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

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)
OA + OD < OB + OC 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)
(Angle opposite to larger side is greater) ∠BCA + ∠ACD < ∠BAC + ∠ADC
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
∠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.

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