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Triangles: Class 10 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 10 paper design. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1MCQ1 mark

In △ABC, XY ∥ BC where X is on AB and Y is on AC. If AX = 4 cm, XB = 6 cm and AY = 5 cm, then the length of AC is:

Diagram for question 1: Triangles
Show answer
Option (A) is correct.

Explanation: By the Basic Proportionality Theorem, if XY ∥ BC with X on AB and Y on AC, then AX/XB = AY/YC ⟹ 4/6 = 5/YC ⟹ YC = (5 × 6)/4 = 7.5 cm ∴ AC = AY + YC = 5 + 7.5 = 12.5 cm.
Q2MCQ1 mark

If △DEF ~ △PQR, ∠D = 47° and ∠Q = 58°, then the measure of ∠F is:

Show answer
Option (A) is correct.

Explanation: Since △DEF ∼ △PQR, corresponding angles are equal, giving ∠D = ∠P, ∠E = ∠Q, and ∠F = ∠R. Given ∠D = 47° and ∠Q = 58°, we have ∠E = 58°. By the angle sum property of △DEF: 47° + 58° + ∠F = 180° ⟹ ∠F = 75°.
Q3MCQ1 mark

In △PQR and △XYZ, ∠Q = ∠Y = 90°, PQ = 6 cm, QR = 8 cm, and XY = 9 cm. If △PQR ~ △XYZ, what is the length of YZ?

Show answer
Option (B) is correct.

Explanation: Since △PQR ~ △XYZ with ∠Q = ∠Y = 90°, the corresponding sides are in proportion: PQ/XY = QR/YZ. Substituting, 6/9 = 8/YZ ⟹ YZ = (8 × 9)/6 = 12 cm.
Q4MCQ1 mark

In △PQR and △STU, PQ = 6 cm, QR = 8 cm, PR = 10 cm, ST = 3 cm, TU = 4 cm, and ∠Q = ∠T = 90°. The length of side SU is:

Show answer
Option (A) is correct.

Explanation: By the SAS Similarity Criterion, since ∠Q = ∠T = 90° and ST/PQ = TU/QR = 3/6 = 4/8 = 1/2, △STU ∼ △PQR. ∴ SU/PR = 1/2 ⟹ SU = (1/2) × 10 = 5 cm.
Q5MCQ1 mark

In triangles ABC and DEF, if ∠A = ∠D and AB/DE = AC/DF, then the two triangles are similar by which criterion?

Show answer
Option (A) is correct.

Explanation: By the SAS Similarity Criterion, if one pair of corresponding angles is equal and the sides including those angles are proportional, then the two triangles are similar. Here, ∠A = ∠D and AB/DE = AC/DF, where ∠A and ∠D are the angles included between the proportional sides AB/DE and AC/DF respectively. ∴ △ABC ∼ △DEF by the SAS similarity criterion.
Q6MCQ1 mark

A flagpole of height 6 m casts a shadow of 4 m on the ground. At the same time, a building casts a shadow of 34 m. What is the height of the building?

Diagram for question 6: Triangles
Show answer
Option (C) is correct.

Explanation: By the AA Similarity Criterion, a vertical pole and its shadow form a triangle similar to the building and its shadow (same sun angle, both perpendicular to the ground), so the ratio of height to shadow length is constant. ⟹ 6/4 = h/34 ⟹ h = (6 × 34)/4 = 204/4 = 51 m.
Q7MCQ1 mark

The perimeters of two similar triangles △MNP and △XYZ are 72 cm and 48 cm respectively. If XY = 16 cm, then the length of MN is:

Show answer
Option (A) is correct.

Explanation: For similar triangles, the ratio of corresponding sides equals the ratio of their perimeters. ⟹ MN/XY = Perimeter of △MNP / Perimeter of △XYZ ⟹ MN/16 = 72/48 ⟹ MN = (72 × 16)/48 = 24 cm.
Q8MCQ1 mark

In triangles PQR and XYZ, ∠Q = ∠Y, ∠R = ∠Z and PQ = 5 XY. Then the two triangles are:

Show answer
Option (B) is correct.

Explanation: By the AA similarity criterion, since ∠Q = ∠Y and ∠R = ∠Z, △PQR ∼ △XYZ. However, PQ = 5XY gives a side ratio of 5 : 1 ≠ 1 : 1, so the corresponding sides are not equal and the triangles are not congruent. Hence the two triangles are similar but not congruent.
Q9MCQ1 mark

In a figure, CD ∥ MN. If CD = 8 cm, MN = 4 cm and OD = 6 cm, then the length of ON is:

Diagram for question 9: Triangles
Show answer
Option (B) is correct.

Explanation: Since CD ∥ MN, by AA similarity criterion △OCD ∼ △OMN (corresponding angles are equal). By the property of similar triangles, ON/OD = MN/CD ⟹ ON/6 = 4/8 = 1/2 ⟹ ON = 3 cm.
Q10MCQ1 mark

In triangles PQR and XYZ, . Which of the following conditions makes the two triangles similar?

Diagram for question 10: Triangles
Show answer
Option (B) is correct.

Explanation: By the SAS Similarity Criterion, two triangles are similar if two pairs of corresponding sides are proportional and the included angle between those sides is equal. Here, the given ratio is PQ/XY = PR/XZ, so the sides forming the ratio are PQ, PR in △PQR and XY, XZ in △XYZ. The angle included between PQ and PR is ∠P, and the angle included between XY and XZ is ∠X. ∴ ∠P = ∠X is the required condition ⟹ △PQR ∼ △XYZ by SAS similarity.
Q11MCQ1 mark

In triangle PQR, a line DE is drawn such that D lies on PQ and E lies on PR, with DE || QR. If PD = 3.5 cm, DQ = 2.5 cm and PE = 4.2 cm, what is the length of ER?

Diagram for question 11: Triangles
Show answer
Option (A) is correct.

Explanation: By the Basic Proportionality Theorem, if DE ∥ QR with D on PQ and E on PR, then PD/DQ = PE/ER ⟹ 3.5/2.5 = 4.2/ER ⟹ ER = (4.2 × 2.5)/3.5 = 10.5/3.5 = 3.0 cm.
Q12MCQ1 mark

In a figure, PQ ∥ RS. If PQ = 9 cm, RS = 3 cm and QT = 6 cm, then the length of ST is:

Diagram for question 12: Triangles
Show answer
Option (A) is correct.

Explanation: Since PQ ∥ RS, by the AA similarity criterion (alternate interior angles ∠TPQ = ∠TRS and ∠TQP = ∠TSR), △TPQ ∼ △TRS. By the property of similar triangles, corresponding sides are proportional: ST/QT = RS/PQ ⟹ ST/6 = 3/9 = 1/3 ⟹ ST = 2 cm.
Q13MCQ1 mark

In △PQR ~ △LMN, if PQ = 8 cm, QR = 6 cm, LM = 12 cm and MN = y cm, then the value of y is:

Show answer
Option (A) is correct.

Explanation: Since △PQR ∼ △LMN, corresponding sides are proportional ⟹ PQ/LM = QR/MN ⟹ 8/12 = 6/y ⟹ y = (6 × 12)/8 = 72/8 = 9 cm.
Q14MCQ1 mark

A flagpole of height 9 m casts a shadow 6 m long on the ground. At the same time, a building casts a shadow 42 m long. What is the height of the building?

Show answer
Option (A) is correct.

Explanation: By the AA Similarity criterion, the flagpole and the building each form similar right triangles with their respective shadows (same angle of elevation of the sun). ⟹ height/shadow length is constant for both. ⟹ 9/6 = H/42 ⟹ H = (9 × 42)/6 = 63 m.
Q15MCQ1 mark

In triangles ABC and DEF, ∠A = ∠D = 50° and ∠B = ∠E = 70°. Which of the following statements is correct about the two triangles?

Show answer
Option (A) is correct.

Explanation: By the AA Similarity Criterion, two triangles are similar if two pairs of corresponding angles are equal. In △ABC: ∠A = 50°, ∠B = 70° ⟹ ∠C = 180° − 50° − 70° = 60°. In △DEF: ∠D = 50°, ∠E = 70° ⟹ ∠F = 60°. Since ∠A = ∠D and ∠B = ∠E, two pairs of corresponding angles are equal, so △ABC ∼ △DEF by the AA similarity criterion. Side lengths are not given, so SSS or SAS similarity cannot be applied.
Q16MCQ1 mark

In △ABC, D is a point on AB and E is a point on AC such that DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 2.4 cm, then the length of EC is:

Diagram for question 16: Triangles
Show answer
Option (A) is correct.

Explanation: By the Basic Proportionality Theorem, if DE ∥ BC, then AD/DB = AE/EC ⟹ 3/5 = 2.4/EC ⟹ EC = (2.4 × 5)/3 = 12/3 = 4 cm.
Q17MCQ1 mark

If △KLM ~ △XYZ, KL = 9 cm, XY = 12 cm and the perimeter of △XYZ is 48 cm, then the perimeter of △KLM is:

Show answer
Option (A) is correct.

Explanation: Since △KLM ∼ △XYZ, the ratio of their perimeters equals the ratio of their corresponding sides. ⟹ KL/XY = 9/12 = 3/4 ⟹ Perimeter of △KLM / Perimeter of △XYZ = 3/4 ⟹ Perimeter of △KLM = (3/4) × 48 = 36 cm.
Q18MCQ1 mark

In triangles LMN and PQR, . Which of the following conditions makes the two triangles similar by SAS similarity criterion?

Diagram for question 18: Triangles
Show answer
Option (A) is correct.

Explanation: By the SAS Similarity Criterion, two triangles are similar if two pairs of corresponding sides are in proportion and the angles included between those sides are equal. The given ratio LM/QR = MN/PQ shows that sides LM and MN (meeting at vertex M in △LMN) are proportional to sides QR and PQ (meeting at vertex Q in △PQR). The included angle between LM and MN is ∠M, and the included angle between QR and PQ is ∠Q. ∴ the required condition is ∠M = ∠Q, giving △LMN ∼ △QRP by SAS similarity.
Q19MCQ1 mark

In △LMN, XY ∥ MN. It is given that LX = 1.8 cm, XM = 2.7 cm and MN = 6.0 cm. What is the length of XY?

Diagram for question 19: Triangles
Show answer
Option (A) is correct.

Explanation: Since XY ∥ MN in △LMN, by the Basic Proportionality Theorem △LXY ∼ △LMN, giving LX/LM = XY/MN. Here LM = LX + XM = 1.8 + 2.7 = 4.5 cm ⟹ XY = (1.8 × 6.0)/4.5 = 10.8/4.5 = 2.4 cm.
Q20MCQ1 mark

In the given figure, DE || BC. If AD = 3.6 cm, DB = 2.4 cm and AE = 5.4 cm, then the value of EC is:

Diagram for question 20: Triangles
Show answer
Option (A) is correct.

Explanation: By the Basic Proportionality Theorem, if DE ∥ BC, then AD/DB = AE/EC ⟹ 3.6/2.4 = 5.4/EC ⟹ EC = (5.4 × 2.4)/3.6 = 12.96/3.6 = 3.6 cm.
Q21MCQ1 mark

In triangles ABC and DEF, ∠A = ∠D and ∠C = ∠F, and AC = 4 EF. Then the two triangles are:

Show answer
Option (B) is correct.

Explanation: By the AA (Angle–Angle) Similarity Criterion, since ∠A = ∠D and ∠C = ∠F, △ABC ∼ △DEF. However, AC = 4EF gives the ratio of corresponding sides as 4 : 1 ≠ 1 : 1, so the triangles are not congruent.
Q22MCQ1 mark

In the given figure, ∠P = ∠R, PQ = 8 cm, PS = 16 cm, RS = 6 cm. Then the length of RT is:

Diagram for question 22: Triangles
Show answer
Option (A) is correct.

Explanation: Since ∠P = ∠R (given) and ∠PSQ = ∠RST (vertically opposite angles), by AA similarity criterion △PSQ ∼ △RST. ⟹ PQ/RT = PS/RS ⟹ 8/RT = 16/6 ⟹ RT = (8 × 6)/16 = 48/16 = 3 cm.
Q23MCQ1 mark

In triangles ABC and PQR, . Which of the following additional conditions makes the two triangles similar by SAS similarity criterion?

Show answer
Option (B) is correct.

Explanation: The SAS Similarity Criterion states that if two sides of one triangle are proportional to the corresponding two sides of another triangle and the included angles are equal, then the triangles are similar. Here, AB/PQ = AC/PR gives the two proportional side-pairs; the angle included between sides AB and AC in △ABC is ∠A, and the angle included between the corresponding sides PQ and PR in △PQR is ∠P. ∴ the additional condition required is ∠A = ∠P.
Q24MCQ1 mark

If △LMN ~ △XYZ, LM = 10 cm, XY = 15 cm and the perimeter of △LMN is 45 cm, then the perimeter of △XYZ is:

Show answer
Option (B) is correct.

Explanation: For similar triangles, the ratio of their perimeters equals the ratio of their corresponding sides. Since △LMN ∼ △XYZ, Perimeter of △XYZ / Perimeter of △LMN = XY / LM ⟹ Perimeter of △XYZ / 45 = 15 / 10 = 3 / 2 ⟹ Perimeter of △XYZ = 45 × 3/2 = 67.5 cm.
Q25Short Answer2 marks

In the given figure, △ABC is right-angled at C. D is a point on AB such that CD ⊥ AB. Show that BC² = BD × AB.

Diagram for question 25: Triangles
Show answer
In △BDC and △BCA:

∠BDC = ∠BCA = 90° (given CD ⊥ AB, and ∠BCA = 90°)

∠B is common to both triangles.

⟹ By AA Similarity Criterion, △BDC ∼ △BCA.

Since corresponding sides of similar triangles are proportional:

BC/BA = BD/BC

⟹ BC² = BD × BA = BD × AB

Hence proved.
Q26Short Answer2 marks

In △ABC, D is a point on side AB and E is a point on side AC such that DE ∥ BC. If AD = 2 cm, DB = 6 cm and AE = 1.5 cm, find the length of EC.

Diagram for question 26: Triangles
Show answer
Since DE ∥ BC, by the Basic Proportionality Theorem,

AD/DB = AE/EC

⟹ 2/6 = 1.5/EC

⟹ EC = (1.5 × 6)/2 = 9/2

∴ EC = 4.5 cm
Q27Short Answer2 marks

In △ABC, a point F is taken on side AB such that ∠AFC = ∠ACB. Show that AC² = AF · AB.

Diagram for question 27: Triangles
Show answer
In △AFC and △ACB:

∠AFC = ∠ACB (given) ...(i)

∠A = ∠A (common) ...(ii)

By AA Similarity Criterion, △AFC ∼ △ACB.

⟹ Corresponding sides are proportional:

AF/AC = AC/AB

⟹ AC² = AF · AB

Hence proved.
Q28Short Answer2 marks

In △ABC, ∠ABC = 90°. D is a point on AC such that BD ⊥ AC. Show that BC² = DC × AC.

Diagram for question 28: Triangles
Show answer
In △ABC and △BDC:

∠BCA = ∠BCD (common angle)

∠ABC = ∠BDC = 90°

By AA Similarity Criterion, △ABC ∼ △BDC

⟹ BC/DC = AC/BC

⟹ BC² = DC × AC

Hence proved.
Q29Short Answer2 marks

The perimeters of two similar triangles △DEF and △LMN are 56 cm and 35 cm respectively. If one side of △LMN is 10 cm, find the length of the corresponding side of △DEF.

Show answer
Since △DEF ∼ △LMN, the ratio of corresponding sides equals the ratio of their perimeters.

⟹ (Corresponding side of △DEF) / (Corresponding side of △LMN) = Perimeter of △DEF / Perimeter of △LMN

⟹ (Side of △DEF) / 10 = 56 / 35 = 8 / 5

⟹ Side of △DEF = 10 × (8/5) = 16

∴ The length of the corresponding side of △DEF is 16 cm.
Q30Short Answer2 marks

BX and QY are altitudes of triangles ABC and PQR respectively such that △ABX ~ △PQY. Prove that △ABC ~ △PQR.

Diagram for question 30: Triangles
Show answer
Since BX and QY are altitudes of △ABC and △PQR respectively,

∠BXA = ∠QYP = 90°

Since △ABX ∼ △PQY (given), corresponding angles are equal.

⟹ ∠BAX = ∠QPY, i.e., ∠BAC = ∠QPR …(i)

⟹ ∠ABX = ∠PQY, i.e., ∠ABC = ∠PQR …(ii)

In △ABC and △PQR:

∠BAC = ∠QPR [from (i)]

∠ABC = ∠PQR [from (ii)]

By the AA (Angle-Angle) similarity criterion,

∴ △ABC ∼ △PQR. Hence proved.

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Triangles — Class 10 Maths Practice Questions