Class 10 — Similar Triangles

Chapter Overview & Weightage

Similar Triangles (Triangles chapter in NCERT) is a high-weightage chapter in Class 10 Maths. It contributes 7–9 marks every year through theorem-based proofs, side-ratio calculations, and applications in real-life geometry.

CBSE Class 10 Weightage (Year-by-Year)

Year	Marks	Question Types
2024	8	1 MCQ + Theorem proof + Numerical
2023	9	Basic Proportionality + 5-mark application
2022	7	Pythagoras-derived problem + similarity
2021	8	Theorem proof + numerical

The 5-mark question is almost always either Basic Proportionality Theorem (BPT) proof or its converse, or a triangle similarity application.

Key Concepts You Must Know

Definition of similar triangles: Two triangles are similar if (i) corresponding angles are equal, AND (ii) corresponding sides are in the same ratio.

Criteria for similarity:

AA (or AAA): two pairs of corresponding angles equal
SSS: all three pairs of corresponding sides in proportion
SAS: one pair of angles equal AND the including sides proportional

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

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

Areas of similar triangles: Ratio of areas = (ratio of corresponding sides)² = (ratio of corresponding altitudes)² = (ratio of corresponding medians)².

Pythagoras’ theorem and its converse — proved using similarity in NCERT.

Important Theorems and Results

In $\triangle ABC$ , if $DE \parallel BC$ where $D$ is on $AB$ and $E$ is on $AC$ :

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

Equivalent form: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ .

If $\triangle ABC \sim \triangle PQR$ :

$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2$

The squaring is what makes this “scaled” — areas scale as the square of linear dimensions.

In a right-angled triangle:

$(\text{hypotenuse})^2 = (\text{base})^2 + (\text{perpendicular})^2$

Proved using similarity: the altitude from the right angle creates two triangles similar to the original.

Solved Previous Year Questions

PYQ 1 — CBSE 2024, 5 Marks

State and prove the Basic Proportionality Theorem.

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

Proof outline (use the area method): in $\triangle ABC$ , draw $DE \parallel BC$ . Join $BE$ and $CD$ . Show that $\triangle ADE$ and $\triangle BDE$ have equal heights, so their areas are in the ratio $AD:DB$ . Similarly for $\triangle ADE$ and $\triangle CDE$ . Since $\triangle BDE$ and $\triangle CDE$ have equal areas (same base $DE$ between parallel lines), the ratios are equal.

PYQ 2 — CBSE 2023, 3 Marks

In a triangle ABC, $DE \parallel BC$ . If $AD = 4$ cm, $DB = 6$ cm, $AE = 3$ cm, find $EC$ .

By BPT: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , so $\dfrac{4}{6} = \dfrac{3}{EC}$ . Solving: $EC = 18/4 = 4.5$ cm.

PYQ 3 — CBSE 2022, 4 Marks

Two triangles are similar with sides in ratio 3:5. If the area of the smaller is 27 sq.cm, find the area of the larger.

Ratio of areas = $(3/5)^2 = 9/25$ . So $\dfrac{27}{\text{larger}} = \dfrac{9}{25}$ , giving larger area $= 27 \times 25/9 = 75$ sq.cm.

Difficulty Distribution

Difficulty	% of Marks	Sub-topics
Easy	30%	Direct ratio problems, criteria identification
Medium	50%	BPT applications, area ratios
Hard	20%	Theorem proofs, multi-step similarity chains

Expert Strategy

Week 1 — Master BPT and its converse. Practice the proof until you can write it from memory. Solve at least 10 ratio numericals.

Week 2 — Areas of similar triangles. Memorise the squaring rule. Practice 5-mark application problems.

Week 3 — Pythagoras + its converse. The proof using similarity is a classic 5-mark question. Practice solid geometry word problems.

Diagram first, formula second. Always draw a clean labelled diagram before applying any theorem. CBSE awards a portion of marks just for the diagram. A wrong diagram makes the rest of the answer collapse.

Common Traps

Trap 1: Wrong correspondence of vertices.

If $\triangle ABC \sim \triangle PQR$ , then $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ . The order matters! $AB/PQ = BC/QR = CA/RP$ , NOT $AB/QR$ .

Trap 2: Forgetting to square for area ratios.

Side ratio 2:3 means area ratio 4:9, not 2:3. CBSE deducts a full mark for this oversight.

Trap 3: Using BPT without verifying parallel.

BPT requires the dividing line to be parallel to the third side. If parallelism isn’t given or proven, you cannot apply BPT. Use the converse if you need to prove parallelism.

Trap 4: Confusing similarity with congruence.

Similar triangles have equal angles and proportional sides — same shape, different size possible. Congruent triangles also have equal sides — same shape AND size. Don’t confuse the two.

Trap 5: Skipping the criterion statement in proofs.

In a similarity proof, always state which criterion you’re using (AA, SSS, SAS) before concluding similarity. CBSE expects this explicit statement for full marks.