These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4

Question 1.

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

Solution:

In right angled triangle ABC

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

or, ∠A + ∠C + 90 = 180° (∵ ∠B = 90)

or, ∠A + ∠C = 90°

∴ ∠B > ∠A

Therefore, AC > BC …….(i)

(Side opposite to greater angle is greater)

Again, ∠B > ∠C

∴ AC > AB ……(ii)

(Side opposite to greater angle is greater larger)

From equation (i) and (ii)

AC is longer then both AB and BC.

Therefore AC is longes side of ∆ABC.

Question 2.

In Fig. 7.48 side AB and AC of ∆ABC are extended to point P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:

In ∆ABC, we have given,

∠PBC < ∠QCB ……(i)

Now, ∠PBC + ∠ABC = 180° …..(ii) (Linear pair)

Again, ∠QCB + ∠ACB = 180° ……(iii) (Linear pair)

From equation (ii) and (iii)

∠PBC + ∠ABC = ∠QCB + ∠ACB

But ∠PBC < ∠QBC Therefore, ∠ABC > ∠ACB

∴ AC > AB (Side opposite to greater angle is greater)

Question 3.

In Fig. 7.49, ∠B > ∠A and ∠C < ∠D. Show thatn AD < BC.

Solution:

In ∆BAO,

∠B < ∠A

∴ OA < OB ……(i)

(Single opposite to greater angle is greater)

Again, In ∆CDO

∠C < ∠D

∴ OD < OC ……(ii)

(Single opposite to greater angle is greater)

Add equation (i) and (ii)

OA + OD < OB + OC

AD < BC

Question 4.

AB and CD are respectively the smallest and longest sides of quadrilateral ABCD (se Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.

Solution:

Given ABCD is a quadrilateral in which AB is smallest and CD is longest side.

To prove that:

(i) ∠A > ∠C

(ii) ∠B > ∠D

Construction: Join A and C.

Proof: In ∆ABC

AB < BC (∵ AB is smallest side)

∴ ∠BCA < ∠BAC …..(i)

(∵ Angle opposite to larger side is greater)

Again, In ∆ACD,

AD < CD (∵ CD is largest side)

∴ ∠ACD < ∠ADC …….(ii)

(Angle opposite to larger side is greater)

Adding equation (i) and (ii)

∠BCA + ∠ACD < ∠BAC + ∠ADC

∠BCD < ∠BAD

or, ∠C < ∠A Similarly by joining B and D we can prove that ∠B > ∠D

Question 5.

In Fig. 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Solution:

In ∆PQR

PR > PQ

∠Q > ∠R

Adding ∠1 both side,

∠Q + ∠1 > ∠R + ∠1

or, ∠Q + ∠1 > ∠R + ∠2 (∵ ∠1 = ∠2)

∠PSR > ∠PSQ

(∵ Exterior angle is equal to the sum of opposite interior angles)

Question 6.

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

Solution:

Given: A straight line l and a point P not lying on l. PM ⊥ l and N is any point on l other than M.

To prove that: PM < PN

Proof: In ∆PMN, we have

∠M = 90

∠N < 90

(∵ ∠M = 90

⇒ ∠MPN + ∠PNM = 90

⇒ ∠P + ∠N = 90

⇒ ∠N < 90)

⇒∠N< ∠M

⇒ PM < PN (Side opposite to greater angle is larger)

Hence, PM is shortest of all line segments from P to AB.