NCERT Solutions for Class 10 Maths Chapter 6 : Triangles

Q: 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.

Q: 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.

Q: 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

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)

Medium

Solution

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Answer

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.

Answer

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.

Medium

Solution

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Answer

(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.

Answer

(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.

Question 3

In Fig., (i) and (ii),

D E ∣ ∣ B C

. Find

E C

in (i) and

A D

in (ii).

Medium

Solution

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Answer

(i) Given :

D E ∥ B C

in

△

ABC,

Using Basic proportionality theorem,

∴ \frac{A D}{D B} = \frac{A E}{E C}

\Rightarrow \frac{1 . 5}{3} = \frac{1}{E C}

\Rightarrow E C = \frac{3}{1 . 5}

E C = 3 \times \frac{1 0}{1 5} = 2

cm

E C = 2

cm.

(ii) In

△ A B C, D E ∥ B C

(Given)

Using Basic proportionality theorem,

∴ \frac{A D}{D B} = \frac{A E}{E C}

\Rightarrow \frac{A D}{7 . 2} = \frac{1 . 8}{5 . 4}

\Rightarrow A D = 1.8 \times \frac{7 . 2}{5 . 4} = \frac{1 8}{1 0} \times \frac{7 2}{1 0} \times \frac{1 0}{5 4} = \frac{2 4}{1 0}

\Rightarrow A D = 2.4

cm

So,

A D = 2.4

cm

Answer

(i) Given :

D E ∥ B C

in

△

ABC,

Using Basic proportionality theorem,

∴ \frac{A D}{D B} = \frac{A E}{E C}

\Rightarrow \frac{1 . 5}{3} = \frac{1}{E C}

\Rightarrow E C = \frac{3}{1 . 5}

E C = 3 \times \frac{1 0}{1 5} = 2

cm

E C = 2

cm.

(ii) In

△ A B C, D E ∥ B C

(Given)

Using Basic proportionality theorem,

∴ \frac{A D}{D B} = \frac{A E}{E C}

\Rightarrow \frac{A D}{7 . 2} = \frac{1 . 8}{5 . 4}

\Rightarrow A D = 1.8 \times \frac{7 . 2}{5 . 4} = \frac{1 8}{1 0} \times \frac{7 2}{1 0} \times \frac{1 0}{5 4} = \frac{2 4}{1 0}

\Rightarrow A D = 2.4

cm

So,

A D = 2.4

cm

Question 4

E

and

F

are points on the sides

P Q

and

P R

respectively of a

△ P Q R

. For each of the following cases, state whether

E F ∣ ∣ Q R

:
(i)

P E = 3.9

cm,

E Q = 3

cm,

P F = 3.6

cm and

F R = 2.4

cm
(ii)

P E = 4

cm,

Q E = 4.5

cm,

P F = 8

cm and

R F = 9

cm
(iii)

P Q = 1.28

cm,

P R = 2.56

cm,

P E = 0.18

cm and

P F = 0.36

cm

Medium

Solution

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Answer

E

and

F

are two points on side

P Q

and

P R

in

△ P Q R .

(i)

P E = 3.9

cm,

E Q = 3

cm and

P F = 3.6

cm,

F R = 2.4

cm

Using Basic proportionality theorem,

∴ \frac{P E}{E Q} = \frac{3 . 9}{3} = \frac{3 9}{3 0} = \frac{1 3}{1 0} = 1.3

\frac{P F}{F R} = \frac{3 . 6}{2 . 4} = \frac{3 6}{2 4} = \frac{3}{2} = 1.5

\frac{P E}{E Q} \neq = \frac{P F}{F R}

So,

E F

is not parallel to

Q R .

(ii)

P E = 4

cm,

Q E = 4.5

cm,

P F = 8

cm,

R F = 9

cm

Using Basic proportionality theorem,

∴ \frac{P E}{Q E} = \frac{4}{4 . 5} = \frac{4 0}{4 5} = \frac{8}{9}

\frac{P F}{R F} = \frac{8}{9}

\frac{P E}{Q E} = \frac{P F}{R F}

So,

E F

is parallel to

Q R .

(iii)

P Q = 1.28

cm,

P R = 2.56

cm,

P E = 0.18

cm,

P F = 0.36

cm

Using Basic proportionality theorem,

E Q = P Q - P E = 1.28 - 0.18 = 1.10

cm

F R = P R - P F = 2.56 - 0.36 = 2.20

cm

\frac{P E}{E Q} = \frac{0 . 1 8}{1 . 1 0} = \frac{1 8}{1 1 0} = \frac{9}{5 5}

... (i)

\frac{P E}{F R} = \frac{0 . 3 6}{2 . 2 0} = \frac{3 6}{2 2 0} = \frac{9}{5 5}

... (ii)

∴ \frac{P E}{E Q} = \frac{P F}{F R .}

So,

E F

is parallel to

Q R .

Answer

E

and

F

are two points on side

P Q

and

P R

in

△ P Q R .

(i)

P E = 3.9

cm,

E Q = 3

cm and

P F = 3.6

cm,

F R = 2.4

cm

Using Basic proportionality theorem,

∴ \frac{P E}{E Q} = \frac{3 . 9}{3} = \frac{3 9}{3 0} = \frac{1 3}{1 0} = 1.3

\frac{P F}{F R} = \frac{3 . 6}{2 . 4} = \frac{3 6}{2 4} = \frac{3}{2} = 1.5

\frac{P E}{E Q} \neq = \frac{P F}{F R}

So,

E F

is not parallel to

Q R .

(ii)

P E = 4

cm,

Q E = 4.5

cm,

P F = 8

cm,

R F = 9

cm

Using Basic proportionality theorem,

∴ \frac{P E}{Q E} = \frac{4}{4 . 5} = \frac{4 0}{4 5} = \frac{8}{9}

\frac{P F}{R F} = \frac{8}{9}

\frac{P E}{Q E} = \frac{P F}{R F}

So,

E F

is parallel to

Q R .

(iii)

P Q = 1.28

cm,

P R = 2.56

cm,

P E = 0.18

cm,

P F = 0.36

cm

Using Basic proportionality theorem,

E Q = P Q - P E = 1.28 - 0.18 = 1.10

cm

F R = P R - P F = 2.56 - 0.36 = 2.20

cm

\frac{P E}{E Q} = \frac{0 . 1 8}{1 . 1 0} = \frac{1 8}{1 1 0} = \frac{9}{5 5}

... (i)

\frac{P E}{F R} = \frac{0 . 3 6}{2 . 2 0} = \frac{3 6}{2 2 0} = \frac{9}{5 5}

... (ii)

∴ \frac{P E}{E Q} = \frac{P F}{F R .}

So,

E F

is parallel to

Q R .

Question 5

In Fig., if

L M ∣ ∣ C B

and

L N ∣ ∣ C D

, prove that

\frac{A M}{A B} = \frac{A N}{A D}

.

Medium

Solution

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Answer

In

△

ABC,

L M ∥ B C

∴

By proportionality theorem,

\frac{A M}{A B} = \frac{A L}{A C} . . . . . . . . . . . . (1)

Similarly,
In

△

ADC,

L N ∥ C D

∴

By proportionality theorem,

\frac{A N}{A D} = \frac{A L}{A C} . . . . . . . . . . . . (2)

∴

from

(1)

and

(2)

,

\frac{A M}{A B} = \frac{A N}{A D}

Answer

In

△

ABC,

L M ∥ B C

∴

By proportionality theorem,

\frac{A M}{A B} = \frac{A L}{A C} . . . . . . . . . . . . (1)

Similarly,
In

△

ADC,

L N ∥ C D

∴

By proportionality theorem,

\frac{A N}{A D} = \frac{A L}{A C} . . . . . . . . . . . . (2)

∴

from

(1)

and

(2)

,

\frac{A M}{A B} = \frac{A N}{A D}

Question 6

In fig.

D E ∣ ∣ A C

and

D F ∣ ∣ A E .

prove that

\frac{B F}{F E} = \frac{B E}{E C}

Medium

Solution

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Answer

In

Δ A B C

D E ∣ ∣ A C

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

\frac{B E}{E C} = \frac{B D}{D A}

__(i)

In

Δ A E B

D F ∣ ∣ A E

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

\frac{B F}{F E} = \frac{B D}{D A}

__(ii)

From (i) & (ii)

\frac{B E}{E C} = \frac{B F}{F E}

∴

Hence proved.

Answer

In

Δ A B C

D E ∣ ∣ A C

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

\frac{B E}{E C} = \frac{B D}{D A}

__(i)

In

Δ A E B

D F ∣ ∣ A E

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

\frac{B F}{F E} = \frac{B D}{D A}

__(ii)

From (i) & (ii)

\frac{B E}{E C} = \frac{B F}{F E}

∴

Hence proved.

Question 7

In Figure, $D E ∥ O Q$ and $D F ∥ O R,$ Show that $E F ∥ Q R$

Medium

Solution

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Answer

In

Δ P O Q,

we have

D E ∥ O Q

(Given)

By Basic proportionality Theorem, we have

\frac{P E}{E Q} = \frac{P D}{D O} \dots (i)

Similarly,

In

Δ P O R

, we have

D F ∥ O R

(Given)

∴ \frac{P D}{D O} = \frac{P F}{F R} \dots (i i)

Now, from

(i)

and

(i i)

, we have

\frac{P E}{E Q} = \frac{P F}{F R} \Rightarrow E F ∥ Q R

[Applying the converse of Basic proportionality Theorem in

Δ P Q R]

Answer

In

Δ P O Q,

we have

D E ∥ O Q

(Given)

By Basic proportionality Theorem, we have

\frac{P E}{E Q} = \frac{P D}{D O} \dots (i)

Similarly,

In

Δ P O R

, we have

D F ∥ O R

(Given)

∴ \frac{P D}{D O} = \frac{P F}{F R} \dots (i i)

Now, from

(i)

and

(i i)

, we have

\frac{P E}{E Q} = \frac{P F}{F R} \Rightarrow E F ∥ Q R

[Applying the converse of Basic proportionality Theorem in

Δ P Q R]

Question 8

In A, B and C are points on OP, OQ and OR respectively such that AB

∣ ∣

PQ and AC

∣

PR. Show that BC

∣ ∣

QR.

Medium

Solution

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Answer

In

△ O P Q

, we have

A B ∥ P Q

Therefore, by using basic proportionality theorem , we have

\frac{O A}{A P} = \frac{O B}{B Q} . . . . . . . . . . . . . . . . . (i)

IN

△ O P R

, we have

A C ∥ P R

Therefore, by using basic proportionality theorem , we have

\frac{O C}{C R} = \frac{O A}{A P} . . . . . . . . . . . . . . . . . (i i)

Comparing (i)&(ii), we get

\frac{O B}{B Q} = \frac{O C}{C R}

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

B C ∥ Q R

Answer

In

△ O P Q

, we have

A B ∥ P Q

Therefore, by using basic proportionality theorem , we have

\frac{O A}{A P} = \frac{O B}{B Q} . . . . . . . . . . . . . . . . . (i)

IN

△ O P R

, we have

A C ∥ P R

Therefore, by using basic proportionality theorem , we have

\frac{O C}{C R} = \frac{O A}{A P} . . . . . . . . . . . . . . . . . (i i)

Comparing (i)&(ii), we get

\frac{O B}{B Q} = \frac{O C}{C R}

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

B C ∥ Q R

Question 9

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.

Medium

Solution

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Answer

Given:

In

△ A B C, D

is midpoint of

A B

and

D E

is parallel to

B C .

∴

A D = D B

To prove:

A E = E C

Proof:

Since,

D E ∥ B C

∴

By Basic Proportionality Theorem,

\frac{A D}{D B} = \frac{A E}{E C}

Since,

A D = D B

∴

\frac{A E}{E C} = 1

∴

A E = E C

Answer

Given:

In

△ A B C, D

is midpoint of

A B

and

D E

is parallel to

B C .

∴

A D = D B

To prove:

A E = E C

Proof:

Since,

D E ∥ B C

∴

By Basic Proportionality Theorem,

\frac{A D}{D B} = \frac{A E}{E C}

Since,

A D = D B

∴

\frac{A E}{E C} = 1

∴

A E = E C

Question 10

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.

Medium

Solution

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Answer

Given: In

△ A B C, D

and

E

are the mid-points of sides

A B

and

A C

respectively.

∴ A D = D B and A E = E C

\Rightarrow \frac{A D}{D B} = 1 and \frac{A E}{E C} = 1

\Rightarrow \frac{A D}{D B} = \frac{A E}{E C}

Therefore, by the converse of Basic proportionality Theorem,

D E ∥ B C

Answer

Given: In

△ A B C, D

and

E

are the mid-points of sides

A B

and

A C

respectively.

∴ A D = D B and A E = E C

\Rightarrow \frac{A D}{D B} = 1 and \frac{A E}{E C} = 1

\Rightarrow \frac{A D}{D B} = \frac{A E}{E C}

Therefore, by the converse of Basic proportionality Theorem,

D E ∥ B C

NCERT Solutions for Class 10 Maths Chapter 6 : Triangles

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NCERT Solutions for Class 10 Maths Chapter 6 : Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Triangles – Brief Overview

1. Triangles

2. Similar Figures

Similarity of Triangles

3. Basic Proportionality Theorem

4. Criteria for Similarity of Triangles

5. Areas of Similar Triangles

6. Pythagoras Theorem

Frequently Asked Questions on NCERT Class 10 Maths Chapter 6 : Triangles