NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Triangles |

Exercise |
Ex 5.4 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

**Question 1.**

**Show that in a right angled triangle, the hypotenuse is the longest side.**

**Solution:**

Let ABC be a right angled triangle, such that ∠ ABC = 90°

We know that,

∠ABC + ∠BCA + ∠CAB = 180° (By A property)

⇒ 90° + ∠BCA + ∠CAB = 180°

⇒ ∠BCA + ∠CAB = 90°

From above, we have ∠ BCA and ∠ CAB are acute angles.

⇒ ∠BCA < 90°

and ∠CAB < 90°

⇒ ∠BCA < ∠ABC

and ∠CAB < ∠ABC

⇒ AB < AC and BC < AC (∵ Side opposite to greater angle is longer)

Hence, the hypotenuse (AC) is the longest side.

**Question 2.**

**In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.**

**Solution:**

We know that,

∠ACB + ∠QCB = 180° (Linear pair)…(i)

and ∠ABC+ ∠PBC = 180° (Linear pair)…(ii)

From Eqs. (i) and (ii), we have

∠ABC + ∠PBC = ∠ACB + ∠QCB ….(iii)

But ∠PBC < ∠QCB (Given)…(iv) From Eqs. (iii) and (iv), we have ∠ ABC > ∠ ACB

⇒ AC > AB

(∵ Side opposite to greater angle is longer)

**Question 3.**

**In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.**

**Solution:**

**Question 4.**

**AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A> ∠C and ∠B > ∠D.**

**Solution:**

Given ABCD is a quadrilateral. AB is the smallest side and CD is the longest side.

To prove ∠A > ∠C and ∠B > ∠D

Construction Join A to C and B to D.

Proof In ∆ABC, we have AB is the smallest side.

∴ AB < BC

⇒ ∠5 < ∠1

(∵ Angle opposite to longer side is greater) …(i)

In ∆ADC, we have CD is the largest side.

**Question 5.**

**In figure, PR > PQ and PS bisect ∠QPR. Prove that**

**∠PSR >∠PSQ.**

**Solution:**

In ∆ PQR, we have

PR > PQ (Given)

⇒ ∠ PQR > ∠ PRQ …(i)

(∵ Angle opposite to longer side is greater)

**Question 6.**

**Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**

**Solution:**

Given x is a line and A is a point not lying on x. AB ⊥ x, Cis any point on x other than B.

To prove AB∠C

⇒ AC> AB

(∵ the Side opposite to greater angle is longer)

⇒ AB < AC

Hence, the perpendicular line segment is the shortest.

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