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Triangles – Class 10 Mathematics notes, important questions and MCQs

NCERT Class 10 Mathematics, chapter 6. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Similar triangles have the same shape: equal corresponding angles and proportional corresponding sides. The Basic Proportionality Theorem and similarity criteria are used to find unknown lengths.

Similarity

  • Two triangles are similar if corresponding angles are equal and corresponding sides are in the same ratio.
  • All congruent triangles are similar, but similar triangles need not be congruent.
  • All equilateral triangles are similar; all circles and all squares are similar figures.

Basic Proportionality Theorem (Thales)

  • A line parallel to one side of a triangle divides the other two sides in the same ratio.
  • In △ABC with DE ∥ BC: AD/DB = AE/EC.
  • Converse: if a line divides two sides in the same ratio, it is parallel to the third side.

Similarity criteria

  • AAA (or AA): two angles equal.
  • SSS: all three sides in the same ratio.
  • SAS: one angle equal and the sides including it in the same ratio.

Key terms

Similar figures
figures with the same shape but not necessarily the same size
BPT
Basic Proportionality Theorem, also called Thales theorem
Corresponding sides
sides opposite equal angles in similar triangles

Formulas

  • DE ∥ BC ⇒ AD/DB = AE/EC
  • △ABC ~ △PQR ⇒ AB/PQ = BC/QR = CA/RP

Important questions with answers

In △ABC, DE ∥ BC with AD = 2 cm, DB = 3 cm and AE = 4 cm. Find EC.2 marks

By BPT, AD/DB = AE/EC, so 2/3 = 4/EC and EC = 6 cm.

State and explain the AA similarity criterion with an example.3 marks

If two angles of one triangle are equal to two angles of another, the triangles are similar (the third angles are then also equal). Example: a person's shadow and a pole's shadow at the same time of day form similar right triangles, since the sun's angle and the right angle are common.

Prove the Basic Proportionality Theorem: a line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides them in the same ratio.5 marks

Given △ABC with DE ∥ BC, D on AB, E on AC. To prove AD/DB = AE/EC. Join BE and CD; draw DM ⟂ AC and EN ⟂ AB. ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN, so ar(ADE)/ar(BDE) = AD/DB. Similarly ar(ADE)/ar(DEC) = AE/EC. Triangles BDE and DEC are on the same base DE between the same parallels DE and BC, so ar(BDE) = ar(DEC). Hence AD/DB = AE/EC.

Quick revision

  • Similar = equal angles + proportional sides.
  • BPT: a line parallel to a side divides the other two sides proportionally.
  • Criteria: AA, SSS, SAS.

Exam tips

  • Write the vertices of similar triangles in matching order (△ABC ~ △PQR means A ↔ P, B ↔ Q, C ↔ R).
  • Draw and label a neat figure before applying BPT.

Practice MCQs

  1. All ____ triangles are similar.

    1. equilateral
    2. right
    3. isosceles
    4. scalene
    Show answer

    Answer: (a) equilateral. Every equilateral triangle has angles 60°, 60°, 60°.

  2. In △ABC, DE ∥ BC, AD = 2, DB = 3, AE = 4. Then EC =

    1. 5
    2. 8
    3. 6
    4. 3
    Show answer

    Answer: (c) 6. 2/3 = 4/EC ⇒ EC = 6.

  3. Two figures with the same shape but not necessarily the same size are

    1. congruent
    2. equal
    3. similar
    4. parallel
    Show answer

    Answer: (c) similar. That is the definition of similar figures.

  4. If △ABC ~ △DEF, AB = 4, DE = 6 and BC = 8, then EF =

    1. 10
    2. 12
    3. 16
    4. 9
    Show answer

    Answer: (b) 12. 4/6 = 8/EF ⇒ EF = 12.

  5. Which is NOT a similarity criterion?

    1. AA
    2. SSS
    3. ASS
    4. SAS
    Show answer

    Answer: (c) ASS. ASS (SSA) is not a criterion.

  6. A 6 m pole casts a 4 m shadow. At the same time a tower casts a 28 m shadow. The tower's height is

    1. 48 m
    2. 36 m
    3. 18.7 m
    4. 42 m
    Show answer

    Answer: (d) 42 m. 6/4 = h/28 ⇒ h = 42 m.

  7. In △ABC, DE ∥ BC with AD/DB = 3/5. If AC = 5.6 cm, then AE =

    1. 1.8 cm
    2. 3.5 cm
    3. 2.4 cm
    4. 2.1 cm
    Show answer

    Answer: (d) 2.1 cm. AE/AC = 3/8 ⇒ AE = 2.1 cm.

  8. In trapezium ABCD with AB ∥ DC, the diagonals meet at O. Then AO/OC equals

    1. AB/AD
    2. BO/OD
    3. DC/AB
    4. OD/OB
    Show answer

    Answer: (b) BO/OD. △AOB ~ △COD gives AO/OC = BO/OD = AB/CD.

All chapters of Class 10 Mathematics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.