# Triangles Chapter 6 Class 10 Maths NCERT Textbook With Solutions PDF

‘NCERT Solutions for Class 10 Maths Chapter 6‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 10 Maths Chapter 6 Exercise Solution’ using the download button.

### Chapter 6: Triangles

#### 6.1 Introduction

You are familiar with triangles and many of their properties from your earlier classes. In Class IX, you have studied the congruence of triangles in detail.

Recall that two figures are said to be congruent if they have the same shape and the same size.

In this chapter, we shall study those figures which have the same shape but not necessarily the same size.

Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of the Pythagoras Theorem learned earlier.

#### 6.3 Similarity of Triangles

What can you say about the similarity between the two triangles? You may recall that triangle is also a polygon. So, we can state the same conditions for the similarity of the two triangles.

That is: Two triangles are similar, if
(i) their corresponding angles are equal and
(ii) their corresponding sides are in the same ratio (or proportion).

Note that if the corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which are as follows:

The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the Basic Proportionality Theorem (now known as the Thales Theorem) for the same.

### NCERT Solutions Class 10 Maths Chapter 6 Triangles

1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

(i) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]

⇒1.5/3 = 1/EC

⇒EC = 3/1.5

EC = 3×10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]

⇒ AD/7.2 = 1.8 / 5.4

⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10

2. E and F are points on the sides PQ and PR respectively of a ΔPQR. 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 = 0.63 cm

Solution:

Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, we get, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

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

Therefore, by using Basic proportionality theorem, we get,

PE/QE = 4/4.5 = 40/45 = 8/9

And, PF/RF = 8/9

So, we get here,

PE/QE = PF/RF

Hence, EF is parallel to QR.

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

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)

So, we get here,

PE/EQ = PF/FR

Hence, EF is parallel to QR.

3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

Solution:

In the given figure, we can see, LM || CB,

By using the basic proportionality theorem, we get,

AM/AB = AL/AC……………………..(i)

Similarly, given, LN || CD and using basic proportionality theorem,

From equations (i) and (ii), we get,

Hence, proved.

4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC

Solution:

In ΔABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BE/EC ………………………………………………(i)

In  ΔBAE, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BF/FE ………………………………………………(ii)

From equations (i) and (ii), we get

BE/EC = BF/FE

Hence, proved.

5. In the figure, DE||OQ and DF||OR, show that EF||QR.

Solution:

Given,

In ΔPQO, DE || OQ

So by using the Basic Proportionality Theorem,

PD/DO = PE/EQ……………… ..(i)

Again given, in ΔPOR, DF || OR,

So by using the Basic Proportionality Theorem,

PD/DO = PF/FR………………… (ii)

From equations (i) and (ii), we get,

PE/EQ = PF/FR

Therefore, by the converse of the Basic Proportionality Theorem,

EF || QR, in ΔPQR.