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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

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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

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 >∠ SQP.

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

1. Chapter 7. Triangles | Exercise 7.1

2. Chapter 7. Triangles | Exercise 7.2

3. Chapter 7. Triangles | Exercise 7.3

4. Chapter 7. Triangles | Exercise 7.4

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