NCERT Solutions for Class 9 Maths Exercise 7.4

NCERT Solutions for Class 9 Maths Exercise 7.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Triangles Download as PDF

NCERT Solutions for Class 9 Mathematics Triangles

1. Show that in a right angles triangle, the hypotenuse is the longest side.

Ans. Given: Let ABC be a right angled triangle, right angled at B.

To prove: Hypotenuse AC is the longest side.

Proof: In right angled triangle ABC,

A + B + C =

A + + C = [B = ]

A + C =

And B =

B > C and B > A

Since the greater angle has a longer side opposite to it.

AC > AB and AC > AB

Therefore B being the greatest angle has the longest opposite side AC, i.e. hypotenuse.

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NCERT Solutions for Class 9 Maths Exercise 7.4

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

Ans. Given: In ABC, PBC < QCB

To prove: AC > AB

Proof: In ABC,

4 > 2 [Given]

Now 1 + 2 = 3 + 4 = [Linear pair]

1 > 3 [ 4 > 2]

AC > AB [Side opposite to greater angle is longer]

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NCERT Solutions for Class 9 Maths Exercise 7.4

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

Ans. In AOB,

B < A [Given]

OA < OB ……….(i) [Side opposite to greater angle is longer]

In COD,

C < D [Given]

OD < OC ……….(ii) [Side opposite to greater angle is longer]

Adding eq. (i) and (ii),

OA + OD < OB + OC

AD < BC

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NCERT Solutions for Class 9 Maths Exercise 7.4

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

Ans. Given: ABCD is a quadrilateral with AB as smallest and CD as longest side.

To prove: (i) A > C (ii) B > D

Construction: Join AC and BD.

Proof: (i) In ABC, AB is the smallest side.

4 < 2 ……….(i)

[Angle opposite to smaller side is smaller]

In ADC, DC is the longest side.

3 < 1 ……….(ii)

[Angle opposite to longer side is longer]

Adding eq. (i) and (ii),

4 + 3 < 1 + 2 C < A

(ii) In ABD, AB is the smallest side.

5 < 8 ……….(iii)

[Angle opposite to smaller side is smaller]

In BDC, DC is the longest side.

6 < 7 ……….(iv)

[Angle opposite to longer side is longer]

Adding eq. (iii) and (iv),

5 + 6 < 7 + 8

D < B

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NCERT Solutions for Class 9 Maths Exercise 7.4

5. In figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ.

Ans. In PQR, PR > PQ [Given]

PQR > PRQ …..(i) [Angle opposite to longer side is greater]

Again 1 = 2 …..(ii) [ PS is the bisector of P]

PQR + 1 > PRQ + 2 ……….(iii)

But PQS + 1 + PSQ = PRS + 2 + PSR = [Angle sum property]

PQR + 1 + PSQ = PRQ + 2 + PSR ………(iv)

[PRS = PRQ and PQS = PQR]

From eq. (iii) and (iv),

PSQ < PSR

Or PSR > PSQ

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NCERT Solutions for Class 9 Maths Exercise 7.4

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

Ans. Given: is a line and P is point not lying on PM

N is any point on other than M.

To prove: PM < PN

Proof: In PMN M is the right angle.

N is an acute angle. (Angle sum property of )

M > N

PN > PM [Side opposite greater angle]

PM < PN

Hence of all line segments drawn from a given point not on it, the perpendicular is the shortest.

NCERT Solutions for Class 9 Maths Exercise 7.4

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