Exercise 7.4

NCERT solutions for Class 9 Maths Triangles 

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.

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]

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

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

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

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.