Triangles (Exercise 7.4)

Triangles


Chapter 7


Exercise 7.4


EX 7.4 QUESTION 1.


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

 

Solution:

Let us consider ∆ABC such that ∠B = 90°
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 90°+ ∠C = 180°
⇒ ∠A + ∠C = 90°
⇒∠A + ∠C = ∠B
∴ ∠B > ∠A and ∠B > ∠C


EX 7.4 QUESTION 2.


 In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.

Solution:

∠ABC + ∠PBC = 180°            [Linear pair]
and ∠ACB + ∠QCB = 180°     [Linear pair]
But ∠PBC < ∠QCB                   [Given]

⇒ 180° – ∠PBC > 180° – ∠QCB
⇒ ∠ABC > ∠ACB
The side opposite to ∠ABC > the side opposite to ∠ACB
⇒ AC > AB.


EX 7.4 QUESTION 3.


In Fig. 7.49, B < A and C < D. Show that AD < BC.

Solution:

Solution: Since ∠A > ∠B [Given]
∴ OB > OA …(1)
[Side opposite to greater angle is longer]
Similarly, OC > OD …(2)
Adding (1) and (2), we have
OB + OC > OA + OD
⇒ BC > AD


EX 7.4 QUESTION 4.


AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).

Show that A > C and B > D.

Solution:

In ΔABD,

AB < AD < BD

So, ADB < ABD — (i) [angle opposite to longer side is always larger]

Now, in ΔBCD,

BC < DC < BD

BDC < CBD — (ii)

Adding both the equations (i) and  (ii),

ADB + BDC < ABD + CBD

ADC < ABC

B > D

Similarly, In triangle ABC,

ACB < BAC — (iii) [Since the angle opposite to the longer side is always larger]

Now, In ΔADC,

DCA < DAC — (iv)

Adding both the  equation (iii) and  (iv)

ACB + DCA < BAC+DAC

⇒ BCD < BAD

∴ A > C


EX 7.4 QUESTION 5.


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

Solution:

In ∆PQR,
PR > PQ
∠P > ∠R           [In a triangle, the greater angle has longer side opposite to it.]
Adding ∠RPS on both sides
∠Q + ∠RPS > ∠R + ∠RPS
⇒ ∠Q + ∠RPS > ∠PSQ     [∠PSQ is an exterior angle of triangle PSR ]
∠Q + ∠RPS > ∠PSQ     [∠RPS = ∠QPS]
∠PSR > ∠PSQ    [∠PSR is an exterior angle of triangle PQS]


EX 7.4 QUESTION 6.


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

Solution:

BF is a line and A is s point outside of BF.
In ∆ABC,  ∠B = 90°   [Given]
∴  ∠BAC < 90°  and ∠ACB < 90°   [∠BAC + ∠ACB = 90°]
Hence,In ∆ABC, ∠B = ∠ACB   [ ∠B = 90° and ∠ACB < 90°]
AC = AB                   [In a triangle, the greater angle has longer side opposite to it.]
Similarly, AD > AB, AE > AB and AF > AB
Hence, AB is the smallest line.