △ABD and △CDB:

AB = CD $\quad$[Given]

AD = CB $\quad$[Given]

BD = DB $\quad$[Common side]

∴ △ABD ≅ △CDB $\quad$[By SSS condition of congruence]

(ii) ∠A = ∠C

Since △ABD ≅ △CDB, corresponding angles are equal.

∴ ∠A = ∠C $\quad$[CPCT]

⇒ Hence proved

In the given figure, ∠SPR = ∠QRP and ∠RSP = ∠PQR. Prove that :

(i) PQ = RS

(ii) PS = QR

Answer

To prove that the sides are equal, we first show that the two triangles formed by diagonal PR are congruent.

In △SPR and △QRP:

∠SPR = ∠QRP $\quad$[Given]

∠RSP = ∠PQR $\quad$[Given]

∴ ∠SRP = ∠QPR $\quad$[Third angles of triangles]

PR = RP $\quad$[Common side]

△SPR ≅ △QRP $\quad$[By ASA condition of congruence]

Therefore, by CPCT:

(i) PQ = RS

(ii) PS = QR

⇒ Hence proved

In the given figure, we have AO = BO and CO = DO. Prove that :

(i) △AOC ≅ △BOD

(ii) AC = BD

Answer

(i) To prove △AOC ≅ △BOD

In △AOC and △BOD:

AO = BO $\quad$[Given]

∠AOC = ∠BOD $\quad$[Vertically opposite angles are always equal]

CO = DO $\quad$[Given]

△AOC ≅ △BOD $\quad$[By SAS condition of congruence]

(ii) To prove AC = BD

Since △AOC ≅ △BOD, we have:

AC = BD $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

In the given figure, AB ⊥ BD, CD ⊥ BD and AB = CD. Prove that :

(i) △ABD ≅ △CDB

(ii) AD = CB

Answer

(i) To prove △ABD ≅ △CDB

In △ABD and △CDB:

AB = CD $\quad$[Given]

∠ABD = ∠CDB = 90° $\quad$[Right-angle]

BD = BD $\quad$[Common side to both triangles]

△ABD ≅ △CDB $\quad$[By SAS condition of congruence]

(ii) To prove AD = CB

Since △ABD ≅ △CDB, we have

AD = CB $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

In the given figure, PL ⊥ OA and PM ⊥ OB such that OL = OM. Prove that :

(i) △OLP ≅ △OMP

(ii) PL = PM

(iii) ∠LOP = ∠MOP

Answer

(i) To prove △OLP ≅ △OMP

In △OLP and △OMP:

OL = OM $\quad$[Given]

∠OLP = ∠OMP = 90° $\quad$[Given PL ⊥ OA and PM ⊥ OB]

OP = OP $\quad$[Common side to both triangles, acting as hypotenuse]

△OLP ≅ △OMP $\quad$[By RHS condition of congruence]

(ii) To prove PL = PM

Since △OLP ≅ △OMP, we have:

PL = PM $\quad$[Corresponding Parts of Congruent Triangles]

(iii) To prove ∠LOP = ∠MOP

Again, using the property of CPCT from congruence proof in part (i):

∠LOP = ∠MOP

⇒ Hence proved

In the given figure, we have PQ = SR and PR = SQ. Prove that :

(i) △PQR ≅ △SRQ

(ii) ∠PQR = ∠SRQ

Answer

(i) To prove △PQR ≅ △SRQ

In △PQR and △SRQ:

PQ = SR $\quad$[Given]

PR = SQ $\quad$[Given]

QR = QR $\quad$[Common base for both triangles]

△PQR ≅ △SRQ $\quad$[By SSS condition of congruence]

(ii) To prove ∠PQR = ∠SRQ

Since △PQR ≅ △SRQ, we can conclude that all their corresponding parts are equal.

By the property of CPCT (Corresponding Parts of Congruent Triangles):

∠PQR = ∠SRQ

⇒ Hence proved

In the given figure, we have AC ⊥ CD, BC ⊥ CD and DA = DB. Prove that : CA = CB.

Answer

To prove CA = CB, first prove △DCA ≅ △DCB

In △DCA and △DCB:

DA = DB $\quad$[Given]

∠DCA = ∠DCB = 90° $\quad$[Given AC ⊥ CD and BC ⊥ CD]

CD = CD $\quad$[Common side]

△DCA ≅ △DCB $\quad$[By RHS condition of congruence]

Since △DCA ≅ △DCB, we have:

CA = CB $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

In the adjoining figure, △ABC is an isosceles triangle in which AB = AC and AD is a median. Prove that :

(i) △ADB ≅ △ADC

(ii) ∠BAD = ∠CAD

Answer

(i) To prove △ADB ≅ △ADC

In △ADB and △ADC:

AB = AC $\quad$[Given]

BD = CD $\quad$[Given that AD is a median, which by definition bisects the base BC]

AD = AD $\quad$[Common side to both triangles]

△ADB ≅ △ADC $\quad$[By SSS condition of congruence]

(ii) To prove ∠BAD = ∠CAD

Since △ADB ≅ △ADC, we have:

∠BAD = ∠CAD $\quad$[Corresponding angles of congruent triangles]

⇒ Hence proved

In the adjoining figure, △ABC is an isosceles triangle in which AB = AC and AD is the bisector of ∠A. Prove that :

(i) △ADB ≅ △ADC

(ii) ∠B = ∠C

(iii) BD = DC

(iv) AD ⊥ BC

Answer

(i) To prove △ADB ≅ △ADC

In △ADB and △ADC:

AB = AC $\quad$[Given]

∠BAD = ∠CAD $\quad$[Given, as AD is the bisector of ∠A]

AD = AD $\quad$[Common side to both triangles]

∴ △ADB ≅ △ADC $\quad$[By SAS condition of congruence]

(ii) To prove ∠B = ∠C

Since △ADB ≅ △ADC, we have:

∠B = ∠C $\quad$[Corresponding angles of congruent triangles]

(iii) To prove BD = DC

Since △ADB ≅ △ADC, we have:

BD = DC $\quad$[Corresponding parts of congruent triangles]

(iv) To prove AD ⊥ BC

From CPCT, we know that:

∠ADB = ∠ADC

Since BC is a straight line, these two angles form a linear pair:

∴ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + ∠ADB = 180° $\quad$[∵∠ADB = ∠ADC]

⇒ 2∠ADB = 180°

⇒ ∠ADB = $\dfrac{180^\circ}{2}$

⇒ ∠ADB = 90°

Since the angle is 90°,

AD ⊥ BC

⇒ Hence proved

In the adjoining figure, △ABC is an isosceles triangle in which AB = AC. If BM ⊥ AC and CN ⊥ AB, prove that :

(i) △BMC ≅ △CNB

(ii) BM = CN

Answer

(i) To prove △BMC ≅ △CNB

Since AB = AC, the base angles of isosceles triangle ABC are equal.

∴ ∠ABC = ∠ACB

As N lies on AB and M lies on AC,

∠CBN = ∠BCM

Also,

∠BMC = ∠CNB = 90° $\quad$[Given, BM ⊥ AC and CN ⊥ AB]

∴ ∠MBC = ∠BCN $\quad$[Third angles of triangles]

In △BMC and △CNB:

∠MBC = ∠BCN $\quad$[Angle]

BC = CB $\quad$[Common side]

∠BCM = ∠CBN $\quad$[Angle]

∴ △BMC ≅ △CNB $\quad$[By ASA condition of congruence]

(ii) To prove BM = CN

Since △BMC ≅ △CNB, we have:

BM = CN $\quad$[Corresponding parts of congruent triangles]

⇒ Hence proved