Triangles — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Triangles" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Triangles" — 8 important questions with detailed answers for CBSE board exam prepar…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 9 Mathematics — Triangles. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Triangles — All 8 Questions Solved
Q1. State the SSS (Side-Side-Side) congruence criterion and apply it: If in triangle ABC, AB = 3 cm, BC = 4 cm, CA = 5 cm and in triangle PQR, PQ = 3 cm, QR = 4 cm, RP = 5 cm, prove that triangle ABC ≅ triangle PQR.
Step 1: SSS congruence criterion states that if all three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are congruent.
Step 2: In triangle ABC:
AB = 3 cm
BC = 4 cm
CA = 5 cm
Step 3: In triangle PQR:
PQ = 3 cm
QR = 4 cm
RP = 5 cm
Step 4: Comparing corresponding sides:
AB = PQ = 3 cm
BC = QR = 4 cm
CA = RP = 5 cm
Step 5: Since all three sides of triangle ABC are equal to corresponding sides of triangle PQR, by SSS criterion, triangle ABC ≅ triangle PQR.
Answer: Triangle ABC is congruent to triangle PQR (triangle ABC ≅ triangle PQR).
Q2. Apply the SAS (Side-Angle-Side) congruence criterion: In triangles ABD and CBD, AB = CB, angle ABD = angle CBD, and BD is common. Prove that triangle ABD ≅ triangle CBD.
Step 1: SAS congruence criterion states that if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.
Step 2: In triangle ABD:
AB = given
BD = common side
angle ABD = given
Step 3: In triangle CBD:
CB = given
BD = common side
angle CBD = given
Step 4: Comparing the triangles:
AB = CB (given)
angle ABD = angle CBD (given)
BD = BD (common side)
Step 5: The included angle between sides AB and BD in triangle ABD is angle ABD.
The included angle between sides CB and BD in triangle CBD is angle CBD.
Step 6: Since two sides and the included angle of triangle ABD are equal to corresponding two sides and included angle of triangle CBD, by SAS criterion, triangle ABD ≅ triangle CBD.
Answer: Triangle ABD is congruent to triangle CBD (triangle ABD ≅ triangle CBD).
Q3. Apply the ASA (Angle-Side-Angle) congruence criterion: In triangles PQR and XYZ, angle Q = angle Y, QR = YZ, and angle R = angle Z. Prove that triangle PQR ≅ triangle XYZ.
Step 1: ASA congruence criterion states that if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, then the triangles are congruent.
Step 2: In triangle PQR:
angle Q = given
QR = given (included side between angles Q and R)
angle R = given
Step 3: In triangle XYZ:
angle Y = given
YZ = given (included side between angles Y and Z)
angle Z = given
Step 4: Comparing the triangles:
angle Q = angle Y (given)
QR = YZ (given)
angle R = angle Z (given)
Step 5: QR is the side included between angles Q and R in triangle PQR.
YZ is the side included between angles Y and Z in triangle XYZ.
Step 6: Since two angles and the included side of triangle PQR are equal to the corresponding two angles and included side of triangle XYZ, by ASA criterion, triangle PQR ≅ triangle XYZ.
Answer: Triangle PQR is congruent to triangle XYZ (triangle PQR ≅ triangle XYZ).
Q4. Apply the RHS (Right angle-Hypotenuse-Side) congruence criterion: In right triangle ABC with right angle at B and right triangle DEF with right angle at E, if AC = DF (hypotenuses) and AB = DE (one side), prove that triangle ABC ≅ triangle DEF.
Step 1: RHS congruence criterion states that if the hypotenuse and one side of a right triangle are equal to the hypotenuse and corresponding side of another right triangle, then the triangles are congruent.
Step 2: In triangle ABC:
Angle B = 90° (given)
AC = hypotenuse
AB = one side
Step 3: In triangle DEF:
Angle E = 90° (given)
DF = hypotenuse
DE = one side
Step 4: Comparing the triangles:
Angle B = Angle E = 90° (both right angles)
AC = DF (given hypotenuses are equal)
AB = DE (given, corresponding sides are equal)
Step 5: By RHS criterion, when the hypotenuse and one side of a right triangle equal the hypotenuse and corresponding side of another right triangle, the triangles are congruent.
Step 6: Therefore, triangle ABC ≅ triangle DEF.
Answer: Triangle ABC is congruent to triangle DEF (triangle ABC ≅ triangle DEF).
Q5. In triangle ABC, if AB = AC, prove that the angles opposite to these sides are equal, i.e., angle C = angle B.
Step 1: Given: AB = AC in triangle ABC. To prove: angle C = angle B (angles opposite equal sides are equal).
Step 2: Draw the angle bisector of angle A, and let it meet BC at point D.
Step 3: In triangle ABD and triangle ACD:
AB = AC (given)
AD = AD (common side)
angle BAD = angle CAD (AD is angle bisector of angle A)
Step 4: By SAS congruence criterion, triangle ABD ≅ triangle ACD.
Step 5: From the congruence, corresponding angles are equal:
angle ABD = angle ACD
angle ADB = angle ADC
BD = CD
Step 6: angle ABD is the angle at B in triangle ABC, which is angle B.
angle ACD is the angle at C in triangle ABC, which is angle C.
Step 7: Therefore, angle B = angle C.
Answer: In a triangle, angles opposite to equal sides are equal (angle B = angle C).
Q6. In triangle ABC, angle A = 60°, angle B = 80°. Calculate angle C. What is the sum of all angles in the triangle?
Step 1: By the angle sum property of a triangle, the sum of all three angles equals 180°.
Step 2: Given:
angle A = 60°
angle B = 80°
angle C = ?
Step 3: Using angle sum property:
angle A + angle B + angle C = 180°
Step 4: Substitute known values:
60° + 80° + angle C = 180°
140° + angle C = 180°
Step 5: Solve for angle C:
angle C = 180° - 140°
angle C = 40°
Step 6: Verify: 60° + 80° + 40° = 180° ✓
Answer: angle C = 40°; Sum of all angles in the triangle = 180°.
Q7. In an isosceles triangle PQR, PQ = PR = 5 cm, and QR = 6 cm. If the altitude from P to QR has length h, find h using the property that the altitude bisects the base.
Step 1: Given isosceles triangle PQR with PQ = PR = 5 cm and QR = 6 cm.
Step 2: The altitude from P to QR bisects the base QR (property of isosceles triangles).
Step 3: Let the altitude from P meet QR at point M.
Then QM = MR = QR/2 = 6/2 = 3 cm
Step 4: In right triangle PQM:
angle PMQ = 90° (altitude is perpendicular)
PQ = 5 cm (hypotenuse)
QM = 3 cm (base)
PM = h (altitude/height)
Step 5: By Pythagoras theorem:
PQ² = PM² + QM²
5² = h² + 3²
25 = h² + 9
h² = 25 - 9
h² = 16
h = 4 cm
Step 6: Verify with triangle PMR:
PR² = PM² + MR²
5² = 4² + 3²
25 = 16 + 9
25 = 25 ✓
Answer: h = 4 cm.
Q8. Prove that the sum of any two sides of a triangle is greater than the third side. Use this to check if sides 3 cm, 4 cm, and 8 cm can form a triangle.
Step 1: Triangle Inequality Theorem: The sum of any two sides of a triangle must be greater than the third side.
Step 2: For a valid triangle with sides a, b, c:
a + b > c
b + c > a
a + c > b
Step 3: Check if sides 3 cm, 4 cm, and 8 cm form a triangle:
Let a = 3 cm, b = 4 cm, c = 8 cm
Step 4: Check condition 1:
a + b > c
3 + 4 > 8
7 > 8 (FALSE)
Step 5: Since the first condition itself fails, these sides cannot form a triangle.
Step 6: Verification with other conditions:
Condition 2: b + c > a → 4 + 8 > 3 → 12 > 3 (TRUE)
Condition 3: a + c > b → 3 + 8 > 4 → 11 > 4 (TRUE)
Step 7: For a valid triangle, ALL three conditions must be satisfied. Since 3 + 4 = 7 is not greater than 8, these three measurements cannot form the sides of a triangle.
Answer: Sides 3 cm, 4 cm, and 8 cm CANNOT form a triangle because 3 + 4 is not greater than 8.
Triangles — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Two triangles are congruent if:
- A. they have the same area.
- B. they have the same perimeter.
- C. they are exact copies of each other and can be made to superimpose.
- D. their corresponding angles are equal.
Answer: C. they are exact copies of each other and can be made to superimpose. — Congruent figures are exact copies of each other, meaning they have the same shape and size and can perfectly overlap.
MCQ 2. If three sides of one triangle are equal to three sides of another triangle, then the two triangles are congruent by which criterion?
- A. SAS
- B. ASA
- C. SSS
- D. RHS
Answer: C. SSS — The SSS (Side-Side-Side) congruence criterion states that if three sides of one triangle are respectively equal to the three sides of another triangle, then the triangles are congruent.
MCQ 3. In triangle ABC, if AB = AC, then which of the following is true?
- A. ∠B = ∠C
- B. ∠A = ∠B
- C. ∠A = ∠C
- D. ∠A + ∠B = 90°
Answer: A. ∠B = ∠C — In an isosceles triangle, the angles opposite to the equal sides are equal. Since AB = AC, the angles opposite to these sides, ∠C and ∠B respectively, must be equal.
MCQ 4. In ΔPQR, it is given that PQ = QR and ∠P = 50°. What is the measure of ∠R?
- A. 50°
- B. 80°
- C. 65°
- D. 40°
Answer: A. 50° — Since PQ = QR, ΔPQR is an isosceles triangle. In an isosceles triangle, angles opposite to equal sides are equal. Therefore, ∠R (opposite to PQ) must be equal to ∠P (opposite to QR), which is 50°.
MCQ 5. Given ΔABC and ΔDEF. If AB = DE, ∠B = ∠E, and BC = EF, which congruence criterion applies?
- A. SSS
- B. SAS
- C. ASA
- D. AAS
Answer: B. SAS — The SAS (Side-Angle-Side) congruence criterion states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.
MCQ 6. Which congruence criterion is specific to right-angled triangles?
- A. SSS
- B. SAS
- C. ASA
- D. RHS
Answer: D. RHS — The RHS (Right angle-Hypotenuse-Side) congruence criterion is used specifically for right-angled triangles, comparing the hypotenuse and one side.
MCQ 7. In ΔXYZ, if XY = XZ and ∠X = 70°, what are the measures of ∠Y and ∠Z?
- A. ∠Y = 55°, ∠Z = 55°
- B. ∠Y = 70°, ∠Z = 40°
- C. ∠Y = 40°, ∠Z = 70°
- D. ∠Y = 60°, ∠Z = 50°
Answer: A. ∠Y = 55°, ∠Z = 55° — Since XY = XZ, ΔXYZ is an isosceles triangle, so ∠Y = ∠Z. The sum of angles in a triangle is 180°, so 70° + 2∠Y = 180°, leading to ∠Y = 55°. Thus, ∠Y = ∠Z = 55°.
MCQ 8. In ΔABC, AB = 5 cm, BC = 8 cm, and CA = 6 cm. Which angle is the largest?
- A. ∠A
- B. ∠B
- C. ∠C
- D. Cannot be determined
Answer: A. ∠A — In a triangle, the angle opposite to the longest side is the largest. Here, the longest side is BC = 8 cm, and the angle opposite to BC is ∠A, so ∠A is the largest angle.
MCQ 9. If ΔABC ≅ ΔPQR, which of the following statements is NOT necessarily true?
- A. AB = PQ
- B. ∠A = ∠P
- C. AC = PR
- D. BC = PR
Answer: D. BC = PR — If two triangles are congruent, their corresponding parts are equal (CPCTC). Here, BC corresponds to QR, not PR. Therefore, BC = QR must be true, but BC = PR is not necessarily true.
MCQ 10. An altitude of a triangle is a line segment from a vertex perpendicular to:
- A. the opposite side.
- B. another vertex.
- C. the midpoint of the opposite side.
- D. the angle bisector.
Answer: A. the opposite side. — An altitude of a triangle is a line segment from a vertex to the opposite side, forming a right angle (perpendicular) with that side.
MCQ 11. In ΔABC, AD is the median to BC. If BD = 5 cm, what is the length of BC?
- A. 5 cm
- B. 10 cm
- C. 7.5 cm
- D. Cannot be determined
Answer: B. 10 cm — A median connects a vertex to the midpoint of the opposite side. If D is the midpoint of BC, then BD = DC. Since BD = 5 cm, DC = 5 cm, so BC = BD + DC = 10 cm.
MCQ 12. In quadrilateral ABCD, AC = AD and AB bisects ∠A. Which congruence criterion proves ΔABC ≅ ΔABD?
- A. SSS
- B. SAS
- C. ASA
- D. AAS
Answer: B. SAS — Given AC = AD and AB bisects ∠A (so ∠CAB = ∠DAB). AB is a common side. By SAS (Side-Angle-Side) criterion (AC = AD, ∠CAB = ∠DAB, AB = AB), ΔABC ≅ ΔABD.
MCQ 13. In ΔPQR, PQ = PR and the exterior angle at R is 120°. What is the measure of ∠P?
- A. 60°
- B. 80°
- C. 40°
- D. 50°
Answer: A. 60° — If the exterior angle at R is 120°, then ∠PRQ = 180° - 120° = 60°. Since PQ = PR, ΔPQR is isosceles, so ∠PQR = ∠PRQ = 60°. Thus, ∠P = 180° - (60° + 60°) = 60°.
MCQ 14. In ΔABC and ΔDEF, given ∠A = ∠D, ∠B = ∠E, and BC = EF. Which congruence criterion proves ΔABC ≅ ΔDEF?
- A. SSS
- B. SAS
- C. ASA
- D. AAS
Answer: D. AAS — We are given two angles (∠A, ∠B) and a non-included side (BC) corresponding to two angles (∠D, ∠E) and a non-included side (EF). This directly applies the AAS (Angle-Angle-Side) congruence criterion.
MCQ 15. Which set of lengths can form the sides of a triangle?
- A. 2 cm, 3 cm, 6 cm
- B. 4 cm, 5 cm, 9 cm
- C. 3 cm, 4 cm, 5 cm
- D. 1 cm, 2 cm, 3 cm
Answer: C. 3 cm, 4 cm, 5 cm — For three lengths to form a triangle, the sum of any two sides must be greater than the third side. Only 3 cm, 4 cm, 5 cm satisfies this (3+4 > 5, 3+5 > 4, 4+5 > 3).
MCQ 16. If a point P lies on the perpendicular bisector of a line segment AB, then...
- A. PA = PB
- B. PA > PB
- C. PA < PB
- D. P is the midpoint of AB
Answer: A. PA = PB — Any point on the perpendicular bisector of a line segment is equidistant from the two endpoints of the segment. Thus, PA = PB.
MCQ 17. AD is an altitude of an isosceles triangle ABC in which AB = AC. Which of the following is true about AD?
- A. AD bisects BC only.
- B. AD bisects ∠A only.
- C. AD bisects BC and also bisects ∠A.
- D. AD does not bisect BC nor ∠A.
Answer: C. AD bisects BC and also bisects ∠A. — In an isosceles triangle, the altitude drawn from the vertex angle to the base also acts as the median to the base and the angle bisector of the vertex angle. So, AD bisects BC and ∠A.
MCQ 18. In ΔLMN, if ∠L = 60°, ∠M = 70°, and ∠N = 50°, which side is the shortest?
- A. LM
- B. MN
- C. LN
- D. Cannot be determined
Answer: B. MN — In a triangle, the side opposite to the smallest angle is the shortest. Here, the smallest angle is ∠N = 50°. The side opposite to ∠N is MN, making it the shortest side.
MCQ 19. Two line segments AB and CD intersect at a point O such that AO = BO and CO = DO. Which of the following is true?
- A. ΔAOC ≅ ΔBOD by SSS
- B. ΔAOC ≅ ΔBOD by SAS
- C. ΔAOC ≅ ΔBOD by ASA
- D. ΔAOC ≅ ΔBOD by AAS
Answer: B. ΔAOC ≅ ΔBOD by SAS — We are given AO = BO and CO = DO. Also, ∠AOC and ∠BOD are vertically opposite angles, so ∠AOC = ∠BOD. Therefore, by SAS (Side-Angle-Side) congruence criterion, ΔAOC ≅ ΔBOD.
MCQ 20. If the lengths of two sides of a triangle are 7 cm and 10 cm, what is the possible range for the length of the third side (x)?
- A. 3 cm < x < 17 cm
- B. x = 3 cm or x = 17 cm
- C. x < 3 cm or x > 17 cm
- D. 3 cm ≤ x ≤ 17 cm
Answer: A. 3 cm < x < 17 cm — For a triangle to be formed, the sum of any two sides must be greater than the third side, and the difference of any two sides must be less than the third side. So, (10 - 7) < x < (10 + 7), which simplifies to 3 cm < x < 17 cm.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.