Chapter 6

Class 10 Mathematics - Chapter 6: Triangles
Chapter 6: Triangles
1. Similar Figures & Basic Proportionality Theorem (BPT)
Similar Triangles: Two triangles are similar ($\sim$) if:
  1. Their corresponding angles are equal.
  2. Their corresponding sides are in the same ratio (proportional).
Basic Proportionality Theorem (Thales Theorem): 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. $$DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC}$$
A B C D E
Example 1.1: Applying BPT

In $\triangle ABC$, $DE \parallel BC$. If $AD = 1.5\text{ cm}$, $DB = 3\text{ cm}$, and $AE = 1\text{ cm}$, find $EC$.

Solution:
Since $DE \parallel BC$, by BPT: $$\frac{AD}{DB} = \frac{AE}{EC}$$ $$\frac{1.5}{3} = \frac{1}{EC}$$ $$\frac{1}{2} = \frac{1}{EC}$$ $$EC = 2\text{ cm}$$ Therefore, $EC = \mathbf{2\text{ cm}}$.

2. Criteria for Similarity of Triangles
1. AAA (Angle-Angle-Angle) Similarity: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio, and the two triangles are similar (AA similarity is also valid if two pairs of angles are equal).

2. SSS (Side-Side-Side) Similarity: If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and the two triangles are similar.

3. SAS (Side-Angle-Side) Similarity: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
Criterion Condition Similarity Statement
AAA / AA $\angle A = \angle D, \quad \angle B = \angle E$ $\triangle ABC \sim \triangle DEF$
SSS $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$ $\triangle ABC \sim \triangle DEF$
SAS $\frac{AB}{DE} = \frac{AC}{DF}, \quad \angle A = \angle D$ $\triangle ABC \sim \triangle DEF$
Example 2.1: Proving Triangle Similarity

In $\triangle ABC$ and $\triangle PQR$, $AB = 2\text{ cm}$, $BC = 4\text{ cm}$, $AC = 3\text{ cm}$, and $PQ = 4\text{ cm}$, $QR = 8\text{ cm}$, $PR = 6\text{ cm}$. Check if they are similar.

Solution:
Calculate ratios of corresponding sides: $$\frac{AB}{PQ} = \frac{2}{4} = \frac{1}{2}$$ $$\frac{BC}{QR} = \frac{4}{8} = \frac{1}{2}$$ $$\frac{AC}{PR} = \frac{3}{6} = \frac{1}{2}$$ Since: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$$ By SSS Similarity Criterion: $$\triangle ABC \sim \triangle PQR$$

3. Areas of Similar Triangles & Pythagoras Theorem
Area Ratio Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: $$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$$
Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. $$AC^2 = AB^2 + BC^2$$ Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
A B C
Example 3.1: Applying Pythagoras Theorem

A ladder $10\text{ m}$ long reaches a window $8\text{ m}$ above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution:
Let $AC$ be the ladder ($10\text{ m}$), $AB$ be the height of the window ($8\text{ m}$), and $BC$ be the distance from the wall.
In right $\triangle ABC$: $$AC^2 = AB^2 + BC^2$$ $$10^2 = 8^2 + BC^2$$ $$100 = 64 + BC^2$$ $$BC^2 = 100 - 64$$ $$BC^2 = 36$$ $$BC = \sqrt{36} = 6\text{ m}$$ The distance of the foot of the ladder from the wall is $6\text{ m}$.