Class 9 Maths Chapter 7 Triangles NCERT Solutions: All Exercises SAS ASA SSS RHS Solved Step-by-Step

March 16, 2026

Class 9 Maths Chapter 7 Triangles NCERT Solutions: All Exercises SAS ASA SSS RHS Solved Step-by-Step

Introduction

Triangles are everywhere in geometry, and Chapter 7 dives into what makes them congruent or similar using SAS, ASA, SSS, and RHS rules. You'll prove sides/angles equal and tackle inequalities like longer side opposite larger angle.

Two exercises: congruence criteria and inequalities. Builds proof skills for boards. Let's solve them all with clear steps.

Key Formulas and Theorems

SAS: Two sides + included angle equal → congruent triangles.
ASA: Two angles + included side.
SSS: Three sides equal.
RHS: Right triangle hypotenuse + one side.
Inequalities: Larger side opposite larger angle; sum any two sides > third.

Exercise 7.1 Solved Questions (Congruence)

Q1: Fill blanks: Two figures congruent if coincide when superimposed.

Solution: Yes, matching shapes/sizes.

Q2: Given triangles congruent, corresponding parts equal (CPCT).

Solution: Sides opposite equal angles equal.

Q3: Consider ΔABC ≅ ΔDEF. Write parts.

Solution: AB=DE, BC=EF, CA=FD; ∠A=∠D etc.

Q4: SAS: Two sides included angle.

Solution: ΔABC, AB=DE, AC=DF, ∠A=∠D → congruent.

Q5: Is SSS valid? Yes.

Q6: Fill: Two figures congruent when they coincide by superposition.

Solution: Correct. Same shape and size.

Q7: ΔABC ≅ ΔPQR. Fill corresponding vertices: A-P, B-Q, C-R.

Solution: Order matters: AB=PQ, BC=QR, CA=RP.

Q8: If ΔABC ≅ ΔDEF, write equal pairs.

Solution: AB=DE, BC=EF, CA=FD; ∠A=∠D, ∠B=∠E, ∠C=∠F.

Q9: SAS example: AB=XY, ∠A=∠X, AC=XZ → ΔABC ≅ ΔXYZ.

Solution: Two sides + included angle.

Q10: Is SSS criteria valid? Yes.

Exercise 7.2 Solved Questions (Inequalities)

Q1: Sides 3cm,4cm,5cm form triangle? Yes (3+4>5 etc).

Solution: Check inequalities: 3+4>5, 3+5>4, 4+5>3.

Q2: AB=AC, prove ∠B=∠C.

Solution: Angle bisector AD to BC. ΔABD=ΔACD (SAS). CPCT ∠ABD=∠ACD → ∠B=∠C.

Q3: ∠B>∠C, prove AB>AC.

Solution: Assume AB≤AC contradicts. Use exterior angle.

Q4: D on BC, AD⊥BC, AB>AC prove BD>CD.

Solution: From RHS or inequality theorem.

Q5: Shortest from point to line is perpendicular.

Solution: PM⊥l, PN>l other N. In ΔPMN, ∠PMP=90>∠NPN → PM<PN.

Exercise 7.2 Solved Questions (Inequalities)

Q1: Can sides 3cm, 4cm, 8cm form triangle? No.

Solution: 3+4=7<8. Violates triangle inequality.

Q2: AB=AC (isosceles). Prove ∠ABC=∠ACB.

Solution: Draw angle bisector AD to BC. ΔABD=ΔACD (SAS: AB=AC, AD common, ∠BAD=∠CAD). CPCT: ∠ABC=∠ACB.

Q3: In ΔABC, ∠B>∠C. Prove AB>AC.

Solution: Assume AB≤AC (contradiction). Draw BD perp AC at D. In ΔABD, ∠ADB=90°>∠C (in ΔABC). So AB>AC.

Q4: D on BC, AD perp BC, AB>AC. Prove BD>CD.

Solution: In ΔABD, ΔACD (both right-angled). AB>AC → BD>CD (larger hypotenuse opp larger leg).

Q5: Shortest distance from point P to line l is perpendicular PM.

Solution: For any N on l, in ΔPMN: ∠MPN=90°>∠MNP → PM<PN.

Extra Practice Questions Solved

Extra Q1: ΔABC, ΔDEF. AB=DE=5cm, BC=EF=6cm, ∠B=∠E=50°. Prove congruent.

Solution: SAS (AB=DE, ∠B=∠E, BC=EF).

Extra Q2: Sides 5,12,13 vs 6,8,10. Congruent? No.

Solution: SSS fails: 5≠6,12≠8,13≠10.

Extra Q3: Right ΔABC (∠C=90°), ΔDEF (∠F=90°). AC=DF=8cm, BC=EF=15cm. Prove congruent.

Solution: RHS (hypotenuse BC=EF, leg AC=DF).

Extra Q4: ΔABC isosceles AB=CB. D midpoint AB. Prove ∠CDA=∠CDB.

Solution: CD common, AD=DB (midpoint), AC=BC → SSS ΔCDA=ΔCDB.

Extra Q5: Sides 7cm,8cm,9cm possible? Yes.

Solution: 7+8>9, 7+9>8, 8+9>7 all true.

Extra Q6: ΔABC, ∠A=40°, ∠B=60°. Find ∠C.

Solution: 180-40-60=80°.

Extra Q7: Prove: In ΔABC, if AB>BC>CA, then ∠C>∠A>∠B.

Solution: Largest side opp largest angle.

15 FAQs

Congruent triangles mean? Same size/shape.
SAS criteria? Two sides + included angle.
ASA? Two angles + included side.
SSS? All three sides.
RHS for? Right triangles hypotenuse+side.
CPCT? Corresponding parts congruent triangles.
Triangle inequality? Two sides sum > third.
Larger angle opposite? Longer side.
Angle bisector in isosceles? Proves base angles equal.
Perpendicular shortest? Yes to line.
Can 2,3,6 sides triangle? No, 2+3<6.
Exterior angle? > interior opposite.
Proof SAS? Superposition.
AAS valid? Yes, derives from ASA.
Sum two sides > third? Always in triangle.

These proofs tricky at first but click with practice. Drop by www.fuzymathacademy.com for my video walkthroughs and live Q&A on triangles. You'll crush Chapter 7 exams easy.