# NCERT CBSE Class 10th Mathematics Chapter 6: Triangles

Safalta Expert Published by: Sylvester Updated Wed, 22 Jun 2022 06:45 PM IST

## Highlights

NCERT CBSE Class 10th Mathematics Chapter 6: Triangles

The sixth chapter in Mathematics textbook is 'Triangles'.

### Similar Figures

• Two figures having the same shape but not necessary the same size are called similar figures.
• All congruent figures are similar but all similar figures are not congruent.

### Similar Polygons

Two polygons are said to be similar to each other, if:
(i) their corresponding angles are equal, and
(ii) the lengths of their corresponding sides are proportional

### Criterion for Similarity of Triangles

Two triangles are similar if either of the following three criterion’s are satisfied:
• AAA similarity Criterion. If two triangles are equiangular, then they are similar.
• Corollary(AA similarity). If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
• SSS Similarity Criterion. If the corresponding sides of two triangles are proportional, then they are similar.
• SAS Similarity Criterion. If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

### Results in Similar Triangles based on Similarity Criterion:

1. Ratio of corresponding sides = Ratio of corresponding perimeters

2. Ratio of corresponding sides = Ratio of corresponding medians

3. Ratio of corresponding sides = Ratio of corresponding altitudes

4. Ratio of corresponding sides = Ratio of corresponding angle bisector segments.

### The topics discussed in this chapter are as follows:

• Types of Triangles
• Basic Proportionality Theorem
• Areas of Similar Triangles
• Proof of Pythagoras Theorem

Equilateral

√3/4 a2

30 cm

## If triangles ABC and DEF are similar and AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm, the perimeter of triangle is

ABC ~ DEF

AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6 cm

AC = (12.4)/6 = 8 cm

Perimeter = AB+BC+AC

= 4+6+8

=18 cm

## Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81