## 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.