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NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Triangles covers similar figures and similar triangles, criteria for similarity of triangles (AA, SAS, SSS), the Basic Proportionality Theorem, areas of similar triangles, and the Pythagoras Theorem.

Overview of Topics & Subtopics (Click to Jump)

1. Similar Triangles

Short Note:

• Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional).

• AA Similarity: if two angles of one triangle are equal to two angles of another, the triangles are similar. SSS Similarity and SAS Similarity criteria also apply.

2. Basic Proportionality Theorem (Thales Theorem)

Short Note:

• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

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

3. Areas of Similar Triangles

Short Note:

• The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

4. Pythagoras Theorem

Short Note:

• In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: (hypotenuse)² = (base)² + (perpendicular)².

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

NCERT Textbook Exercise Questions

In ΔABC, DE || BC and AD/DB = 3/5, AC = 5.6 cm. Find AE.
Answer: By the Basic Proportionality Theorem, AD/DB = AE/EC. So AE/EC = 3/5, meaning AE = (3/8) × AC = (3/8) × 5.6 = 2.1 cm.
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.
Answer: Using Pythagoras theorem: (10)² = (8)² + d² ⇒ 100 = 64 + d² ⇒ d² = 36 ⇒ d = 6 m.
Diagonals of a trapezium ABCD with AB || DC intersect each other at O. Prove that AO/OC = BO/OD.
Answer: Since AB || DC, in ΔAOB and ΔCOD, ∠AOB = ∠COD (vertically opposite) and ∠OAB = ∠OCD (alternate angles), so ΔAOB ~ ΔCOD by AA similarity, giving AO/OC = BO/OD.
Sides of two similar triangles are in the ratio 4:9. Find the ratio of the areas of these triangles.
Answer: Ratio of areas = (ratio of sides)² = (4/9)² = 16/81.
Two poles of heights 6 m and 11 m stand vertically on a plane ground, 12 m apart. Find the distance between their tops.
Answer: The horizontal distance is 12 m and vertical difference in heights is 11−6 = 5 m. Using Pythagoras theorem: distance = √(12²+5²) = √(144+25) = √169 = 13 m.
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Class 10 › Mathematics › Triangles
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Similar TrianglesBasic Proportionality Theorem (Thales Theorem)Areas of Similar TrianglesPythagoras Theorem