How to Use This Page
Complete NCERT Solutions for CBSE Class 10 Maths Chapter 6 — Triangles: the Basic Proportionality Theorem (BPT/Thales), similarity criteria, the area-ratio theorem, and the Pythagoras theorem.
Key Concepts (Quick Recap)
- BPT (Thales): DE ∥ BC ⟹ AD/DB = AE/EC.
- Similarity criteria: AA, SSS, SAS.
- Area theorem: ar(△1)/ar(△2) = (corresponding side ratio)².
- Pythagoras: in a right triangle, (hypotenuse)² = (sum of squares of the other two sides).
Exercise 6.2 — Basic Proportionality Theorem
Q1. In △ABC, DE ∥ BC. If AD = 1.5 cm, DB = 3 cm, AE = 1 cm, find EC. By BPT: AD/DB = AE/EC → 1.5/3 = 1/EC → EC = (3 × 1)/1.5 = 2 cm.
Q2. In △ABC, DE ∥ BC. If AD = 4 cm, DB = 4.5 cm, AE = 8 cm, find AC. AD/DB = AE/EC → 4/4.5 = 8/EC → EC = (4.5 × 8)/4 = 9 cm. AC = AE + EC = 8 + 9 = 17 cm.
Q3. E and F lie on sides AB and AC of △ABC. For which case is EF ∥ BC? AB = 5.6 cm, AE = 1.4 cm, AC = 7.2 cm, AF = 1.8 cm. EB = 5.6 − 1.4 = 4.2; FC = 7.2 − 1.8 = 5.4. AE/EB = 1.4/4.2 = 1/3; AF/FC = 1.8/5.4 = 1/3. Since AE/EB = AF/FC, EF ∥ BC (converse of BPT).
Q4. In △ABC, DE ∥ BC with AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x. AD/DB = AE/EC → x/(x − 2) = (x + 2)/(x − 1) → x(x − 1) = (x + 2)(x − 2) → x² − x = x² − 4 → −x = −4 → x = 4.
Exercise 6.3 — Similarity Criteria
Pythagoras Theorem (Applications)
Q1. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall. base² = 10² − 8² = 100 − 64 = 36 → base = 6 m.
Q2. A right triangle has legs 5 cm and 12 cm. Find the hypotenuse. hyp² = 5² + 12² = 25 + 144 = 169 → hyp = 13 cm.
Q3. Two poles of heights 6 m and 11 m stand on level ground. If the distance between their feet is 12 m, find the distance between their tops. Vertical gap = 11 − 6 = 5 m; horizontal = 12 m → distance = √(5² + 12²) = √169 = 13 m.
Common Mistakes
- Applying BPT when the line is not parallel to a side.
- Writing the similarity ratio with sides in the wrong correspondence (match equal angles first).
- Using side ratio for areas — areas use the square of the side ratio.
Exam Tips
- For "prove similar," state the criterion (AA/SSS/SAS) and the equal angles/proportional sides explicitly.
- Area ratio = (side ratio)²; conversely side ratio = √(area ratio).
- Pythagoras problems often hide a right triangle in a real-world setup — draw it first.