CBSE Notes Class 10 Maths Chapter 6 Triangles

1.0Introduction to triangles

A triangle is a type of polygon with three sides. Triangles are of different types like Equilateral triangles, Isosceles triangles, and Right angle triangles. 

2.0Download CBSE Notes for Class 10 Maths Chapter 6: Triangles - Free PDF!!

Ready to ace the chapter on Triangles? Grab these free CBSE Class 10 Maths notes in PDF format and make understanding concepts like similarity and theorems a breeze!

3.0CBSE Class 10 Maths Chapter 6 Triangles - Revision Notes

What are similar figures? 

Similar figures mean that two or more figures have the same shape but not necessarily the same size. For example: All equilateral triangles are similar but not necessarily congruent. Similarly, all squares are similar but not congruent. Here, we say that all congruent figures are similar but the similar figures need not be congruent. 

Similarity in Triangles

Two triangles are similar if: 

• Their corresponding angles are equal. 

That is Angle A = D, B = E, C = F. 

• Their corresponding sides are in the same ratio (or proportion).

$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$

Properties of Triangles

Theorem 1: 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. (Thales or basic proportionality theorem). 

$\text{ To prove: }\frac{AP}{BP}=\frac{AQ}{CQ}$

Given: PQ is parallel to BC. 

Construction: Join P with C and B with Q. and construct $QN\perp AP$ and $PM\perp AQ$

Solution:

$\text{ Area of a triangle }=\frac{1}{2}\times \text{ Base }\times \text{ Height }$

$\text{ So, in }△APQ\text{ and }△BPQ$

$\mathrm{ar}(△APQ)=\frac{1}{2}\times AP\times QN$

$\mathrm{ar}(△BPQ)=\frac{1}{2}\times BP\times QN$

$△APQ\text{ and }△CPQ$

$\mathrm{ar}(△APQ)=\frac{1}{2}\times AQ\times PM$

$\mathrm{ar}(△CPQ)=\frac{1}{2}\times CQ\times PM$

Now, 

$\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(BPQ)}=\frac{\frac{1}{2}\times AP\times QN}{\frac{1}{2}\times BP\times QN}=\frac{AP}{BP}$

and,

$\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(CPQ)}=\frac{\frac{1}{2}\times AP\times PM}{\frac{1}{2}\times CQ\times PM}=\frac{AP}{CQ}$

We know in maths, that the area of two triangles with the same base and between two parallel lines is equal. Hence, ar(BPQ) = ar(CPQ). 

Hence, $\frac{AP}{BP}=\frac{AP}{CQ}$

Theorem 2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of Thales theorem) 

To Prove: PQ is parallel to BC.

Given: $\frac{AP}{BP}=\frac{AP}{CQ}$

Construction: Join P with C and B with Q. and construct $QN\perp AP$ and $PM\perp AQ$

Solution: From above we know that, 

$\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(BPQ)}=\frac{\frac{1}{2}\times AP\times QN}{\frac{1}{2}\times BP\times QN}=\frac{AP}{BP}$

and,

$\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(CPQ)}=\frac{\frac{1}{2}\times AP\times PM}{\frac{1}{2}\times CQ\times PM}=\frac{AP}{CQ}$

$\frac{AP}{BP}=\frac{AP}{CQ}(\text{ Given })$

So, 

$\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(BPQ)}=\frac{\mathrm{ar}(APQ)}{\mathrm{ar}(CPQ)}$

Hence, 

ar(BPQ) = ar(CPQ). 

If the area of two triangles between two lines with equal bases is equal to each other then the lines will be parallel. Hence, PQ is parallel to BC. 

Criteria for Similarity

• In Maths, if only two angles are equal then the third angle will automatically be equal. 
• SAS (Side-Angle-Side Theorem): Meaning if two triangles are similar then the rest of the corresponding parts will also be equal. 

AAA stands for Angle-Angle-Angle meaning when all angles of two triangles are equal. 

Example: In the given figure: Prove that $POQ\sim SOR$ given that PQ is parallel to RS. 

$\text{ Solution: in }△POQ\text{ and }△SOR$

$\angle \mathrm{P}=\angle \mathrm{S}$

$\angle \mathrm{Q}=\angle \mathrm{R}$

$\angle \mathrm{POQ}=\angle \mathrm{SOR}\text{ (Vertically opposite angle) }$

$△POQ\sim △SOR(\mathrm{AAA})$

SAS: Stands for Side-angle-Side which means the ratio of corresponding sides is equal to an angle between these sides. 

Example: In the given figure $\frac{DB}{AD}=\frac{CE}{AE}$ . Prove that $△ADE\sim △ABC.$

Solution:

$\text{ In }△ADE\text{ and }△ABC\text{. }$

$\angle \mathrm{A}=\angle \mathrm{A}(\text{ Common })$

$\frac{DB}{AD}=\frac{CE}{AE}(\text{ Given })$

Add 1 on both sides, 

$\frac{DB}{AD}+1=\frac{CE}{AE}+1$

$\frac{DB+AD}{AD}=\frac{CE+AE}{AE}$

$\frac{AB}{AD}=\frac{AC}{AE}$

$△ADE\sim △ABC\text{ (SAS) }$

SSS: Stands for Side-Side-Side which means the ratio of all the corresponding sides is equal. 

Example: In the given figures BC = EF, and AB.DF = AC.DE Prove that $△DEF\sim △ABC$ .

Solution:

$\text{ In }△DEF\text{ and }△ABC$

EF = BC (Given) 

AB.DF = AC.DE (Given) 

$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$

$△DEF\sim △ABC(\mathbf{SSS})$

4.0Key Features of CBSE Maths Notes for Class 10 Chapter 6

• The notes are aligned with the latest curriculum suggested by CBSE. 
• Step-by-step guides along with examples are given to provide a better understanding of concepts. 
• The language used is easy hence ideal for self-learning.