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Triangles : Exercise 7.4 (Mathematics NCERT Class 9th)

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

Let ABC be a right angled $\Delta \,$ in which $\angle ABC =$ 90º
But $\angle ABC + \angle BCA + \angle CAB =$ 180º       [By angle-sum property] $\Rightarrow$ 90º + $\angle BCA + \angle CAB =$ 180º
$\Rightarrow$ $\angle BCA + \angle CAB =$ 90º
$\Rightarrow$ $\angle BCA\,\,and\,\,\angle CAB$ are acute angles
$\Rightarrow$ $\angle BCA\, < {90^o}\,and\,\,\angle CAB < {90^o}$
$\Rightarrow$ $\angle BCA\, < \angle ABC\,and\,\angle CAB < \angle ABC$
$\Rightarrow$ AC > AB and AC > BC [Since side opp. to greater angle is larger]
Hence , in a right triangle , the hypotenuse is the  longest side.

Q.2     In figure, sides AB and AC of $\Delta$ ABC are extended to points P and Q respectively. Also $\angle PBC\, < \angle QCB$. Show that AC > AB.

Since $\angle PBC\, < \angle QCB$    (given)
$\Rightarrow$ $- \angle PBC\, > - \angle QCB$ (Both sides multiply by - )
$\Rightarrow$ 180º $- \angle PBC\, > {180^o} - \angle QCB$ (Adding 180º on both sides)
[Since $\angle PBC\,$ and $\angle ABC\,$  as well as, $\angle QCB\,$ and $\angle ACB\,$ are linear pair]
$\Rightarrow$ $\angle ABC\, > \angle ACB$
$\Rightarrow$ AC > AB [Since side opp. to greater angle is larger]

Q.3     In figure , $\angle B\, < \angle A\,and\,\angle C < \angle D.$ Show that AD < BC.

Since $\angle B\, < \angle A\,and\,\angle C < \angle D$  (Given)
Therefore AO < BO ..................(1)
and OD < OC.......................(2)
[Since side opp. to greater angle is larger]

Adding  (1) and (2) , we get
AO + OD < BO + OC
$\Rightarrow$ AD < BC.

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

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 $\Delta$ ABC, we have
BC > AB
$\Rightarrow$     $\angle 8\, > \;\angle 3$ ... (1)
[Since angle opp. to longer side is greater]
Since CD is the longest side of quadrilateral ABCD.
Now , in  $\Delta$ ACD, we have
$\Rightarrow$       $\angle 7 > \angle 4$ ... (2)
[Since angle opp. to longer side is greater]
Adding (1) and (2), we get
$\angle 8 + \angle 7 > \angle 3 + \angle 4$
$\Rightarrow$      $\angle A > \angle C$

Again , in $\Delta$ ABD , we have
AD > AB [Since AB is the shortest side]
$\Rightarrow$     $\angle 1 > \angle 6$ ... (3)
In $\Delta$ BCD, we have
CD > BC [Since CD is the longest side]
$\Rightarrow$       $\angle 2 > \angle 5$ ... (4)
Adding (3) and (4), we get
$\angle 1 + \angle 2 > \angle 5 + \angle 6$
$\Rightarrow$ $\angle B > \angle D$
Thus , $\angle A > \angle C\,\,and\,\,\angle B > \angle D$

Q.5      In figure PR > PQ and PS bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$

In $\Delta$ PQR ,  we have
PR > PQ [Given]
$\Rightarrow$      $\angle PQR > \angle PRQ$  [Since angle opp. to larger  side is greater] $\Rightarrow$  $\angle PQR + \angle 1 > \angle PRQ + \angle 1$    [Adding $\angle 1$ on both sides]
$\Rightarrow$ $\angle PQR + \angle 1 > \angle PRQ + \angle 2$    [Since PS is the bisector of $\angle P$ since $\angle 1 = \angle 2$]
Now , in $\Delta s\,$ PQS and PSR ,we have
$\angle PQS + \angle 1 + \angle PSQ =$ 180º
and  $\angle PRS + \angle 2 + \angle PSR =$ 180º

$\Rightarrow$  $\angle PQS + \angle 1 = {180^o} - \angle PSQ$
and  $\angle PRS + \angle 2 = {180^o} - \angle PSR$

Therefore 180º – $\angle PSQ > {180^o} - \angle PSR$                  [From (1)]
$\Rightarrow$  $- \angle PSQ > - \angle PSR$
$\Rightarrow$  $\angle PSQ < \angle PSR$ i.e. $\angle PSR > \angle PSQ$

Q.6    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 $\bot$ l and N is any point on l other than M.
In $\Delta$ PMN , we have
$\angle M = {90^o}$
$\Rightarrow$ $\angle N < {90^o}$ [Since $\angle M = {90^o}$ $\Rightarrow$$\angle MPN + \angle PNM = {90^o}$$\Rightarrow$    $\angle P + \angle N = {90^o} \Rightarrow \angle N < {90^o}]$
$\Rightarrow$    $\angle N\, < \angle M$
$\Rightarrow$    PM <  PN                   [Since side opp. to greater angle is larger]
Hence, PM is the shortest of all line segments from P to AB.