CBSE Class 10 Maths — Triangles

Chapter Overview & Weightage

Triangles is Chapter 6 of CBSE Class 10 Maths. It covers similar triangles, criteria for similarity, the Basic Proportionality Theorem (BPT), Pythagoras theorem, and area relationships of similar triangles. This chapter builds on Class 9 Triangles (congruence) and is foundational for trigonometry and coordinate geometry.

Exam Year	Marks Allocated	Question Types
2024	12–15 marks	1 MCQ + 2 short + 2 long + 1 case study
2023	10–12 marks	1 MCQ + 1 short + 2 long + 1 proof
2022	12 marks	2 short + 2 long + 1 proof
2021	10 marks	1 short + 2 long + 1 proof
2020	12 marks	2 short + 2 long

Triangles is one of the highest-weightage chapters in Class 10 Maths — consistently 10–15 marks. Proof questions appear every year. The most tested theorems are: BPT (Thales’ theorem), AA similarity criterion, and Pythagoras theorem. Learn these proofs by heart — they’re scoring and predictable.

Key Concepts You Must Know

Similar triangles: Two triangles are similar if their corresponding angles are equal AND corresponding sides are proportional. Similar triangles have the same shape but may differ in size.

Criteria for similarity:

AA (Angle-Angle): If two angles of one triangle are equal to two angles of another, the triangles are similar. (Third angle is automatically equal since angle sum = 180°.)
SAS (Side-Angle-Side) similarity: If one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the triangles are similar.
SSS similarity: If corresponding sides of two triangles are in the same ratio, the triangles are similar.

Basic Proportionality Theorem (BPT / Thales’ Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

If DE || BC in △ABC with D on AB and E on AC:

\frac{AD}{DB} = \frac{AE}{EC}

Converse of BPT: If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.

Area of similar triangles: If two triangles are similar with ratio of corresponding sides $k:1$ , then the ratio of their areas is $k^2:1$ .

\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle PQR} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides.

\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2

Converse of Pythagoras: If in a triangle, the square of one side equals the sum of squares of the other two sides, the angle opposite the first side is a right angle.

Important Formulas

BPT: If DE || BC, then $\frac{AD}{DB} = \frac{AE}{EC}$

Also: $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (full ratio form)

Area ratio for similar triangles: $\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{k_1}{k_2}\right)^2$ where $k_1:k_2$ is the ratio of corresponding sides

In △ABC with right angle at C:

AB^2 = AC^2 + BC^2

Common Pythagorean triples: 3-4-5 (9+16=25), 5-12-13 (25+144=169), 8-15-17, 7-24-25

Multiples work too: 6-8-10, 10-24-26, etc.

Solved Previous Year Questions

PYQ 1: (CBSE 2024, 3 marks)

Q: In △ABC, DE || BC. If AD = 3 cm, DB = 5 cm, and AE = 6 cm, find EC.

Solution: By BPT (since DE || BC):

\frac{AD}{DB} = \frac{AE}{EC}

\frac{3}{5} = \frac{6}{EC}

EC = \frac{6 \times 5}{3} = 10 \text{ cm}

PYQ 2: (CBSE 2023, 5 marks)

Q: State and prove the Basic Proportionality Theorem.

Solution: Statement: If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

Given: △ABC with DE || BC, where D is on AB and E is on AC.

To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$

Construction: Draw DM ⊥ AC and EN ⊥ AB. Join BE and CD.

Proof:

$\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EN$ (base AD, height EN)

$\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$ (base DB, height EN)

$\therefore \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{AD}{DB}$ …(1)

Similarly, $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{AE}{EC}$ …(2)

Now, △BDE and △CDE share the same base DE and lie between the same parallel lines DE and BC.

$\therefore \text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$ …(3)

From (1), (2), and (3): $\frac{AD}{DB} = \frac{AE}{EC}$ (Proved)

PYQ 3: (CBSE 2022, 4 marks)

Q: Two similar triangles have areas in the ratio 9:16. If one side of the smaller triangle is 6 cm, find the corresponding side of the larger triangle.

Solution: Ratio of areas = 9:16

Since area ratio = (side ratio)², side ratio = $\sqrt{9/16} = 3/4$

So smaller : larger = 3 : 4

If smaller side = 6 cm: $\frac{6}{\text{larger side}} = \frac{3}{4}$

Larger side = $\frac{6 \times 4}{3} = 8$ cm

Difficulty Distribution

Difficulty	%	Question Type
Easy (30%)	Apply BPT to find unknown lengths, identify similar triangles	2 mark
Medium (40%)	Area ratios, Pythagoras applications, prove using AA	3–4 mark
Hard (30%)	Prove BPT or Pythagoras theorem, multi-step problems	5 mark + proof

Expert Strategy

For proof questions: The BPT proof and Pythagoras theorem proof are fixed 5-mark questions. Memorize both proofs word-for-word in the standard format (Given, To Prove, Construction, Proof, Hence Proved). The examiner follows a fixed marking scheme — each step earns specific marks. Missing “Construction” step costs 1 mark even if your proof is otherwise correct.

For similarity problems: Always establish which similarity criterion applies before writing proportions. State “△ABC ~ △PQR (by AA)” before writing $\frac{AB}{PQ} = \frac{BC}{QR}$ . Writing proportions without stating the criterion costs 1 mark. Also, write the vertices in the correct matching order — △ABC ~ △PQR means ∠A = ∠P, AB/PQ = BC/QR = AC/PR. Getting the matching order wrong makes the proportions incorrect.

For finding missing sides: identify which sides are corresponding (same position in both triangles), set up the proportion, cross-multiply, and solve. Always check your answer using the Pythagoras theorem if it’s a right triangle.

Common Traps

Trap 1: Confusing congruence and similarity. Similar triangles have equal angles and proportional sides — but corresponding sides may be different lengths. Congruent triangles have equal angles AND equal sides. An equilateral triangle of side 3 cm is similar to (but not congruent to) an equilateral triangle of side 5 cm.

Trap 2: Using area ratio instead of side ratio (or vice versa) in proportion problems. Area ratio = (side ratio)². If triangles have sides in ratio 3:4, their areas are in ratio 9:16. If areas are in ratio 25:36, sides are in ratio 5:6. Students sometimes directly equate area ratio to side ratio, giving wrong answers.

Trap 3: Applying BPT incorrectly when the transversal doesn’t pass through the vertex. BPT applies when DE is parallel to BC and D is on AB (not extended). If the line intersects the sides when extended (external division), the ratio holds but with a sign convention. At Class 10 level, all BPT problems use internal division — be careful if the problem setup seems unusual.