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NCERT Solutions Class 10 Maths Chapter 6 Triangles

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NCERT Solutions for Class 10 Chapter 6 Maths Triangles - FREE PDF Download

NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF is profound. The PDF of Class 10 Maths Chapter 6 NCERT Solutions has been prepared by expert mathematicians at Vedantu after thorough research on the subject matter. All the solutions for Triangles Class 10 NCERT Solutions provided here are written in a simple and lucid manner. With the aid of these NCERT Solutions for Class 10 Chapter 6 of Maths, students can not only improve their knowledge but also aspire to score better in their examinations.

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Table of Content
1. NCERT Solutions for Class 10 Chapter 6 Maths Triangles - FREE PDF Download
2. Glance of NCERT Solutions for Class 10 Maths Chapter 6 Triangles | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 6 Maths Class 10
4. Exercises under NCERT Solutions for Class 10 Maths Chapter 6 Triangles
5. Access NCERT Solutions for Class 10 Maths Chapter 6 Triangles
    5.1Exercise 6.1
    5.2Exercise 6.2
    5.3Exercise 6.3
6. Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 6 Triangles
7. Class 10 Maths Chapter 6: Exercises Breakdown
8. Other Study Material for CBSE Class 10 Maths Chapter 6
9. Chapter-Specific NCERT Solutions for Class 10 Maths
10. NCERT Study Resources for Class 10 Maths
FAQs


Glance of NCERT Solutions for Class 10 Maths Chapter 6 Triangles | Vedantu

  • In Class 10 Maths Ch 6, we will learn about triangles, Similarity of Triangles, types of Triangles.

  • Cover terms that are related to triangles along with that we will learn how to calculate the area of a triangle by using a simple formula 

  • This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 6 - Triangles, which you can download as PDFs.

  • There are three exercises (29 fully solved questions) in class 10th Maths chapter 6 Triangles.


Access Exercise Wise NCERT Solutions for Chapter 6 Maths Class 10

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NCERT Solutions Class 10 Maths Chapter 6 Triangles
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Exercises under NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6, "Triangles," is a chapter that deals with the properties and classification of triangles. The chapter contains six exercises, each covering a different aspect of the topic. Below is a brief explanation of each exercise:


Exercise 6.1: In this exercise, you will be introduced to the basic concepts of triangles, including the definition, elements, types, and angles. You will also learn about congruent triangles and the criteria for their congruence.


Exercise 6.2: This exercise focuses on the properties of triangles, such as the angle sum property, the exterior angle property, and the inequality theorem. You will also learn about the Pythagorean theorem and its applications.


Exercise 6.3: In this exercise, you will learn about the similarity of triangles, including the criteria for similarity, the theorem of basic proportionality, and the application of similarity in practical situations.


Access NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Exercise 6.1

1. Fill in the Blanks Using Correct Word Given in the Brackets:

i. All circles are ______. (congruent, similar)

Ans: Similar.


ii. All squares are ______. (similar, congruent)

Ans: Similar.


iii. All ______ triangles are similar. (isosceles, equilateral)

Ans: Equilateral.


iv. Two polygons of the same number of sides are similar, if their corresponding angles are ______ (equal, proportional)

Ans: Equal


And their corresponding sides are ______. (equal, proportional)

Ans: Proportional.


2. Give Two Different Examples of Pair of -

i. Similar Figures

Ans: The two examples for similar figures are 

a. Two equilateral triangles having sides \[\text{2cm}\] and $\text{4cm}$.


Two equilateral triangles

                                

b. Two squares having sides \[\text{2cm}\] and $\text{4cm}$.


Two squares having sides


ii. Non-Similar Figures

Ans: The two examples for non - similar figures are

a. A Trapezium and Square


A Trapezium and Square


b. A triangle and a Paralellogram


A triangle and a Paralellogram


3. State whether the following quadrilaterals are similar or not.


The given quadrilaterals PQRS and ABCD


Ans: The given quadrilaterals $PQRS$ and $ABCD$ are not similar because their corresponding sides are proportional, that is, $1:2$ but their corresponding angles are not equal.


Exercise 6.2

1. (i) From the figure (i) , if \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]. Find \[\text{EC}\].

basic proportionality theorem



Ans: Let us assume that \[\text{EC = x cm}\]

Given that $\,\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}$

But from basic proportionality theorem, we know that

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

$\dfrac{\text{1}\text{.5}}{\text{3}}$ $=$ $\dfrac{\text{1}}{\text{x}}$

\[\text{x = }\dfrac{\text{3 x 1}}{\text{1}\text{.5}}\]

\[x\text{ }=\text{ }2\]

\[\therefore \]\[\text{EC = 2 cm}\]

(ii) From the figure (ii) , if \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]. \[\text{AD}\] in (ii).

assume that.



Ans: 

Let us assume that \[\text{AD = x cm}\]

Given that \[\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\text{.}\]

But from basic proportionality theorem we know that

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

$ \dfrac{\text{x}}{\text{7}\text{.2}}\text{ = }\dfrac{\text{1}\text{.8}}{\text{5}\text{.4}} $

$ \text{x = }\dfrac{\text{1}\text{.8 x 7}\text{.2}}{\text{5}\text{.4}} $

$ \text{x = 2}\text{.4} $

\[\therefore \text{AD = 2}\text{.4}\]$\text{cm}$

2. (i) In a $\text{ }\!\!\Delta\!\!\text{ PQR,}$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PE = 3}\text{.9 cm, EQ = 3 cm, PF = 3}\text{.6 cm}\] and \[\text{FR = 2}\text{.4 cm}\]

Ans:

points on the sides



Given, \[\text{PE = 3}\text{.9 cm, EQ = 3 cm, PF = 3}\text{.6 cm}\],\[\text{FR = 2}\text{.4 cm}\]

$\dfrac{\text{PF}}{\text{EQ}}$ $\text{=}$$\dfrac{\text{3}\text{.9}}{\text{3}}$\[\text{ }\!\!~\!\!\text{ = 1}\text{.3}\]

$\dfrac{\text{PF}}{\text{FR}}$ \[\text{=}\] $\dfrac{\text{3}\text{.6}}{\text{2}\text{.4}}$ \[\text{= 1}\text{.5}\]

Hence, $\dfrac{\text{PE}}{\text{EQ}}$ \[\ne \] $\dfrac{\text{PF}}{\text{FR}}$

Therefore , \[\text{EF}\] is parallel to \[\text{QR}\].

(ii) In a $\text{ }\!\!\Delta\!\!\text{ PQR,}$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PE = 4 cm, QE = 4}\text{.5 cm, PF = 8 cm}\] and \[\text{RF = 9 cm}\]

 Ans:

points on the sides 2



\[\text{PE = 4 cm,QE = 4}\text{.5 cm,PF = 8 cm,RF = 9 cm}\]

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{4}}{\text{4}\text{.5}}\text{ = }\dfrac{\text{8}}{\text{9}} $

$ \dfrac{\text{PF}}{\text{FR}}\text{ = }\dfrac{\text{8}}{\text{9}} $

Hence, $\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

(iii) In a $\Delta PQR,$ \[\text{E}\] and \[\text{F}\] are any two points on the sides \[\text{PQ}\] and \[\text{PR}\] respectively. State whether \[\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\] for \[\text{PQ = 1}\text{.28 cm, PR = 2}\text{.56 cm, PE = 0}\text{.18 cm}\] and \[\text{PF = 0}\text{.63 cm}\]

Ans:

any two points on the sides PQ



\[\text{PQ = 1}\text{.28 cm,PR = 2}\text{.56 cm,PE = 0}\text{.18 cm,PF = 0}\text{.36 cm}\]

$\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{0}\text{.18}}{\text{1}\text{.28}}\text{ = }\dfrac{\text{18}}{\text{128}}\text{ = }\dfrac{\text{9}}{\text{64}} $

$\dfrac{\text{PF}}{\text{PR}}\text{ = }\dfrac{\text{0}\text{.36}}{\text{2}\text{.56}}\text{ = }\dfrac{\text{9}}{\text{64}} $

Hence, $\dfrac{\text{PE}}{\text{PQ}}\text{ = }\dfrac{\text{PF}}{\text{PR}}$

Therefore, \[\text{EF}\] is parallel to \[\text{QR}\].

3. In the figure given below, if sides \[\text{LM  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CB}\] and \[\text{LN  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD,}\]Show that $\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AN}}{\text{AD}}$

the figure given below, if sides LM



Ans:

Given that in the figure, \[\text{LM  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CB}\]

But from basic proportionality theorem, we know that

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

Also, \[\text{LN  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

$\therefore \dfrac{\text{AN}}{\text{AD}}\text{ = }\dfrac{\text{AL}}{\text{AC}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (ii)}$

From (i) and (ii), we get

$\dfrac{\text{AM}}{\text{AB}}\text{ = }\dfrac{\text{AN}}{\text{AD}}$

4. In the figure given below, if sides $\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AC}$ and $\text{DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AE}\text{.}$Show that $\dfrac{\text{BF}}{\text{FE}}\text{ = }\dfrac{\text{BE}}{\text{EC}}$

the figure given below, if sides DE



Ans:

In

$\text{ }\!\!\Delta\!\!\text{ ABC,DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AC}$ 

$\therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{EC}} $

(By Basic proportionality theorem)

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ BAE,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AE} $

$ \therefore \dfrac{\text{BD}}{\text{DA}}\text{ = }\dfrac{\text{BE}}{\text{FE}} $

By Basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{BE}}{\text{EC}}\text{ = }\dfrac{\text{BF}}{\text{FE}}$

5. In the figure given below, if sides $\text{DE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OQ}$ and $\text{DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OR}$, Show that $\text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}$

if sides DE



Ans:

$\text{In}$ 

$ \text{ }\!\!\Delta\!\!\text{ POQ,DE }\!\!|\!\!\text{  }\!\!|\!\!\text{ OQ} $

$ \therefore \dfrac{\text{PE}}{\text{EQ}}\text{=}\dfrac{\text{PD}}{\text{DO}} $              ……………………(i) By basic proportionality theorem$\text{In}$

$ \text{ }\!\!\Delta\!\!\text{ POR,DF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  OR} $

$ \therefore \dfrac{\text{PF}}{\text{FR}}\text{=}\dfrac{\text{PD}}{\text{DO}}$

……………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{PE}}{\text{EQ}}\text{ = }\dfrac{\text{PF}}{\text{FR}} $

$ \therefore \text{EF  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR} $                             Converse of Basic proportionality theorem

6.In the figure given below, \[\text{A, Band C}\] are points on \[\text{OP, OQ and OR}\] respectively such that \[\text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PQ}\] and \[\text{AC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PR}\]. Prove that \[\text{BC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  QR}\].

the figure given below, A, B and C



Ans:

In

$\text{ }\!\!\Delta\!\!\text{ POQ,AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PQ} $

$\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OB}}{\text{PQ}} $

$……………………(i) By basic proportionality theorem

$\text{In}$ 

$\text{ }\!\!\Delta\!\!\text{ POR,AC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  PR} $

\[\therefore \dfrac{\text{OA}}{\text{OP}}\text{ = }\dfrac{\text{OC}}{\text{CR}}\]  ………………(ii) By basic proportionality theorem

From (i) and (ii),we get

$\dfrac{\text{OB}}{\text{BQ}}\text{ = }\dfrac{\text{OC}}{\text{CR}} $

$ \therefore \text{BC  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CR} $

Converse of Basic proportionality theorem

7. By using Basic proportionality theorem, Show that a line passing through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). 

Ans:

one side of a triangle parallel



Let us assume in the given figure in which \[\text{PQ}\] is a line segment passing through the mid-point \[\text{P}\] of line \[\text{AB}\], such that \[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\].

From basic proportionality theorem, we know that

$\dfrac{\text{AQ}}{\text{QC}}\text{ = }\dfrac{\text{AP}}{\text{PB}} $

$ \dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

As \[\text{P}\] is the midpoint of \[\text{AB}\] ,\[\text{AP  =  PB}\]

\[\Rightarrow \text{AQ = QC}\]

Or

\[\text{Q}\] is the midpoint of \[\text{AC}\]

8. By using Converse of basic proportionality theorem, Show that the line joined by the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX). 

Ans:

the midpoints of any two sides of a triangle is parallel to the third side



Let us assume that the given figure in which \[\text{PQ}\] is a line segment joined by the mid-points \[\text{P and Q}\] of lines \[\text{AB and AC}\] respectively. 

i.e., \[\text{AP  =  PB and AQ  =  QC}\]

Also it is clear that

$\dfrac{\text{AP}}{\text{PB}}\text{ = 1}$ and

$\dfrac{\text{AQ}}{\text{QC}}\text{ = 1} $

$ \therefore \dfrac{\text{AP}}{\text{PB}}\text{ = }\dfrac{\text{AQ}}{\text{QC}} $ 

Hence, using basic proportionality theorem, we get 

\[\text{PQ  }\!\!|\!\!\text{  }\!\!|\!\!\text{  BC}\]

9. If \[\text{ABCD}\] is a trapezium where \[\text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  DC}\] and its diagonals intersect each other at the point \[\text{O}\]. Prove that $\dfrac{\text{AO}}{\text{BO}}\text{ = }\dfrac{\text{CO}}{\text{DO}}$

Ans:


its diagonals intersect


Draw a line  \[\text{EF}\] through point \[\text{O}\] , such that 

In \[\text{ }\!\!\Delta\!\!\text{ ADC}\], \[\text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD}\]

Using basic proportionality theorem, we get

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}}$____________________(i)

In \[\text{ }\!\!\Delta\!\!\text{ ABD}\]\[\text{, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\] 

So, using basic proportionality theorem, we get

\[\frac{\text{AE}}{\text{ED}}\ =\ \frac{\text{BO}}{\text{DO}}\ \] ___________________(ii)

From equation (i) and (ii), we get 

$\frac{\text{AO}}{\text{CO}}\text{= }\frac{\text{BO}}{\text{DO}}$

$\therefore \ \frac{\text{AO}}{\text{BO}}\text{= }\frac{\text{CO}}{\text{DO}}$

10.  The diagonals of a quadrilateral \[\text{ABCD}\] intersect each other at the point \[\text{O}\] such that $\dfrac{\text{AO}}{\text{BO}}\text{ = }\dfrac{\text{CO}}{\text{DO}}$ Prove that \[\text{ABCD}\] is a trapezium. 

Ans: 

Let us assume the following figure for the given question.

Draw a line \[\text{OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

Let us assume the following figure



In \[\text{;ABD, OE  }\!\!|\!\!\text{  }\!\!|\!\!\text{  AB}\]

Using basic proportionality theorem, we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{BO}}{\text{OD}}\text{ }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{  }\!\!\_\!\!\text{ (i)}$

However, it is given that 

$\frac{\text{AO}}{\text{BO}}\text{ = }\frac{\text{CO}}{\text{DO}} $

$ \therefore \text{ }\frac{\text{AO}}{\text{CO}}\text{ = }\frac{\text{BO}}{\text{DO}}\ \text{ }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ }\!\!~\!\!\text{  }\_\text{ (ii)} $

From equations (i) and (ii), we get 

$\dfrac{\text{AE}}{\text{ED}}\text{ = }\dfrac{\text{AO}}{\text{OC}} $

$ \Rightarrow \text{EO  }\!\!|\!\!\text{  }\!\!|\!\!\text{  DC} $

By the converse of basic proportionality theorem

$\Rightarrow \text{ AB }\left| \left| \text{ OE } \right| \right|\text{ DC}  $

$\Rightarrow \text{AB  }\!\!|\!\!\text{  }\!\!|\!\!\text{  CD} $

\[\therefore \text{ ABCD}\] is a trapezium.


Exercise 6.3

1. State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

(i)


the following figure are similar


(ii)


the pairs of similar triangles


(iii)


the pairs of similar triangles iii


(iv)


the pairs of similar triangles iv.png


(v)


the pairs of similar triangles v


(vi)


the pairs of similar triangles vi


Ans: 

I. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$

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

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

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

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$


II. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{QRP}$

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{QP}}=\frac{1}{2}$

$\therefore$ By SSS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{QRP}$


III. In $\triangle \mathrm{LMP}$ and $\triangle \mathrm{DEF}$

$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}, \frac{\mathrm{LP}}{\mathrm{DF}}=\frac{1}{2}$

The sides are not in the equal ratios, Hence the two triangles are not similar.


IV. In $\triangle \mathrm{MNL}$ and $\triangle \mathrm{QPR}$

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

$\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{MNL} \sim \triangle \mathrm{QP} \mathrm{R}$


V. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFD}$

$\angle \mathrm{A}=\angle \mathrm{F}, $

$\frac{AB}{FD}=\frac{BC}{FD}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{EFD}$


VI. In $\triangle \mathrm{DEF}$ and $\triangle \mathrm{PQR}$

Since, sum of angles of a triangle is $180^{\circ}$, Hence, $\angle \mathrm{F}=30^{\circ}$ and $\angle \mathrm{P}=70^{\circ}$

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

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

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

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}$


2. In the following figure, $\Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ}$ and $\angle \mathrm{CDO}=70^{\circ}$. Find $\angle \mathrm{DOC}, \angle \mathrm{DCO}$ and $\angle \mathrm{OAB}$


In the following figure delta ODC


Ans: Given:

$\triangle \mathrm{ODC} \sim \triangle \mathrm{OBA}$

$\angle \mathrm{BOC}=125^{\circ}$

$\angle \mathrm{CDO}=70^{\circ}$

To find: $\angle D O C, \angle D C O$ and $\angle O A B$

Sol: Here, $B D$ is a line,

So, we can apply a linear pair on it.

$\angle B O C+\angle D O C=180^{\circ} (Linear Pair)$

$125^{\circ}+\angle D O C=180^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=55^{\circ}$

Now in $\triangle \mathrm{DCO}$

$\angle C D O+\angle D C O+\angle D O C=180^{\circ}$

$70^{\circ}+\angle D C O+55^{\circ}=180^{\circ}$

$125^{\circ}+\angle D C O=180^{\circ}$

$\angle D C O=180^{\circ}-125^{\circ}$

$\angle D C O=55^{\circ}$

Now it is given that

$\triangle O D C \sim \triangle O B A$

Hence,

$\angle \mathrm{DCO}=\angle \mathrm{OAB}$

$55^{\circ}=\angle \mathrm{OAB}$

$\angle \mathrm{OAB}=55^{\circ}$

Now in $\triangle \mathrm{DCO}$

$\angle \mathrm{CDO}+\angle \mathrm{DCO}+\angle \mathrm{DOC}=180^{\circ} \quad$ (Sum of all angles of triangle is $180^0$ 

$70^{\circ}+\angle \mathrm{DCO}+55^{\circ}=180^{\circ}$ 

$125^{\circ}+\angle \mathrm{DCO}=180^{\circ}$ 

$\angle \mathrm{DCO}=180^{\circ}-125^{\circ}$ $\angle \mathrm{DCO}=55^{\circ}$

Now, it is given that

$\Delta \mathrm{ODC} \sim \triangle \mathrm{OBA}$

Hence.

$\angle D C O=\angle O A B$(Corresponding angles of a similar triangles are equal)

$55^{\circ}=\angle O A B$


3. Diagonals AC and BD of a trapezium ABCD with AB \|DC intersect each other at the point $\mathrm{O}$. Using a similarity criterion for two triangles, show that $\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$

Ans: In $\Delta \mathrm{DOC}$ and $\triangle \mathrm{BOA}$

$\angle C D O=\angle A B O$ (Alternate interior angles as $A B \| C D)$

$\angle \mathrm{DCO}=\angle \mathrm{BAO}$ (Alternate interior angles as $\mathrm{AB} \| \mathrm{CD})$

$\angle \mathrm{DOC}=\angle \mathrm{BOA}$ (Vertically opposite angles $)$

$\therefore \Delta \mathrm{DOC} \sim \Delta \mathrm{BOA}$ (AAA similarity criterion)

$\therefore \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}} \quad($ Corresponding sides are proportional)

$\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$


4.  In the figure, $\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}} \text { and } \angle 1=\angle 2 \text {.Show that } \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}$


In the figure delta(PQR)


Ans: In $\delta \mathrm{PQR}, \angle \mathrm{PQR}=\angle \mathrm{PRQ}$

$\therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{i})$

Given 

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$

Using (i), we obtain


we obtain


$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}$

In $\triangle \mathrm{PQS}$ and $\triangle \mathrm{TQR}$,

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}[\text { [Using (ii) }]$

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

$\therefore \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR} \quad[\text { SAS similarity criterion }]$


5. $\mathrm{S}$ and $\mathrm{T}$ are point on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$. Show that $\triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}$.

Ans: Given: $\Delta P Q R$

and the points $S$ and $T$ on sides PR and QR.

Such that $\angle P=\angle R T S$


the points S and T on sides PR and QR


To Prove: $\triangle \mathrm{RPQ} \sim \Delta$ RTS.

Proof:

In $\triangle \mathrm{RPQ}$ and $\triangle \mathrm{RTS}$.

$\angle P=\angle R T S$

(Given)

And $\angle \mathrm{PRQ}=\angle \mathrm{TRS}=\angle \mathrm{R}$

(Common)

So, $\triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}$.

(AA similarity)

Hence proved


6. In the following figure, if $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$, show that $\triangle \mathrm{ADE} \sim \Delta \mathrm{ABC}$.

Ans: 


We know that the corresponding portions of two triangles that are congruent to each other are equal.png


We know that the corresponding portions of two triangles that are congruent to each other are equal.

The two 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 is known as the SAS (Side - Angle - Side) similarity criteria.

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACD}$

$\mathrm{AD}=\mathrm{AE}(\triangle \mathrm{ABE} \cong \Delta \mathrm{ACD} \text { given }) \ldots \ldots \ldots \text { (1) }$

$\mathrm{AB}=\mathrm{AC}(\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD} \text { given })$

Now Consider $\triangle A D E$ and $\triangle A B C$

and $\angle$ DAE $=\angle B A C$ (Common angle)

Thus, $\triangle$ ADE $\sim A$ ABC (SAS criterion)


7. In the following figure, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\Delta \mathrm{ABC}$ intersect each other at the point, P. Show, that:


two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.png


I. $\triangle \mathrm{AEP} \sim \Delta \mathrm{CDP}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.


For two triangles


In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$

$[\because \mathrm{CE} \perp \mathrm{AB}$ and $\mathrm{AD} \perp \mathrm{BC} ;$ altitudes $]$

$\angle A P E=\angle C P D$ (Vertically opposite angles)

$\Rightarrow \triangle$ AEP $\sim \triangle$ CPD (AA criterion)


II. $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.


For two triangles, this is known as the AA similarity criteria.png


In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$

$\angle \mathrm{ADB}=\angle C E B=90^{\circ}$

$\angle \mathrm{ABD}=\angle C B E \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{ABD} \sim \Delta C B E \text { (AA criterion) }$


III. $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.


the two triangles are said to be comparable.png


In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{ADB}$

$\angle \mathrm{AEP}=\angle \mathrm{ADB}=9 \mathrm{O}^{\circ}$

$\angle \mathrm{PAE}=\angle \mathrm{BAD} \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB} \text { (AA criterion) }$


IV. $\Delta \mathrm{PDC} \sim \Delta \mathrm{BEC}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.


one triangle are equivalent to two angles from another triangle.png


In $\triangle \mathrm{PDC}$ and $\triangle \mathrm{BEC}$

$\angle \mathrm{PDC}=\angle \mathrm{BEC}=9 \mathrm{O}^{\circ}$

$\angle \mathrm{PCD}=\angle \mathrm{BCE} \text { (Common angle) }$

$\Rightarrow \triangle \text { PDC } \sim \triangle \mathrm{BEC} \text { (AA criterion)}$


8. $\mathrm{E}$ is a point on the side AD produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \Delta \mathrm{CFB}$


the side AD produced of a parallelogram.png


Ans:

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{CFB}$,

$\angle \mathrm{A}=\angle \mathrm{C}$ (Opposite angles of a parallelogram)

$\angle \mathrm{AEB}=\angle \mathrm{CBF}$ (Alternate interior angles as $\mathrm{AE} \| \mathrm{BC})$

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$ (By AA similarity criterion)


9. In the following figure, $\mathrm{ABC}$ and AMP are two right triangles, right angled at B and M respectively, prove that:


two right triangles, right angled at B and M respectively.png


I. $\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$

II. $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

Ans: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{AMP}$

$\angle \mathrm{ABC}=\angle \mathrm{AMP}\left(\operatorname{Each} 90^{\circ}\right)$ $\angle \mathrm{A}=\angle \mathrm{A}(\mathrm{Common})$

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$ (By AA similarity criterion)

$\Rightarrow \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$ (Corresponding sides of similar triangles are proportional)


10. $\mathrm{CD}$ and $\mathrm{GH}$ are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides $\mathrm{AB}$ and $\mathrm{FE}$ of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFG}$ respectively. If $\triangle \mathrm{ABC} \sim$ $\Delta \mathrm{FEG}$, Show that

(i) $\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$


lie on sides


Ans: 

$\text { It is given that } \triangle \mathrm{ABC} \sim \Delta \mathrm{FEG}.$

$\therefore \angle \mathrm{A}=\angle \mathrm{F}, \angle \mathrm{B}=\angle \mathrm{E}, \text { and } \angle \mathrm{ACB}=\angle \mathrm{FGE}$

$\text { Since, } \angle \mathrm{ACB}=\angle \mathrm{FGE} $

$\therefore \angle \mathrm{ACD}=\angle \mathrm{FGH} \text { (Angle bisector) } $

$\text { And, } \angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Angle bisector) } $

$\text { In } \triangle \mathrm{ACD} \text { and } \Delta \mathrm{FGH} \text {, }$

$\angle \mathrm{A}=\angle \mathrm{F} \text { (Proved above) }$

$\angle \mathrm{ACD}=\angle \mathrm{FGH} \text { (Proved above) }$

$\therefore \Delta \mathrm{ACD} \sim \Delta \mathrm{FGH} \text { (By AA similarity criterion) }$

$\Rightarrow \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$


(ii) $\triangle \mathrm{DCB} \sim \Delta \mathrm{HGE}$

Ans: $\text { In } \triangle \mathrm{DCB} \text { and } \triangle \mathrm{HGE} \text {, } $

$\angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Proved above) } $

$\angle \mathrm{B}=\angle \mathrm{E} \text { (Proved above) } $

$\therefore \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE} \text { (By AA similarity criterion) }$


(iii) $\triangle \mathrm{DCA} \sim \Delta \mathrm{HGF}$

Ans: In $\Delta \mathrm{DCA}$ and $\Delta \mathrm{HGF}$,

$\angle \mathrm{ACD}=\angle \mathrm{FGH}$ (Proved above)

$\angle A=\angle F$ (Proved above)

$\therefore \Delta \mathrm{DCA} \sim \Delta \mathrm{HGF}$ (By AA similarity criterion)


11. In the following figure, $\mathrm{E}$ is a point on side CB produced of an isosceles triangle $\mathrm{ABC}$ with $\mathrm{AB}=\mathrm{AC}$. If $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{EF} \perp \mathrm{AC}$, prove that $\triangle \mathrm{ABD} \sim$ $\triangle \mathrm{ECF}$


an isosceles triangle ABCDEF.png


Ans: It is given that $\mathrm{ABC}$ is an isosceles triangle.

$\therefore \mathrm{AB}=\mathrm{AC}$

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ECF}$

In $\Delta \mathrm{ABD}$ and $\triangle \mathrm{ECF}$

$\angle \mathrm{ADB}=\angle \mathrm{EFC}\left(\operatorname{Each} 90^{\circ}\right)$

$\angle \mathrm{BAD}=\angle \mathrm{CEF}$ (Proved above)

$\therefore \Delta \mathrm{ABD} \sim \triangle \mathrm{ECF}$ (By using AA similarity criterion)


12. Sides $\mathrm{AB}$ and $\mathrm{BC}$ and median AD of a triangle $\mathrm{ABC}$ are respectively proportional to sides PQ and QR and median PM of $\triangle \mathrm{PQR}$ (see the given figure). Show that $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$.


Median equally divides the opposite side


Ans: Median equally divides the opposite side.

$\therefore \mathrm{BD}=\frac{\mathrm{BC}}{2}$ and $\mathrm{QM}=\frac{\mathrm{QR}}{2}$

Given that,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{PQM}$,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \text { (Proved above) }$

$\therefore \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}$ (By SSS similarity criterion)

$\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{PQM}$ (Corresponding angles of similar triangles)

In $\triangle \mathrm{ABC}$ and $\Delta \mathrm{PQR}$,

$\angle \mathrm{ABD}=\angle \mathrm{PQM} \text { (Proved above) }$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}$

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ (By SAS similarity criterion)


13. $\mathrm{D}$ is a point on the side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ such that $\angle \mathrm{ADC}=\angle \mathrm{BAC}$. Show that $\mathrm{CA}^{2}= \mathrm{CB.CD}$


We know that corresponding sides of similar triangles are in proportion


Ans: $\text { In } \triangle \mathrm{ADC} \text { and } \triangle \mathrm{BAC} \text {, }$

$\angle \mathrm{ADC}=\angle \mathrm{BAC}$ (Given) $\angle \mathrm{ACD}=\angle \mathrm{BCA}$ (Common angle)

$\therefore \Delta \mathrm{ADC} \sim \triangle \mathrm{BAC}$ (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion. $\therefore \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}}$ $\Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \cdot \mathrm{CD}$


14. Sides $\mathrm{AB}$ and AC and median AD of a triangle $\mathrm{ABC}$ are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$


Let us extend AD and PM up to point


Ans: Given that,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

Let us extend AD and PM up to point $E$ and $L$ respectively, such that $A D=$ DE and PM $=$ ML. 

Then, join $B$ to $E, C$ to $E, Q$ to $L$, and $R$ to $L .$


medians divide opposite sides


We know that medians divide opposite sides.

Therefore, $\mathrm{BD}=\mathrm{DC}$ and $\mathrm{QM}=\mathrm{MR}$

Also, $\mathrm{AD}=\mathrm{DE}$ (By construction)

And, $\mathrm{PM}=\mathrm{ML}$ (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram.

$\therefore \mathrm{AC}=\mathrm{BE}$ and $\mathrm{AB}=\mathrm{EC}$ (Opposite sides of a parallelogram are equal) Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, $P Q=L R$

It was given that

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}$

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{\mathrm{AE}}{\mathrm{PL}}$

$\therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{PQL}$ (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

$\therefore \angle \mathrm{BAE}=\angle \mathrm{QPL} \ldots$ (1)

Similarly, it can be proved that $\triangle \mathrm{AEC} \sim \triangle \mathrm{PLR}$ and

$\angle \mathrm{CAE}=\angle \mathrm{RPL} \ldots$ (2)

Adding equation (1) and (2), we obtain

$\angle \mathrm{BAE}+\angle \mathrm{CAE}=\angle \mathrm{QPL}+\angle \mathrm{RPL}$

$\Rightarrow \angle \mathrm{CAB}=\angle \mathrm{RPQ} \ldots$ (3)

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}$ (Given)

$\angle \mathrm{CAB}=\angle \mathrm{RPQ}[\mathrm{Using}$ equation $(3)]$

$\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ (By SAS similarity criterion)


15. A vertical pole of a length $6 \mathrm{~m}$ casts a shadow $4 \mathrm{~m}$ long on the ground and at the same time a tower casts a shadow $28 \mathrm{~m}$ long. Find the height of the tower.


the same time a tower.png


Ans: Let $\mathrm{AB}$ and $\mathrm{CD}$ be a tower and a pole respectively.

Let the shadow of $\mathrm{BE}$ and DF be the shadow of $\mathrm{AB}$ and $\mathrm{CD}$ respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, $\angle \mathrm{DCF}=\angle \mathrm{BAE}$

And, $\angle \mathrm{DFC}=\angle \mathrm{BEA}$

$\angle \mathrm{CDF}=\angle \mathrm{ABE}$ (Tower and pole are vertical to the ground)

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CDF}$ (AAA similarity criterion)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{BE}}{\mathrm{DF}}$

$\Rightarrow \frac{\mathrm{AB}}{6 \mathrm{~cm}}=\frac{28}{4}$

$\Rightarrow \mathrm{AB}=42 \mathrm{~m}$

Therefore, the height of the tower will be 42 metres.


16. If $\mathrm{AD}$ and $\mathrm{PM}$ are medians of triangles $\mathrm{ABC}$ and $\mathrm{PQR}$, respectively where $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ Prove that $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$


the corresponding sides of similar triangles


Ans: It is given that $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$

We know that the corresponding sides of similar triangles are in proportion. 

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}} \ldots(1)$

Also, $\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}, \angle \mathrm{C}=\angle \mathrm{R} \quad \ldots$ (2)

Since AD and PM are medians, they will divide their opposite sides. $\therefore \mathrm{BD}=\frac{\mathrm{BC}}{2}$ and $\mathrm{QM}=\frac{\mathrm{QR}}{2}$

From equations (1) and (3), we obtain

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{PQM}$,

$\angle B=\angle Q$ (Using equation (2))

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

(Using equation (4))

$\therefore \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}$ (By SAS similarity criterion)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$


Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 6 Triangles

Chapter

Dropped Topics

Triangles

Exercise 6.4 

Exercise 6.5

Exercise 6.6

 


Class 10 Maths Chapter 6: Exercises Breakdown

Exercises

Number of Questions

Exercise 6.1

3 Questions & Solutions (3 Short Answers)

Exercise 6.2

10 Questions & Solutions (9 Short Answers, 1 Long Answer)

Exercise 6.3

16 Questions & Solutions (13 Short Answers, 3 Long Answers)



Conclusion

NCERT Solutions for Triangle Chapter Class 10 offers a clear and comprehensive understanding of triangle concepts. It makes studying simple and interesting by simplifying difficult subjects like the Pythagorean theorem, area computation, and trigonometric ratios.  It's crucial for students to focus on understanding fundamental concepts, such as triangle properties and different theorems. Practicing with the provided solutions and solving previous year question papers is essential for exam preparation. In previous year question papers, around 4-5 questions have been typically asked Ch 6 Class 10 Maths.


Other Study Material for CBSE Class 10 Maths Chapter 6



Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

FAQs on NCERT Solutions Class 10 Maths Chapter 6 Triangles

1. How Many Exercises are There in NCERT Solutions for Class 10 Maths Chapter 6 Triangles?

The Class 10 Maths NCERT Solutions for the Chapter Triangles contain exercises corresponding to each topic. The chapter contains a total of 6 exercises with a total of 65 questions. The questions include a mix of long and short type questions. Students should attempt to understand all the concepts and theorems given in the chapter and then solve the questions in the exercises. Solving these questions will definitely give the students a competitive edge in the exams.

2. How Many Marks are Allotted to the Class 10 Maths Chapter 6 Triangles in the Board Exam?

The Class 10 Maths Chapter 6 Triangles is a part of a broader unit ‘Geometry’ in the Board exams. The unit of Geometry comprises a total of 15 marks in the Board exams. The Triangles chapter is an important chapter as per the examination point of view and as such is likely to carry around 5-6 marks in the Class 10 Board exams.

3. Which are the Important Topics to Remember Present in CBSE Class 10 Chapter 6 Triangles?

In the CBSE Class 10 Maths Chapter 6, the topic discussed is Triangles. The topics that are important from this chapter are:

  • Similarity theorems of triangles.

  • Criteria for triangle similarity.

  • Area calculation of similar triangles.

  • Pythagoras theorem and the concept of similar triangles.

Students should make sure that they are thorough with all these topics and should leave no stone unturned to practise as many questions as possible while preparing this topic for the exams.

4. Can the PDF of NCERT Solutions for Class 10 Maths Chapter 6 Triangles be Downloaded for Free?

Yes, at Vedantu you can download the  NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF for absolutely free of cost. The solutions of this chapter have been compiled by some of the best subject experts and provide a clear insight into the various concepts included in the chapter. To download the PDF of the Class 10 Maths Chapter 6, you will just be required to click on the link provided on this page.


You can also choose to take a print out of the PDF and keep it handy for revision purposes. You can also download the Vedantu app n your phone from where you will be able to access the top-notch study material for your Class 10 exam preparation at one go.

5. Do I need to practice all the questions given in the NCERT Solutions Class 10 Maths Triangles?

It is a good idea to practice every question given in the NCERT Solutions for the Class 10 Maths chapter on Triangles. This way you will understand all the topics and concepts clearly and solve all the problems easily. You will also gain confidence about the exam with increased speed and accuracy because the NCERT Solutions provided by Vedantu are curated by subject matter experts. These solutions are therefore guaranteed to help you to clear your concepts easily and effectively for your exam.

6. What are the important topics covered in Class 10 Maths NCERT Solutions Chapter 6?

Chapter 6 of the Class 10 Maths NCERT book deals with Triangles. The most important topics that are covered in this chapter are:

  • Definition of a triangle

  • Similarity of two polygons with an equal number of edges

  • Similarity of triangles

  • Proving the Pythagorean Theorem

The concepts of Class 10 Maths Chapter 6 may be a bit tricky to understand. Therefore it's a good idea to download and study the NCERT solutions for Class 10 Maths. These solutions are prepared by subject matter experts with decades of experience and will help you to understand all concepts thoroughly and easily.

7. How can I score the best in Class 10 Maths Chapter 6 Triangles?

The following points will help you to score well and get to the best of your potential in Class 10 Maths Chapter 6 Triangles:

  • Understand the concept of this chapter.

  • Prioritise NCERT.

  • Refer to extra materials like NCERT Solutions of Class 10 Maths Chapter 6 Triangles available at free of cost on the Vedantu app and on the Vedantu website.

  • Solve model papers.

  • Maintain a separate notebook for formulas and theorems.

  • Practice graphs and diagrams.

If you want to score better than all your peers, then your best shot is definitely to download the NCERT Solutions for Class 10 Maths by Vedantu. Vedantu’s NCERT solutions are prepared by the best Maths teachers in India and written in easy to understand language. 

8. What are the most important theorems that come in Class 10 Chapter 6 Triangles?

The most important theorems in class 10 Chapter 6 Triangles are:

  • Pythagoras Theorem

  • Midpoint Theorem

  • Remainder Theorem

  • Angle Bisector Theorem

  • Inscribed Angle Theorem

To clearly understand the major theorems included in the Class 10 Chapter 6 Triangles, it's best to download the NCERT solutions for Class 10 Maths. These solutions will help you in learning advanced theorems like the ones present in this chapter. This way you can be sure that you will be able to score well in your Class 10 board exams as well. 

9. What is the name of chapter 6 class 10 maths?

The name of chapter 6 in Class 10 is Triangles.

10. What is the formula for the triangles in Ch 6 Maths Class 10?

  • General Formula for area of a triangle is  ½ * base * height

  • Pythagorean Theorem: a² + b² = c² (where a and b are the lengths of the two shorter sides, and c is the hypotenuse - the longest side opposite the right angle)

  • Heron's Formula for Area of a Triangle: A = √(s(s - a)(s - b)(s - c)) where s = (a + b + c) / 2