Triangles Class 10 Notes CBSE Maths Chapter 6 (Free PDF Download)

Triangles

Similar Figures- Look at given figures (i) ,(ii) and (iii) 

The above are similar figures.

• Although all congruent figures are similar, not all comparable forms are congruent.

• A regular polygon with the same number of sides is comparable to another regular polygon with the same number of sides, if

1. Their respective angles are equal and

2. Their corresponding sides are in the same proportion.

Similarity of Triangles:

• Two triangles are said to be similar, if

1. Their respective angles are equal and

2. Their corresponding sides are in the same proportion.

• Criterion of similarity:

In $\Delta {\text{ABC}}$ and $\Delta {\text{DEF}}$, if

1. If $\angle {\text{A}} = \angle {\text{D}},\angle {\text{B}} = \angle {\text{E}},\angle {\text{C}} = \angle {\text{F}}$ and

2. $\dfrac{{AB}}{{DE}} = \dfrac{{BC}}{{EF}} = \dfrac{{CA}}{{FD}}$, then the two triangles will be termed as similar.

Theorems in Similarity (SSS, AA, SAS, BPT) 

AA Criterion of Similarity or AAA Criterion:

When the respective angles in two triangles are identical, their corresponding sides have the same ratio (or percentage), and the two triangles are comparable.

The AAA (Angle-Angle-Angle) criteria of the resemblance of two triangles is the name given to this criterion.

This theorem may be shown by taking two triangles ${\text{ABC}}$ and ${\text{DEF}}$ arranging them in such a way that $\angle {\text{A}} = \angle {\text{D}},\angle {\text{B}} = \angle {\text{E}}$, $\angle {\text{C}} = \angle {\text{F}}.$

Cut ${\text{DP}} = {\text{AB}}$ and ${\text{DQ}} = {\text{AC}}$ and join ${\text{PQ}}$. 

So, $\Delta {\text{ABC}} \cong \vartriangle {\text{DPQ}}$

This gives $\angle {\text{B}} = \angle {\text{P}} = \angle {\text{E}}$ and 

Therefore, $\dfrac{{DP}}{{PE}} = \dfrac{{DQ}}{{QF}}$

i.e. $\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}}$

Similarly, $\dfrac{{AB}}{{DE}} = \dfrac{{BC}}{{EF}}$ and so 

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

SSS Criterion of Similarity:

If the sides of one triangle are proportional to (that is, in the same ratio as) the sides of the second triangle, their corresponding angles are equal, and the two triangles are comparable.

For two triangles, this criteria is known as the SSS (Side-Side-Side) similarity criterion.

This theorem may be shown by taking two triangles and arranging them in such a way that

$\dfrac{{AB}}{{DE}} = \dfrac{{BC}}{{EF}} = \dfrac{{CA}}{{FD}}( < 1)$

Cut ${\text{DP}} = {\text{AB}}$ and ${\text{DQ}} = {\text{AC}}$ and join ${\text{PQ}}$. 

It can be seen that $\dfrac{{DP}}{{PE}} = \dfrac{{DQ}}{{QF}}$ and 

So, $\angle {\text{P}} = \angle {\text{E}}$ and $\angle {\text{Q}} = \angle {\text{F}}$

Therefore, $\dfrac{{DP}}{{DE}} = \dfrac{{DQ}}{{DF}} = \dfrac{{PQ}}{{EF}}$

So, $\dfrac{{DP}}{{DE}} = \dfrac{{DQ}}{{DF}} = \dfrac{{BC}}{{EF}}$

So, $BC = PQ$

SAS Criterion of Similarity:

The triangles are comparable if one of their angles is equal to one of the other triangle's angles, and the sides that include these angles are proportionate.

For two triangles, this criteria is known as the SAS (Side-Angle-Side) similarity criterion.

This theorem may be proved, as previously, by selecting two triangles and arranging them in such a way that

$\dfrac{{AB}}{{DE}} = \dfrac{{AC}}{{DF}}( < 1)$ and $\angle {\text{A}} = \angle {\text{D}}$.

Cut DP $ = {\text{AB}},{\text{DQ}} = {\text{AC}}$ and join ${\text{PQ}}$.

Now,  and $\vartriangle {\text{ABC}} \cong \vartriangle {\text{DPQ}}$

So, $\angle {\text{A}} = \angle {\text{D}},\angle {\text{B}} = \angle {\text{P}}$ and $\angle {\text{C}} = \angle {\text{Q}}$

Therefore, $\vartriangle {\text{ABC}}\sim\Delta {\text{DEF}}$

Basic Proportionality Theorem:

When a line is drawn parallel to one of the triangle's sides in order for the other two sides to meet in separate places, the other two sides are split in the same ratio.

Proof: Assuming a triangle ${\text{ABC}}$ given in which a line parallel to side ${\text{BC}}$ intersects the other two sides ${\text{AB}}$ and ${\text{AC}}$ at ${\text{D}}$ and ${\text{E}}$ respectively.

So we have to prove that $\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}}$.

So, we will assume that we had joined ${\text{BE}}$ and ${\text{CD}}$ and then we will draw ${\text{DM}} \bot {\text{AC}}$ and ${\text{EN}} \bot {\text{AB}}$.

Now, area of $\Delta {\text{ADE}} = \dfrac{1}{2} \times $ base $ \times $ height and that will be equal to 

$ = \dfrac{1}{2} \times {\text{AD}} \times {\text{EN}}$

Since, area of $\Delta {\text{ADE}}$ is represented as ar(ADE). 

So, $\operatorname{ar} ({\text{ADE}}) = \dfrac{1}{2} \times {\text{AD}} \times {\text{EN}}$

Similarly, $\operatorname{ar} ({\text{BDE}}) = \dfrac{1}{2} \times {\text{DB}} \times {\text{EN}}$

$\operatorname{ar} ({\text{ADE}}) = \dfrac{1}{2} \times {\text{AE}} \times {\text{DM}}$ and

$\operatorname{ar} ({\text{DEC}}) - \dfrac{1}{2} \times {\text{EC}} \times {\text{DM}}$

Therefore, $\dfrac{{\operatorname{ar} (ADE)}}{{\operatorname{ar} (BDE)}} = \dfrac{{\dfrac{1}{2} \times AD \times EN}}{{\dfrac{1}{2} \times DB \times EN}}$

$ = \dfrac{{AD}}{{DB}} \ldots (1)$

and $\dfrac{{\operatorname{ar} (ADE)}}{{\operatorname{ar} (DEC)}} = \dfrac{{\dfrac{1}{2} \times AE \times DM}}{{\dfrac{1}{2} \times EC \times DM}}$

$ = \dfrac{{AE}}{{EC}} \ldots (2)$

Here, it can be noted that that $\vartriangle {\text{BDE}}$ and $\vartriangle {\text{DEC}}$ are on the same base ${\text{DE}}$ and between the same parallels ${\text{BC}}$ and ${\text{DE}}$. 

So, $\operatorname{ar} ({\text{BDE}}) = \operatorname{ar} ({\text{DEC}}) \ldots (3)$

Therefore, by using the equation 1, 2 and 3, we get:

$\dfrac{{AD}}{{DB}} = \dfrac{{AE}}{{EC}}$

Areas of Similar Triangles:

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

Proof: Suppose we are with a two triangles ${\text{ABC}}$ and ${\text{PQR}}$ such that $\vartriangle {\text{ABC}}\sim\vartriangle {\text{PQR}}$

And we have to prove that $\dfrac{{\operatorname{ar} (ABC)}}{{\operatorname{ar} (PQR)}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2} = {\left( {\dfrac{{BC}}{{QR}}} \right)^2} = {\left( {\dfrac{{CA}}{{RP}}} \right)^2}$

Therefore, for finding the areas of the two triangles, we draw altitudes AM and PN of the triangles. 

Now, $\operatorname{ar} ({\text{ABC}}) = \dfrac{1}{2} \times {\text{BC}} \times {\text{AM}}$

and $\operatorname{ar} ({\text{PQR}}) = \dfrac{1}{2} \times {\text{QR}} \times {\text{PN}}$

So, $\dfrac{{\operatorname{ar} (ABC)}}{{\operatorname{ar} (PQR)}} = \dfrac{{\dfrac{1}{2} \times BC \times AM}}{{\dfrac{1}{2} \times QR \times PN}} \ldots (1)$

Now, in $\vartriangle {\text{ABM}}$ and $\Delta {\text{PQN}}$, $\angle {\text{B}} = \angle {\text{Q}}\quad $ (As $\vartriangle {\text{ABC}}\sim\Delta {\text{PQR}})$

and $\angle {\text{M}} = \angle {\text{N}}\quad \left( {} \right.$ Each is $\left. {{{90}^\circ }} \right)$

So, $\Delta {\text{ABM}}\sim\Delta {\text{PQN}}\quad $ (AA similarity criterion) 

Therefore, $\dfrac{{AM}}{{PN}} = \dfrac{{AB}}{{PQ}} \ldots $ (2)

Also, $\vartriangle {\text{ABC}}\sim\Delta {\text{PQR}}\quad $ (Given)

So, $\dfrac{{AB}}{{PQ}} = \dfrac{{BC}}{{QR}} = \dfrac{{CA}}{{RP}} \ldots (3)$

Therefore, $\dfrac{{\operatorname{ar} (ABC)}}{{\operatorname{ar} (PQR)}} = \dfrac{{AB}}{{PQ}} \times \dfrac{{AM}}{{PN}}$

From (1) & (3) $ = \dfrac{{AB}}{{PQ}} \times \dfrac{{AB}}{{PQ}} $ From (2)

$ = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2}$

Now using (3) we get:

$\dfrac{{\operatorname{ar} (ABC)}}{{\operatorname{ar} (PQR)}} = {\left( {\dfrac{{AB}}{{PQ}}} \right)^2} = {\left( {\dfrac{{BC}}{{QR}}} \right)^2} = {\left( {\dfrac{{CA}}{{RP}}} \right)^2}$

Proof of Pythagoras Theorem Using Similarity

You may note that in $\vartriangle {\text{ADB}}$ and $\vartriangle {\text{ABC}}$ $\angle {\text{A}} = \angle {\text{A}}$ and $\angle {\text{ADB}} = \angle {\text{ABC}}$

So, $\vartriangle {\text{ADB}}\sim\vartriangle {\text{ABC}}$

Similarly, $\Delta {\text{BDC}}\sim\Delta {\text{ABC}}$

From (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ${\text{ABC}}.$

Also, since $\vartriangle {\text{ADB}}\sim\Delta {\text{ABC}}$

and $\vartriangle BDC\sim\Delta ABC$

So, $\Delta {\text{BDC}}\sim\Delta {\text{BDC}}$

When a perpendicular is traced from the vertices of a right triangle's right angle to the hypotenuse, the triangles on both sides are identical to the complete triangle and to each other.

Now Let us Use this Theorem to Prove Pythagoras' Theorem:

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

Proof: We are given a right triangle ${\text{ABC}}$ right angled at ${\text{B}}$.

We need to prove that ${\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

Let us draw ${\text{BD}} \bot {\text{AC}}$. 

Now, $\vartriangle {\text{ADB}}\sim\vartriangle {\text{ABC}}$

So, $\dfrac{{AD}}{{AB}} = \dfrac{{AB}}{{AC}}$ (Sides are proportional) or ${\text{AD}} \cdot {\text{AC}} = {\text{A}}{{\text{B}}^2} \ldots (1)$

Also, $\vartriangle {\text{BDC}}\sim\vartriangle {\text{ABC}}$

So, $\dfrac{{CD}}{{BC}} = \dfrac{{BC}}{{AC}}$

or ${\text{CD}} \cdot {\text{AC}} = {\text{B}}{{\text{C}}^2} \ldots (2)$

Adding (1) and (2), ${\text{AD}} \cdot {\text{AC}} + {\text{CD}} \cdot {\text{AC}} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

Or ${\text{AC}}({\text{AD}} + {\text{CD}}) = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

Or ${\text{AC}} \cdot {\text{AC}} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

Or ${\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$

Converse of Pythagoras Theorem:

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

Proof: Here we are given a triangle ${\text{ABC}}$ in which ${\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$ 

We need to prove that $\angle {\text{B}} = {90^\circ }$ 

To start with, we construct a $\vartriangle PQR$ right angled at $Q$ such that $PQ = AB$ and $QR = BC$

Now, from $\Delta {\text{PQR}}$, we have:

${\text{P}}{{\text{R}}^2} = {\text{P}}{{\text{Q}}^2} + {\text{Q}}{{\text{R}}^2}\quad $ (Pythagoras Theorem, as $\left. {\angle {\text{Q}} = {{90}^\circ }} \right)$

Or ${\text{P}}{{\text{R}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$ (By construction). (1)

But ${\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2} + {\text{B}}{{\text{C}}^2}$ (Given) ... (2)

So, ${\text{AC}} = {\text{PR}}$ (From (1) and (2) $) \ldots $ (3)

Now, in $\Delta {\text{ABC}}$ and $\Delta {\text{PQR}}$,

${\text{AB}} = {\text{PQ}}$ (By construction)

${\text{BC}} = {\text{QR}}$ (By construction)

${\text{AC}} = $ PR (Proved in (3) above $)$ 

So, $\vartriangle {\text{ABC}} \cong \Delta {\text{PQR}}$ (SSS congruence)

Therefore, $\angle {\text{B}} = \angle {\text{Q}}$ (CPCT)

But $\angle Q = {90^\circ }$ (By construction)

So, $\angle B = {90^\circ }$

Comprehensive Revision Notes for CBSE Class 10 Maths Chapter 6: Triangles

Class 10 Maths Chapter 6 Notes free pdf

Students can get a PDF format after downloading the content free of cost on the official website of Vedantu. Triangles Class 10 Notes is also available for the convenience of students to learn topics easily. These videos help the students store either a soft copy or a hard copy without bothering the internet connection. They can revise all the chapters during the time of examinations. These Notes on Triangles PDF also helps the students during their further education and several entrance tests.

Triangles Chapter Class 10 Notes

Triangles

This chapter has aimed to recall the basic properties and the rules of a triangle along with its types, namely- equilateral, isosceles, scalene triangles based on the sides. Based on angles, another classification includes an acute angle triangle, an obtuse angle triangle, and a right-angle triangle. All these concepts were recalled in the introduction part of Maths Class 10 Triangles Notes.

Similar Triangles

It is the new topic introduced in Chapter 6 of Class 10 notes. The well-experienced faculty of mathematics has given various definitions of similar triangles. In general, if two figures have the same shape irrespective of their sizes, then those figures are said to be similar triangles. To understand these similar triangles clearly, we have several solved and unsolved examples given in the PDF of Maths Class 10 Similar Triangles Notes. 

Similarity of Triangles

If we observe the next topic of Triangles Class 10 Notes, here the similarity of triangles can be defined as if two triangles are said to be similar if and only if their angles should be equal. Their sides should be in equal ratio too. Thales theorem is used, and it is also known as the basic proportionality theorem. Few examples using different figures are available to get strong enough in this particular concept. Chapter 6 of Class 10 Notes has explained the slight differences between similar triangles and the similarity of triangles.

Criteria for the Similarity of Triangles

It is another important topic of Triangles Chapter Class 10 Notes. Along with the concepts of similar triangles and the similarity of triangles, it is also important to understand the criteria for triangles' similarity. It also involves sides and angles of triangles. In some cases, the vertices of triangles are also equal, and those triangles are also said to be similar to triangles. To get a clear view of these triangles, different theorems like if two angles of a triangle are equal to the two angles of another triangle, then those triangles are said to be similar triangles, and this is the AA similarity criterion for triangles. And hence it is proved.

Maths Class 10 Similar Triangles Notes PDF has few more theorems that aim to prove that the triangles are similar. Only if the sides of one triangle are proportionate to the sides of another triangle can those angles be equal. These triangles are said to be similar. It is called the SSS similarity criterion of triangles. Another theorem is about the SAS similarity criterion of triangles. According to this rule, the angle of one triangle is equal to the angle of another triangle where the sides are already proportionate to each other.

Areas of Similar Triangles

It can be found easily by understanding the concept clearly. The formula is also derived from the concept that the ratio of two areas of triangles is equal to the square of their sides. The area finding methodology is one and same; few differences may arise based on the figure. This can be understood only by solving multiple figures as examples. 

Pythagoras Theorem

This is a mandatory theorem in Triangles Class 10 Notes for the students to score better marks in the examination point of view. Here another term called hypotenuse is also explained, and this theorem can be proved using a right-angle triangle. Notes on Triangles have explained that the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of sides of that particular triangle; it is also proved using several examples and unsolved questions that were also given for students' practice.

Importances of Triangles Class 10 Notes CBSE Maths Chapter 6 (Free PDF Download)

The availability of free PDF download notes for CBSE Class 10 Mathematics Chapter 6 - "Triangles" holds significant importance for students. This chapter is a foundational component of geometry, and these notes provide a structured and comprehensive overview of this crucial topic. Understanding the principles of triangles is essential not only for academic success but also for practical applications in various fields, from architecture to engineering. These notes simplify complex geometric concepts, theorems, and properties, making it easier for students to grasp and apply these principles. They lay the groundwork for students to understand the geometric world around them, fostering logical reasoning and problem-solving skills. Overall, these downloadable notes are an invaluable resource for aspiring mathematicians, providing them with the knowledge and tools needed to navigate the world of triangles and geometry effectively.

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