Triangles — Previous Year Questions (Class 10 Mathematics)
Triangles chapter covers similarity, congruence, Pythagoras theorem, and area relationships. CBSE focuses on similarity criteria (AA, SSS, SAS) and proof-b
TL;DR: Triangles chapter covers similarity, congruence, Pythagoras theorem, and area relationships. CBSE focuses on similarity criteria (AA, SSS, SAS) and pr…
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Triangles chapter covers similarity, congruence, Pythagoras theorem, and area relationships. CBSE focuses on similarity criteria (AA, SSS, SAS) and proof-b
Triangles — Previous Year Questions with Solutions
Q (2023, 2 marks): In triangle ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 6 cm, and AE = 3 cm, find EC.
Answer: By Basic Proportionality Theorem (Thales' theorem), if DE || BC:
AD/DB = AE/EC
4/6 = 3/EC
EC = (3 × 6) / 4 = 18/4 = 4.5 cm
Final Answer: EC = 4.5 cm
Q (2022, 5 marks): Prove that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
Answer: Given: Triangle ABC with line DE parallel to BC (D on AB, E on AC).
Prove: AD/DB = AE/EC
Construction: Draw EF || AB, intersecting BC at F.
Proof:
Since DE || BC, quadrilateral DBEF is a parallelogram (DB || EF and DE || BF).
Therefore, DB = EF ... (1)
In triangle ABC, since DE || BC, triangles ADE and ABC are similar.
AD/AB = AE/AC = DE/BC
AD/AB = AE/AC
AD/(AD + DB) = AE/(AE + EC)
Let AD/AB = k
Then AD = k·AB and DB = (1-k)·AB
AE = k·AC and EC = (1-k)·AC
AD/DB = k/(1-k) and AE/EC = k/(1-k)
Therefore, AD/DB = AE/EC
Hence proved.
Q (2021, 3 marks): Two similar triangles have areas 81 cm^2 and 121 cm^2. If a side of the first triangle is 9 cm, find the corresponding side of the second triangle.
Answer: For similar triangles, the ratio of areas = (ratio of corresponding sides)^2
Area_1 / Area_2 = (side_1 / side_2)^2
81 / 121 = (9 / side_2)^2
sqrt(81/121) = 9 / side_2
9/11 = 9 / side_2
side_2 = (9 × 11) / 9 = 11 cm
Final Answer: 11 cm
Q (2023, 3 marks): In right triangle ABC with right angle at C, if AC = 5 cm and BC = 12 cm, find the altitude from C to hypotenuse AB.
Answer: First find AB using Pythagoras theorem:
AB^2 = AC^2 + BC^2 = 5^2 + 12^2 = 25 + 144 = 169
AB = 13 cm
Area of triangle = (1/2) × AC × BC = (1/2) × 5 × 12 = 30 cm^2
Also, Area = (1/2) × AB × h (where h is altitude from C to AB)
30 = (1/2) × 13 × h
h = 60/13 cm
Final Answer: 60/13 cm or ≈ 4.62 cm
Q (2022, 3 marks): Triangles ABC and DEF are similar. If AB = 4 cm, BC = 5 cm, CA = 6 cm, and the perimeter of triangle DEF is 45 cm, find the sides of triangle DEF.
Answer: Perimeter of ABC = 4 + 5 + 6 = 15 cm
Perimeter of DEF = 45 cm
Ratio of perimeters = 45/15 = 3
Since triangles are similar, ratio of corresponding sides = 3
DE = 3 × AB = 3 × 4 = 12 cm
EF = 3 × BC = 3 × 5 = 15 cm
FD = 3 × CA = 3 × 6 = 18 cm
Final Answer: DE = 12 cm, EF = 15 cm, FD = 18 cm
Q (2021, 5 marks): If triangle ABC is right-angled at C, prove that AB^2 = AC^2 + BC^2.
Answer: Given: Triangle ABC with right angle at C.
Prove: AB^2 = AC^2 + BC^2
Construction: Drop perpendicular from C to AB, meeting at D.
Proof:
Triangles ACD and ABC are similar (AA similarity: angle A is common, angle ADC = angle ACB = 90°)
So AC/AB = AD/AC ⇒ AC^2 = AD × AB ... (1)
Triangles BCD and BAC are similar (AA similarity: angle B is common, angle BDC = angle BCA = 90°)
So BC/AB = BD/BC ⇒ BC^2 = BD × AB ... (2)
Adding (1) and (2):
AC^2 + BC^2 = AD × AB + BD × AB = AB(AD + BD) = AB × AB = AB^2
Therefore, AB^2 = AC^2 + BC^2
Hence proved (Pythagoras theorem).
Frequently Asked Questions
What are the three similarity criteria for triangles?
AA (Angle-Angle): two angles equal. SSS (Side-Side-Side): all three sides proportional. SAS (Side-Angle-Side): two sides proportional and included angle equal.
How is the Basic Proportionality Theorem applied?
If a line parallel to one side of a triangle intersects the other two sides, it divides those sides in the same ratio. If DE || BC in triangle ABC, then AD/DB = AE/EC.
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