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

Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. 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

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

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- Importatnt key points with additional Assignment and questions bank solved.

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

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- 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

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## 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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