CG Board Class 9 Mathematics Chapter 10 Solutions: Congruency of Triangles (त्रिभुजों की सर्वांगसमता) – 2026-27

• Solution / :
• Corresponding vertices: A rightarrow D, O rightarrow O, C rightarrow B.
• Correct Option: (b) Δ AOC ≅ Δ DOB.

• Solution / :
• Matching sides: AB = 3.3 cm = QP, BC = 4 cm = PR, AC = 3 cm = QR.
• Correct Option: (b) Δ ABC ≅ Δ QPR.

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• AAA (Angle-Angle-Angle) gives similar triangles, not necessarily congruent triangles.
• Correct Option: (c) AAA.

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• Correct Option: (b) At least one corresponding side is equal.

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• Since ∠ B = ∠ C ⇒ AC = AB ⇒ AB : AC = 1 : 1.
• Correct Option: (c) 1 : 1.

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• Since Δ ABC ≅ Δ PQR ⇒ AC = PR (CPCT).
• 3x + 2 = 6x – 13 ⇒ 6x – 3x = 2 + 13 ⇒ 3x = 15 ⇒ x = 5.
• Since Δ ABC ≅ Δ PQR ⇒ QR = BC = 5x.
• QR = 5(5) = 25.

• Solution / :
• By SSS / ASA congruency rule between the two triangles formed across the intersection point O:
• Distance AB = CD = 160 m.

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• *Yes. The path forms a rectangle where opposite sides and diagonal are equal (AB = CD, AC = DB, BC = CB).

• Solution / :
• In Δ ABC and Δ CDA:
• BC = AD (Opposite sides of parallelogram ABCD)
• AB = CD (Opposite sides of parallelogram ABCD)
• AC = CA (Common side)
• By SSS congruence rule, Δ ABC ≅ Δ CDA.

• Solution / :
• Given: ∠ BAD = ∠ EAC.
• Add ∠ DAC to both sides:
• ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC ⇒ ∠ BAC = ∠ DAE.
• In Δ ABC and Δ ADE:
• AB = AD (Given)
• ∠ BAC = ∠ DAE (Proved above)
• AC = AE (Given)
• By SAS congruence rule, Δ ABC ≅ Δ ADE.
• Therefore, BC = DE (CPCT). Hence proved.

Correct Option: (iii) 60°.

• Solution / :
• ∠ A = ∠ B ⇒ BC = AC ⇒ AC : BC = 1 : 1.
• Correct Option: (i) 1 : 1.

Correct Option: (iv) 80°.

• Solution / :
• ∠ C = ∠ A ⇒ BC = AB = 4 cm.
• Correct Option: (iii) 4 cm.

Correct Option: (iii) 35°.

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• Each interior angle of a regular hexagon = frac{(6-2) × 180°{6} = 120°.
• The angle x°= frac{180°- 120°{2} = {30°.

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• Since AB = BC ⇒ ∠ C = ∠ A = 2x + 8.
• Sum of angles = (2x + 8) + (4x – 20) + (2x + 8) = 180°.
• 8x – 4 = 180° ⇒ 8x = 184° ⇒ {x = 23°.
• Angles: ∠ A = 54°, ∠ C = 54°, ∠ B = 72°.
• Since all angles are less than 90°, it is an *Acute angled triangle.

• Solution / :
• In Δ ABC and Δ DEF:
• BC = EF (Given)
• ∠ B = ∠ E (Corresponding angles since AB ∥ ED)
• ∠ C = ∠ F (Corresponding angles since CA ∥ FD)
• By ASA congruence rule, Δ ABC ≅ Δ DEF. Hence proved.

• Solution / :

1. In Δ ABD and Δ BAC:

• AD = BC (Given)
• ∠ DAB = ∠ CBA (Given)
• AB = BA (Common side)
• By SAS congruence rule, Δ ABD ≅ Δ BAC.

1. Since Δ ABD ≅ Δ BAC ⇒ BD = AC (CPCT).
2. Since Δ ABD ≅ Δ BAC ⇒ ∠ ABD = ∠ BAC (CPCT). Hence proved.

• Solution / :
• Given BE = FC.
• Add EC to both sides: BE + EC = FC + EC ⇒ BC = FE.
• In Δ ABC and Δ DFE:
• AB = DF (Given)
• AC = DE (Given)
• BC = FE (Proved above)
• By SSS congruence rule, Δ ABC ≅ Δ DFE. Hence proved.

• Solution / :
• In Δ AOB: ∠ B < ∠ A ⇒ AO < BO ------ (1) (Side opposite to smaller angle is smaller)
• In Δ COD: ∠ C < ∠ D ⇒ OD < OC ------ (2)
• Adding equations (1) and (2):
• AO + OD < BO + OC ⇒ AD < BC. Hence proved.

• Solution / :
• Interior angle at M: ∠ PMN = 180°- x.
• Interior angle at N: ∠ PNM = 180°- y.
• Since x > y ⇒ -x < -y ⇒ 180°- x < 180°- y ⇒ ∠ PMN < ∠ PNM.
• Side opposite to larger angle ∠ PNM is MP, and side opposite to smaller angle ∠ PMN is NP.
• Therefore, MP > NP. Hence proved.

• Solution / :
• In Δ PQR: PQ > PR ⇒ ∠ PRQ > ∠ PQR.
• Dividing by 2: 1/2∠ PRQ > 1/2∠ PQR ⇒ ∠ SRQ > ∠ SQR.
• In Δ SQR, side opposite to larger angle ∠ SRQ is SQ, and side opposite to smaller angle ∠ SQR is SR.
• Therefore, SQ > SR. Hence proved.

• Solution / :
• In Δ PQR: PQ = PR ⇒ ∠ PQR = ∠ PRQ.
• Exterior angle of Δ PQS at Q: ∠ PQR = ∠ S + ∠ QPS > ∠ S.
• Thus, ∠ PRQ = ∠ PQR > ∠ S.
• In Δ PRS, since ∠ PRQ > ∠ S ⇒ PS > PR.
• Since PR = PQ, we get PS > PQ. Hence proved.