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**Q.1Â Â Â Â Show that in a right angled triangle, the hypotenuse is the longest side.**

**Sol.**

Let ABC be a right angled in which 90Âº

But 180ÂºÂ Â Â Â Â Â [By angle-sum property]

90Âº + 180Âº

90Âº

are acute angles

AC > AB and AC > BC [Since side opp. to greater angle is larger]

Hence , in a right triangle , the hypotenuse is the Â longest side.

**Q.2Â Â Â Â In figure, sides AB and AC of ABC are extended to points P and Q respectively. Also . Show that AC > AB. **

Since Â Â Â (given)

- \angle QCB" /> (Both sides multiply by - )

180Âº {180^o} - \angle QCB" /> (AddingÂ 180Âº on both sides)

[Since and Â as well as, and are linear pair]

\angle ACB" />

AC > AB [Since side opp. to greater angle is larger]

**Q.3Â Â Â Â In figure , Show that AD < BC. **

Since Â (Given)

Therefore AO < BO ..................(1)

and OD < OC.......................(2)

[Since side opp. to greater angle is larger]

AddingÂ (1) and (2) , we get

AO + OD < BO + OC

AD < BC.

**Q.4Â Â Â Â AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \angle C\,and\,\angle B > \angle D" />**

ABCD is a quadrilateral such that AB is its smallest side and CD is its largest side.

Join AC and BD. Since AB is the smallest side of quadrilateral ABCD.

Therefore , in ABC, we have

BC > AB

Â Â \;\angle 3" /> ... (1)

[Since angle opp. to longer side is greater]

Since CD is the longest side of quadrilateral ABCD.

Now , in Â ACD, we have

CD > AD

Â Â Â \angle 4" /> ... (2)

[Since angle opp. to longer side is greater]

Adding (1) and (2), we get

\angle 3 + \angle 4" />

Â Â Â \angle C" />

Again , in ABD , we have

AD > AB [Since AB is the shortest side]

Â Â \angle 6" /> ... (3)

In BCD, we have

CD > BC [Since CD is the longest side]

Â Â Â \angle 5" /> ... (4)

Adding (3) and (4), we get

\angle 5 + \angle 6" />

\angle D" />

Thus , \angle C\,\,and\,\,\angle B > \angle D" />

**Q.5Â Â Â Â Â In figure PR > PQ and PS bisects . Prove that \angle PSQ" />**

In PQR ,Â we have

PR > PQ [Given]

Â Â Â \angle PRQ" />Â [Since angle opp. to larger Â side is greater]

Â \angle PRQ + \angle 1" />Â Â Â [Adding on both sides]

\angle PRQ + \angle 2" />Â Â Â [Since PS is the bisector of since ]

Now , in PQS and PSR ,we have

180Âº

and Â 180Âº

Â

and Â

Therefore 180Âº â€“ {180^o} - \angle PSR" />Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (1)]

Â - \angle PSR" />

Â i.e. \angle PSQ" />

**Q.6Â Â Â Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**

**Sol.**

Let P be any point not on the straight line l. PM l and N is any point on l other than M.

In PMN , we have

[Since Â Â

Â Â

Â Â PM < Â PNÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Since side opp. to greater angle is larger]

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