EX 7.4 QUESTION 1.
Show that in a right-angled triangle, the hypotenuse is the longest side.
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.
∠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: 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.
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.
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.
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.