Class 10 Maths Chapter 6: Triangles

Triangles is Chapter 6 of NCERT Class 10 Maths. It builds on the idea of congruence from Class 9 and moves to similarity: figures with the same shape but not necessarily the same size. Watch the five video lessons below, then use these notes, theorems and solved examples to revise the Basic Proportionality Theorem and the similarity criteria.

What you will learn

• The difference between congruent and similar figures
• The two conditions for similarity of triangles
• The Basic Proportionality Theorem (Thales theorem) and its converse
• The AA, SSS and SAS similarity criteria
• Using similar triangles to find unknown lengths and heights

Video Lessons

Triangles - Part 1

Triangles - Part 2

Triangles - Part 3

Triangles - Part 4

Triangles - Part 5

Key Concepts and Formulas

Similar Figures

Two figures are similar when one is an enlarged or reduced copy of the other, like a photograph and its enlargement.

• Similar figures have the same shape; congruent figures have the same shape and the same size
• All circles, all squares and all equilateral triangles are similar to each other
• Every pair of congruent figures is similar, but similar figures need not be congruent
• Two polygons with the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio

Similarity of Triangles

For triangles, the general rule for polygons applies, and we write ΔABC ~ ΔPQR. The order of the letters matters because it tells you which vertex matches which.

If ΔABC ~ ΔPQR:
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
AB/PQ = BC/QR = CA/RP
Perimeter ratio = AB/PQ

Basic Proportionality Theorem

This is the most important theorem of the chapter and its proof is often asked for 3 marks. The proof uses areas of triangles with the same base and between the same parallels.

In ΔABC, D on AB, E on AC:
BPT:      DE ∥ BC ⇒ AD/DB = AE/EC
Also:     AD/AB = AE/AC, DB/AB = EC/AC
Converse: AD/DB = AE/EC ⇒ DE ∥ BC

• BPT (Thales theorem): if a line is drawn parallel to one side of a triangle to meet the other two sides at distinct points, it divides those two sides in the same ratio
• Converse of BPT: if a line divides any two sides of a triangle in the same ratio, it is parallel to the third side

Criteria for Similarity

You do not need all six conditions to prove two triangles similar; any one of these criteria is enough.

• AAA criterion: if corresponding angles are equal, the triangles are similar
• AA criterion: two pairs of equal angles are enough, since the third pair is then automatically equal (angle sum is 180°)
• SSS criterion: if all three pairs of corresponding sides are in the same ratio, the triangles are similar
• SAS criterion: if one pair of angles is equal and the sides including these angles are in the same ratio, the triangles are similar

Writing a Similarity Proof

Examiners look for clear statements with reasons.

• Name the two triangles and list the matching parts, giving a reason for each (for example, vertically opposite angles or alternate angles)
• State the criterion used, such as "ΔABC ~ ΔDEF (by AA similarity)"
• Then write the proportion of sides you need, in the correct order

Solved Examples

Example 1: In ΔABC, DE ∥ BC with AD = 3 cm, DB = 6 cm and AE = 4 cm. Find EC.

By BPT: AD/DB = AE/EC
3/6 = 4/EC
EC = (4 × 6)/3
∴ EC = 8 cm

Example 2: In ΔABC, DE ∥ BC with AD = x + 1, DB = 3x − 1, AE = x + 3 and EC = 3x + 4. Find x.

By BPT: (x + 1)/(3x − 1) = (x + 3)/(3x + 4)
(x + 1)(3x + 4) = (x + 3)(3x − 1)
3x² + 7x + 4 = 3x² + 8x − 3
x = 7
Check: 8/20 = 10/25 = 0.4
∴ x = 7

Example 3: D and E lie on sides PQ and PR of ΔPQR with PD = 2.5 cm, DQ = 5 cm, PE = 3 cm and ER = 6 cm. Is DE ∥ QR?

PD/DQ = 2.5/5 = 1/2
PE/ER = 3/6 = 1/2
PD/DQ = PE/ER
∴ DE ∥ QR (converse of BPT)

Example 4: A girl 1.5 m tall casts a 2 m shadow, and at the same time a tower casts a 36 m shadow. Find the height of the tower.

Sun's rays make equal angles → triangles similar (AA)
Height of tower / 1.5 = 36/2
Height = 1.5 × 18
∴ Height of tower = 27 m

Example 5: ΔABC has sides 4 cm, 6 cm and 8 cm, and ΔPQR has sides 6 cm, 9 cm and 12 cm. Are they similar? Compare their perimeters.

4/6 = 2/3, 6/9 = 2/3, 8/12 = 2/3
All ratios equal → similar by SSS
Perimeters: 4 + 6 + 8 = 18, 6 + 9 + 12 = 27
18 : 27 = 2 : 3
∴ Similar (SSS), perimeters in ratio 2 : 3

Example 6: If ΔABC ~ ΔDEF with ∠A = 50° and ∠E = 70°, find ∠C.

∠B = ∠E = 70° (corresponding angles)
∠C = 180° − ∠A − ∠B
∠C = 180° − 50° − 70°
∴ ∠C = 60°

Example 7: In ΔPQR and ΔXYZ, PQ = 5 cm, PR = 7.5 cm, ∠P = 40°, and XY = 4 cm, XZ = 6 cm, ∠X = 40°. If QR = 6.5 cm, find YZ.

PQ/XY = 5/4 = 1.25
PR/XZ = 7.5/6 = 1.25
∠P = ∠X = 40° (included angle)
ΔPQR ~ ΔXYZ (SAS)
YZ = QR ÷ 1.25 = 6.5 ÷ 1.25
∴ YZ = 5.2 cm

Common Mistakes to Avoid

• Writing AD/DB = AE/AC, which mixes a part with a whole; keep part/part or part/whole consistently on both sides.
• Writing the similarity statement in the wrong order, such as ΔABC ~ ΔQPR when A matches P and B matches Q; the order decides which sides correspond.
• In the SAS criterion, using an angle that is not included between the two proportional sides.
• Skipping reasons in proofs; every equal angle or ratio needs a short justification for full marks.

Quick Revision

BPT:      DE ∥ BC ⇒ AD/DB = AE/EC
Converse: AD/DB = AE/EC ⇒ DE ∥ BC
ΔABC ~ ΔPQR ⇒ AB/PQ = BC/QR = CA/RP

• Similar: same shape; congruent: same shape and size
• Similarity criteria: AAA (or AA), SSS, SAS
• Perimeters of similar triangles are in the ratio of corresponding sides

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