# NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4

Get here all NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 Triangles with suitable explanations and examples. Questions are well explained in PDF and video format using simplified steps, making class 9 Maths too easy.

Exercise 7.4 of standard 9 Maths solution’s videos are available in Hindi and English Medium free to use or download. Question wise solutions are given in videos with step by step derivation using simple formulae.## Class 9 Maths Exercise 7.4 Solution in Hindi and English

Class: 9 | Mathematics |

Chapter: 7 | Triangles |

Exercise: 7.4 | NCERT Solutions in PDF and Videos |

### NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 in Hindi and English Medium

### Class 9 Maths Exercise 7.4 Solution in Hindi Medium Video

#### Class 9 Maths Chapter 7 Exercise 7.4 Solution in Videos

#### Main Points of Class 9 Maths Chapter 7

- If the three sides of one triangle are equal to the other triangle’s three sides, then the two triangles are congruent (SSS Rule).
- If two right triangles, the hypotenuse and one side of a triangle are equal to the hypotenuse, and the other is on one side of the triangle, then the two triangles are congruent (RHS Rule).
- In a triangle, the angle opposite the longest side is greater (larger).
- In a triangle, the side opposite the largest (largest) angle is longer.
- The sum of any two sides of a triangle is greater than the third side.

## Important Example 1 for Exams

D is a point on side BC of Δ ABC such that AD = AC. Show that AB > AD. Solution: In Δ DAC, AD = AC (Given) So, ∠ ADC = ∠ ACD (Angles opposite to equal sides) Now, ∠ ADC is an exterior angle for ΔABD. So, ∠ ADC > ∠ ABD or, ∠ ACD > ∠ ABD or, ∠ ACB > ∠ ABC So, AB > AC (Side opposite to larger angle in Δ ABC) or, AB > AD (AD = AC)

##### What is the central concept on which exercise 7.4 based?

Class 9 Maths exercise 7.4 is based on the concepts of inequalities. Questions are based on the comparison of two angles or two sides, which are unequal.

##### Which example in exercise 7.4 is important for exams?

There is only one example that is example number 9. This example important not only for exams but to understand some other questions also.

##### Which is the most challenging question of exercise 7.4 of class 9 Maths?

Question number 3 and 5 are a little bit difficult to understand. These questions need more practice to be confident.

## Example 2 with Solutions for School tests

P is a point equidistant from two lines l and m intersecting at point A. Show that the line AP bisects the angle between them. Solution: You are given that lines l and m intersect each other at A. Let PB ⊥ l, PC ⊥ m. It is given that PB = PC. We have to show that ∠ PAB = ∠ PAC. Now, consider Δ PAB and Δ PAC. In these two triangles, PB = PC (Given) ∠ PBA = ∠ PCA = 90° (Already given in the question) PA = PA (as these are Common) So, Δ PAB ≅ Δ PAC (Using RHS congruency rule) So, ∠ PAB = ∠ PAC (Because these are CPCT)