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Triangles Class 10 Maths NCERT Solutions Chapter 6 is concerned with the understanding of various aspects of the shape triangle. Students preparing for class 10 maths chapter 6 will be able to clear all their concepts on triangles at the root level. Class 10 triangles are one of the most fundamental geometrical shapes, and their study is extremely useful in the actual world. The concepts taught in this chapter will also assist students in studying other forms such as circles, trapeziums, rectangles, and so on, as well as understanding the distinction between congruence and similarity by applying the appropriate principles to these diverse figures.

NCERT Solutions for Class 10 Maths chapter 6 concentrates on the essential concepts such as the similar figures, triangle similarities, triangle similarity criterion, area of similar triangles, and Pythagoras theorem This chapter focuses on subjects such as triangle congruence, similarity criteria, and how the Pythagoras theorem can be demonstrated using these concepts. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

Chapter 6 Maths Class 10 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. Triangles Class 10 Solutions Chapter 6 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Exercise 6.1

Question 1

Fill in the blanks using the correct word given in brackets :

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

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

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

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

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

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

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

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

Solution

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Two figures that have the same shape are said to be similar.

When two figures are similar, the ratios of the lengths of their corresponding sides are equal.

(i) All circles are similar.

Since they have same shape.

(ii) All square are similar.

Since the ratios of the lengths of their corresponding sides are equal.

(iii) All equilateral triangles are similar.

Since the ratios of the lengths of their corresponding sides are equal.

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

Question 2

Give two different examples of pair of:

(i) similar figures. (ii) non-similar figures.

(i) similar figures. (ii) non-similar figures.

Solution

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(i) Similar figures :

1. Two equilateral triangles of sides $5$ cm and $6$ cm each.

2. Two circle of different diameter and centre.

(ii) Non-similar figures :

1. A square and a triangle.

2. A circle and a quadrilateral.

This is one of the various possible solutions as this question might have several possible answers.

Exercise 6.2

Question 1

In Fig., (i) and (ii), $DE∣∣BC$. Find $EC$ in (i) and $AD$ in (ii).

Solution

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(i) Given : $DE∥BC$ in $△$ ABC,

Using Basic proportionality theorem,

$∴DBAD =ECAE $

$⇒31.5 =EC1 $

$⇒EC=1.53 $

$EC=3×1510 =2$ cm

$EC=2$ cm.

(ii) In $△ABC,DE∥BC$ (Given)

Using Basic proportionality theorem,

$∴DBAD =ECAE $

$⇒7.2AD =5.41.8 $

$⇒AD=1.8×5.47.2 =1018 ×1072 ×5410 =1024 $

$⇒AD=2.4$cm

So, $AD=2.4$ cm

Question 2

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $△PQR$. For each of the following cases, state whether $EF∣∣QR$ :

(i) $PE=3.9$ cm, $EQ=3$ cm, $PF=3.6$ cm and $FR=2.4$ cm

(ii) $PE=4$ cm, $QE=4.5$ cm, $PF=8$ cm and $RF=9$ cm

(iii) $PQ=1.28$ cm, $PR=2.56$ cm, $PE=0.18$ cm and $PF=0.36$ cm

(i) $PE=3.9$ cm, $EQ=3$ cm, $PF=3.6$ cm and $FR=2.4$ cm

(ii) $PE=4$ cm, $QE=4.5$ cm, $PF=8$ cm and $RF=9$ cm

(iii) $PQ=1.28$ cm, $PR=2.56$ cm, $PE=0.18$ cm and $PF=0.36$ cm

Solution

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$E$ and $F$ are two points on side $PQ$ and $PR$ in $△PQR.$

(i) $PE=3.9$ cm, $EQ=3$ cm and $PF=3.6$ cm, $FR=2.4$ cm

Using Basic proportionality theorem,

$∴EQPE =33.9 =3039 =1013 =1.3$

$FRPF =2.43.6 =2436 =23 =1.5$

$EQPE =FRPF $

So, $EF$ is not parallel to $QR.$

(ii) $PE=4$ cm, $QE=4.5$ cm, $PF=8$ cm, $RF=9$ cm

Using Basic proportionality theorem,

$∴QEPE =4.54 =4540 =98 $

$RFPF =98 $

$QEPE =RFPF $

So, $EF$ is parallel to $QR.$

(iii) $PQ=1.28$ cm, $PR=2.56$ cm, $PE=0.18$ cm, $PF=0.36$ cm

Using Basic proportionality theorem,

$EQ=PQ−PE=1.28−0.18=1.10$ cm

$FR=PR−PF=2.56−0.36=2.20$ cm

$EQPE =1.100.18 =11018 =559 $... (i)

$FRPE =2.200.36 =22036 =559 $ ... (ii)

$∴EQPE =FR.PF $

So, $EF$ is parallel to $QR.$

Question 3

In Fig., if $LM∣∣CB$ and $LN∣∣CD$, prove that $ABAM =ADAN $.

Solution

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In $△$ABC,

$LM∥BC$

$∴$ By proportionality theorem,

$ABAM =ACAL ............(1)$

Similarly,

In $△$ADC,

$LN∥CD$

$∴$ By proportionality theorem,

$ADAN =ACAL ............(2)$

$∴$ from $(1)$ and $(2)$,

$ABAM =ADAN $

$LM∥BC$

$∴$ By proportionality theorem,

$ABAM =ACAL ............(1)$

Similarly,

In $△$ADC,

$LN∥CD$

$∴$ By proportionality theorem,

$ADAN =ACAL ............(2)$

$∴$ from $(1)$ and $(2)$,

$ABAM =ADAN $

Question 4

In fig.

$DE∣∣AC$ and $DF∣∣AE.$ prove that

$FEBF =ECBE $

$FEBF =ECBE $

Solution

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In $ΔABC$

$DE∣∣AC$

Line drawn parallel to one side of triangle, insects the other two sides. It divides the other side in same ratio.

$ECBE =DABD $ __(i)

In $ΔAEB$

$DF∣∣AE$

Line drawn parallel to one side of triangle, intersects the other sides. It divides the other sides in same ratio.

$FEBF =DABD $ __(ii)

From (i) & (ii)

$ECBE =FEBF $

$∴$ Hence proved.

Question 5

In Figure, $DE∥OQ$ and $DF∥OR,$ Show that $EF∥QR$

Solution

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In $ΔPOQ,$ we have

$DE∥OQ$(Given)

By Basic proportionality Theorem, we have

$EQPE =DOPD …(i)$

Similarly,

In $ΔPOR$, we have

$DF∥OR$(Given)

$∴DOPD =FRPF …(ii)$

Now, from $(i)$ and $(ii)$, we have

$EQPE =FRPF ⇒EF∥QR$

[Applying the converse of Basic proportionality Theorem in $ΔPQR]$

Question 6

In A, B and C are points on OP, OQ and OR respectively such that AB$∣∣$PQ and AC$∣$ PR. Show that BC$∣∣$QR.

Solution

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In $△OPQ$, we have

$AB∥PQ$

Therefore, by using basic proportionality theorem , we have

$APOA =BQOB .................(i)$

IN $△OPR$, we have

$AC∥PR$

Therefore, by using basic proportionality theorem , we have

$CROC =APOA .................(ii)$

Comparing (i)&(ii), we get

$BQOB =CROC $

Therefore, by using converse of basic proportionality theorem, we get

$BC∥QR$

Question 7

Using Theorem $6.1,$ prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Theorem 6.1: 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.

Solution

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Given:

In $△ABC,D$ is midpoint of $AB$ and $DE$ is parallel to $BC.$

$∴$ $AD=DB$

$∴$ $AD=DB$

To prove:

$AE=EC$

Proof:

Since, $DE∥BC$

$∴$ By Basic Proportionality Theorem,

$DBAD =ECAE $

$∴$ By Basic Proportionality Theorem,

$DBAD =ECAE $

Since, $AD=DB$

$∴$ $ECAE =1$

$∴$ $AE=EC$

$∴$ $ECAE =1$

$∴$ $AE=EC$

Question 8

Using converse of Basic proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution

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Given: In $△ABC,D$ and $E$ are the mid-points of sides $AB$ and $AC$ respectively.

$∴AD=DBandAE=EC$

$⇒DBAD =1andECAE =1$

$⇒DBAD =ECAE $

Therefore, by the converse of Basic proportionality Theorem, $DE∥BC$

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Practice more questions

A triangle is a three-sided polygon having three edges and three vertices in geometry. It is a fundamental shape in geometry. ∆ABC denotes a triangle with vertices A, B, and C.

Similar figures are two figures that have the same shape but not necessarily the same size.

Congruent figures are two figures that have the same shape and size.

It should be noted that all congruent figures are similar, but comparable figures do not have to be congruent.

Two triangles are similar:

- If their corresponding angles are identical and
- Their corresponding sides have the same ratio (or proportion)

It is important to note that equiangular triangles are formed when the corresponding angles of two triangles are equal. Any two comparable sides in two equiangular triangles have the same ratio.

Thales, a prominent Greek mathematician, revealed an important insight about two equiangular triangles, which is as follows:

The ratio of any two corresponding sides in two equiangular triangles is always the same.

He is thought to have utilized a result known as the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.

Theorem - If a line is drawn parallel to one side of a triangle in order to meet the other two sides at unique spots, the other two sides are divided in the same ratio. As a result, in ∆ABC, if DE || BC, then

(Angle-Angle-Angle) AAA Similarity Criterion states that If corresponding angles in two triangles are equal, then their corresponding sides are in the same ratio, therefore the two triangles are similar.

The (Angle-Angle) AA Similarity Criterion states that If two angles of one triangle are equal to two angles of the other triangle, then the two triangles are similar.

(Side-Side-Side) SSS Similarity Criterion states that If corresponding sides in two triangles have the same ratio, their corresponding angles are equal, and the triangles are similar.

The (Side-Angle-Side) SAS Similarity Criterion states that If one angle of a triangle is equal to one angle of another triangle and the sides containing these angles have the same ratio (proportion), the two triangles are similar.

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

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Pythagoras Theorem - In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Thus in right ∆ABC right angled at B,

AC^{2} = AB^{2} + BC^{2}

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Q1. What are the most important topics in NCERT Class 10 Chapter 6 Triangles?

Answer: All the concepts related to Triangles are covered in Chapter 6 of the Class 10 Maths NCERT book. The NCERT Solutions Class 10 Chapter 6 includes a variety of key topics such as triangle similarity, triangle congruence, triangle area, and Pythagoras theorem demonstration using congruence and similarity theorems. These theorems and postulates must be understood in order to pass the tests. Students should ensure that they are well familiar with all of these topics and should leave no concept overlooked in order to practice as many questions as possible while studying for the tests on this topic.

Q2. How many exercises are there in Chapter 6 of Class 10 Maths?

Answer: Chapter 6 of Class 10 Maths Triangles comprises of 65 questions divided into 6 exercises, 25 of which are easy, 20 of which are moderate, and 20 of which are extended answer type problems. These questions take various forms, such as word problems, fill-in-the-blanks, activity-based sums, and so on. The questions are a mix of long and short answers. Students should try to comprehend all of the concepts and theorems presented in the chapter before attempting to answer the problems in the exercises. The variety of questions accessible makes studying more fascinating and engaging, while also assisting students in receiving the highest possible grade.

Q3. What are the different theorems discussed in the Class 10 Chapter 6 Triangles?

Answer: The following are the most important theorems in Chapter 6 of Class 10:

- Basic Proportionality Theorem
- Pythagoras Theorem
- Midpoint Theorem
- Remainder Theorem
- Angle Bisector Theorem
- Inscribed Angle Theorem

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability