Triangles : Exercise 7.4 (Mathematics NCERT Class 9th)

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

Let ABC be a right angled \Delta \, in which \angle ABC = 90º
But \angle ABC + \angle BCA + \angle CAB = 180º       [By angle-sum property]

28 \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 QCB" /> (Both sides multiply by - )
 \Rightarrow 180º {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 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 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.
11Therefore , in \Delta ABC, we have
 \Rightarrow     \;\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 4" /> ... (2)
[Since angle opp. to longer side is greater]
Adding (1) and (2), we get
\angle 3 + \angle 4" />
 \Rightarrow       \angle C" />

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

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


In \Delta PQR ,  we have
PR > PQ [Given]
 \Rightarrow       \angle PRQ" />  [Since angle opp. to larger  side is greater]

33 \Rightarrow   \angle PRQ + \angle 1" />    [Adding \angle 1 on both sides]
 \Rightarrow \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 – {180^o} - \angle PSR" />                  [From (1)]
 \Rightarrow   - \angle PSR" />
 \Rightarrow  \angle PSQ < \angle PSR i.e. \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.

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}

32[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.

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