Our ncert solutions for Exercise 7.4 Class 9 maths 7. Triangles - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 7.4 Class 9 maths 7. Triangles - ncert solutions - Toppers Study

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## Exercise 7.4 Class 9 maths 7. Triangles - ncert solutions - Toppers Study

CBSE board students who preparing for **class 9 ncert solutions maths and Mathematics** solved exercise **chapter 7. Triangles** available and this helps in upcoming exams
2023-2024.

### You can Find Mathematics solution Class 9 Chapter 7. Triangles

- All Chapter review quick revision notes for chapter 7. Triangles Class 9
- NCERT Solutions And Textual questions Answers Class 9 Mathematics
- Extra NCERT Book questions Answers Class 9 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 7.4 class 9 Mathematics Chapter 7. Triangles

Sure! The following topics will be covered in this article

- Exercise 7.4 Class 9 Maths 7. Triangles - Ncert Solutions - Toppers Study
- Class 9 Ncert Solutions
- Solution Chapter 7. Triangles Class 9
- Solutions Class 9
- Chapter 7. Triangles Exercise 7.4 Class 9

## 7. Triangles

### | Exercise 7.4 |

## Exercise 7.4 Class 9 maths 7. Triangles - ncert solutions - Toppers Study

**Q1: Show that in a right angled triangle, the hypotenuse is the longest side.**

**Solution: **

Given:Let us consider a right-angled triangle ABC, right-angled at B.

To prove: AC is the longest side.

Proof: In ΔABC,

∠ A + ∠ B + ∠ C = 180° (Angle sum property of a

∠ A + 90º + ∠ C = 180°

∠ A + ∠ C = 90°

Hence, the other two angles have to be acute (i.e., less than 90º).

∴ ∠ B is the largest angle in ΔABC.

⟹∠ B > ∠ A and ∠ B > ∠C

⟹ AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]

Therefore, AC is the largest side in ΔABC.

However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

**Q2 : ****In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠ PBC < ∠ QCB. Show that ****AC > AB****.**

**Answer : **Given: AB and AC of ΔABC are extended to points PandQ respectively. Also, ∠PBC < ∠QCB.

To prove: AC > AB

Proof: In the given figure,

∠ ABC + ∠ PBC = 180° (Linear pair)

⇒ ∠ ABC = 180° - ∠ PBC ... (1)

Also,

∠ ACB + ∠ QCB = 180°

∠ ACB = 180° - ∠ QCB … (2)

As ∠ PBC < ∠ QCB,

⇒ 180º - ∠ PBC > 180º - ∠ QCB

⇒ ∠ ABC > ∠ ACB [From equations (1) and (2)]

⇒ AC > AB (Side opposite to the larger angle is larger.)

**Q3 : ****In the given figure,**** ∠ B < ∠ A and ∠ C < ∠ D****. Show that AD < BC.**

**Solution: **

**Given: **∠ B < ∠ A and ∠ C < ∠ D.

**To prove: **AD < BC

**Proof: **In ΔAOB,

∠ B < ∠ A

⇒ AO < BO (Side opposite to smaller angle is smaller... (1)

In ΔCOD,

∠ C < ∠ D

⇒ OD < OC (Side opposite to smaller angle is smaller) ... (2)

On adding equations (1) and (2), we obtain

AO + OD < BO + OC

AD < BC

**Q4 :****AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠ A > ∠ C and ∠ B > ∠ D.**

**Solution :**

**Given:** AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD .

**To prove: **∠ A > ∠ C and ∠ B > ∠ D.

**Construction:** Let us join AC.

**Proof:** In ΔABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠ 2 < ∠ 1 (Angle opposite to the smaller side is smaller) ... (1)

In ΔADC,

AD < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠ 4 < ∠ 3 (Angle opposite to the smaller side is smaller) ... (2)

On adding equations (1) and (2), we obtain

∠ 2 + ∠ 4 < ∠ 1 + ∠ 3

⇒ ∠ C < ∠ A

⇒ ∠ A > ∠ C

Let us join BD.

In ΔABD,

AB < AD (AB is the smallest side of quadrilateral ABCD)

∴ ∠ 8 < ∠ 5 (Angle opposite to the smaller side is smaller) ... (3)

In ΔBDC,

BC < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠ 7 < ∠ 6 (Angle opposite to the smaller side is smaller) ... (4)

On adding equations (3) and (4), we obtain

∠ 8 + ∠ 7 < ∠ 5 + ∠ 6

⇒ ∠ D < ∠ B

⇒ ∠ B > ∠ D

**Q5 : In ****the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR >∠ ****SQ****P.**

**Solution : **

**Given:** ** **PR > PQ and PS bisects ∠ QPR.

**To prove:** ∠ PSR >∠ SQP.

**Proof: **As PR > PQ,

∴ ∠ PQR > ∠ PRQ (Angle opposite to larger side is larger) ... (1)

PS is the bisector of ∠ QPR.

∴∠ QPS = ∠ RPS ... (2)

∠ PSR is the exterior angle of ΔPQS.

∴ ∠ PSR = ∠ PQR + ∠ QPS ... (3)

∠ PSQ is the exterior angle of ΔPRS.

∴ ∠ PSQ = ∠ PRQ + ∠ RPS ... (4)

Adding equations (1) and (2), we obtain

∠ PQR + ∠ QPS > ∠ PRQ + ∠ RPS

⇒ ∠ PSR > ∠ PSQ [Using the values of equations (3) and (4)]

**Q6 : ****Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**

**Solution: **

**Given:** PNM is a right angled triangle at N.

**To prove:** PN < PM.

**Proof:** In ΔPNM,

∠ N = 90º

∠ P + ∠ N + ∠ M = 180º (Angle sum property of a triangle)

∠ P + ∠ M = 90º

Clearly, ∠ M is an acute angle.

∴ ∠ M < ∠ N

⇒ PN < PM (Side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to *l*, it can be proved that PN is smaller in comparison to them.

Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

##### Other Pages of this Chapter: 7. Triangles

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