Chapter 6: Triangles

Mathematics · CBSE Class 10 · NCERT Solutions

📖 Chapter 6 — Triangles (CBSE Class 10 Maths): Detailed notes — similar figures, Basic Proportionality Theorem (Thales), and the criteria for similarity of triangles.

Chapter 6 — Triangles

🔺 1. Similar Figures

Two figures are similar if they have the same shape (not necessarily the same size). For two polygons to be similar: (a) corresponding angles are equal, and (b) corresponding sides are in the same ratio (proportional).

All circles, all squares and all equilateral triangles are similar.

📏 2. Basic Proportionality Theorem (Thales)

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

If DE ∥ BC in △ABC, then AD/DB = AE/EC

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

🔗 3. Criteria for Similarity of Triangles

Criterion	Condition
AAA (or AA)	All corresponding angles equal
SSS	All corresponding sides in the same ratio
SAS	One angle equal & the two including sides proportional

If △ABC ~ △DEF, then ∠A=∠D, ∠B=∠E, ∠C=∠F and AB/DE = BC/EF = CA/FD.

🧠 Remember: Similar = same shape (equal angles + proportional sides). BPT: DE∥BC → AD/DB = AE/EC. Converse: equal ratio → parallel. Similarity: AAA/AA, SSS, SAS. Write similarity in correct order (matching vertices).

⭐ Exam Tips: ① Similar figures & conditions. ② BPT: find an unknown length. ③ Converse of BPT: check if a line is parallel. ④ Prove two triangles similar (state AAA/SSS/SAS). ⑤ Use CPST (corresponding sides of similar triangles are proportional).

📘 EXERCISE 6.1

1Fill in the blanks using the correct word.

(i) All circles are similar. (ii) All squares are similar. (iii) All equilateral triangles are similar. (iv) Two polygons of the same number of sides are similar if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

📗 EXERCISE 6.2 (Basic Proportionality Theorem)

1In the figure DE ∥ BC. Find EC (given AD = 1.5 cm, DB = 3 cm, AE = 1 cm).

DE ∥ BC, AD = 1.5, DB = 3, AE = 1 By BPT: AD/DB = AE/EC. Substitute: 1.5/3 = 1/EC. Cross-multiply: 1.5 × EC = 3 × 1 → EC = 3/1.5. ∴ EC = 2 cm. [In (ii) with AE=1.8, EC=5.4, DB=7.2 → AD = 7.2×1.8/5.4 = 2.4 cm]

2E and F are points on PQ and PR. State whether EF ∥ QR for the given measurements.

Use the converse of BPT: EF ∥ QR only if PE/EQ = PF/FR. (i) PE/EQ = 3.9/3 = 1.3 ; PF/FR = 3.6/2.4 = 1.5 → 1.3 ≠ 1.5 → EF is NOT parallel to QR. (ii) PE/QE = 4/4.5 = 8/9 ; PF/RF = 8/9 → equal → EF ∥ QR. (iii) EQ = 1.28−0.18 = 1.10, FR = 2.56−0.36 = 2.20 → PE/EQ = 0.18/1.10 = 0.164 ; PF/FR = 0.36/2.20 = 0.164 → equal → EF ∥ QR.

📙 EXERCISE 6.3 (Similarity Criteria)

1State which pairs of triangles are similar and write the criterion used.

Compare the triangles: if all three angles match → AAA; if all three sides are in the same ratio → SSS; if one angle is equal and the two including sides are proportional → SAS. Write each pair in correct symbolic order, e.g. △ABC ~ △PQR, matching equal angles/proportional sides. ∴ e.g. △ABC ~ △PQR (by AAA), △ABC ~ △QRP (by SSS), etc. — as the figure data requires.

2△ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

△ODC ~ △OBA, ∠BOC = 125°, ∠CDO = 70° ∠DOC and ∠BOC are a linear pair (BD is a line): ∠DOC = 180° − 125° = 55°. In △ODC, angle sum = 180°: ∠DCO = 180° − ∠DOC − ∠CDO = 180° − 55° − 70° = 55°. Since △ODC ~ △OBA, corresponding angles are equal: ∠OAB = ∠OCD = 55°. ∴ ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.

Chapter 6: Triangles