ICSE Class 10 Maths Chapter 17 Similarity of Triangles

Chapter 17

Similarity of Triangles

Points To Remember

1. Similar Triangles

Triangle ABC and Triangle DEF are said to be similar if their corresponding angles are equal and the corresponding sides are proportional.

i.e., when \(\angle A = \angle D\), \(\angle B = \angle E\), \(\angle C = \angle F\)

and \(\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}\)

Then we write, \(\triangle ABC \sim \triangle DEF\).

The sign '-' is read as 'is similar to'.

2. Three Similarity Axioms For Triangles

(i) SAS-Axiom

If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar.

If in triangle ABC and triangle DEF, we have

\(\angle A = \angle D\) and \(\frac{AB}{DE} = \frac{AC}{DF}\) then,

\(\triangle ABC \sim \triangle DEF\)

(ii) AA-Axiom or AAA-Axiom

If two triangles have two pairs of corresponding angles equal, the triangles are similar.

If in triangle ABC and triangle DEF, we have

\(\angle A = \angle D\) and \(\angle B = \angle E\), then

\(\triangle ABC \sim \triangle DEF\)

(iii) SSS-Axiom

If two triangles have their three pairs of corresponding sides proportional, then the triangles are similar.

If in triangle ABC and triangle DEF, we have

\(\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}\), then \(\triangle ABC \sim \triangle DEF\)

Teacher's Note

Similar triangles appear in everyday life when we look at shadows cast by objects at the same time of day - the shadow of a person and the shadow of a tall building form similar triangles with the sun's rays, allowing us to calculate heights indirectly.

3. Results on Area of Similar Triangles (Theorems)

Theorem 1

The areas of two similar triangles are proportional to have squares on their corresponding sides.

Given: \(\triangle ABC \sim \triangle DEF\)

To Prove: \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}\)

Construction: Draw \(AL \perp BC\) and \(DM \perp EF\).

Proof

Hence, \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}\)

Theorem 2

The areas of two similar triangles are proportional to the squares on their corresponding altitudes.

Teacher's Note

When architects design scaled models of buildings, they use the property that areas scale with the square of linear dimensions - a model that is half the linear size has one-quarter the surface area.

Proof of Theorem 2

Given: \(\triangle ABC \sim \triangle DEF\), \(AL \perp BC\) and \(DM \perp EF\)

To Prove: \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{AL^2}{DM^2}\)

Hence, \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{AL^2}{DM^2}\)

Theorem 3

The areas of two similar triangles are proportional to the squares on their corresponding medians.

Given: \(\triangle ABC \sim \triangle DEF\) and AP, DQ are their medians.

To Prove: \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} = \frac{AP^2}{DQ^2}\)

Proof

Teacher's Note

Medians in triangles are used by engineers to find centers of mass, which is critical in designing balanced structures like bridges and aircraft.

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