# Ncert Solutions For Class 9 Maths Ex 7.4

## Ncert Solutions For Class 9 Maths Chapter 7 Ex 7.4

Question 1:

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

Solution:

ABC is a right angle triangle, right angled at B.

Now A+C=90o$\angle A + \angle C = 90^{o}$

Angles A and C are less than 90o

Now,    B>A$\angle B > \angle A$

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again, B>C$\angle B > \angle C$

AC > AB …(ii)

(Side opposite to greater angle are longer)

Hence, from (i) and (ii), we can say that AC (hypotenuse) is the longest side.

Hence proved.

Question 2:

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

Solution:

ABC+PBC=180o(Linearpair)$\angle ABC + \angle PBC = 180^{o} (Linear pair)$ ABC=180oPBC$\Rightarrow \angle ABC = 180^{o} – \angle PBC$

Similarly, ACB=180oQCB$\angle ACB = 180^{o} – \angle QCB$

It is given that PBC<QCB$\angle PBC < \angle QCB$

180oQCB<180oPBC$180^{o} – \angle QCB < 180^{o} – \angle PBC$

Or ACB<ABC$\angle ACB < \angle ABC$ [From (i) and (ii)]

AB<AC$\Rightarrow AB < AC$ AC>AB$\Rightarrow AC > AB$

Hence proved.

Question 3:

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

Solution:

B<A$\angle B < \angle A$ (Given)

BO > AO …(i)

(Side opposite to greater angle is longer)

C<D$\angle C < \angle D$ (Given)

CO > DO …(ii)

(Same reason)

BO + CO > AO + DO

Hence Proved.

Question 4:

AB and CD are respectively the smallest and longest side of a quadrilateral ABCD (see fig.) Show that A>C andB>D$\angle A > \angle  C\ and \angle B > \angle D$ .

Solution:

Join AC.

Mark the angles as shown in the figure..

In ABC$\triangle ABC$ ,

BC > AB (AB is the shortest side)

2>4$\angle 2 > \angle 4$ …(i)

[Angle opposite to longer side is greater]

In ADC$\triangle ADC$ ,

CD > AD (CD is the longest side)

1>3$\angle 1 > \angle 3$ …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

2+1>4+3$\angle 2 + \angle 1 > \angle 4 + \angle 3$ A>C$\Rightarrow \angle A > \angle C$

Similarly, by joining BD, we can prove that

B>D$\angle B > \angle D$

Question 5:

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

Solution:

PR > PQ

PQR>PRQ$\angle PQR > \angle PRQ$ …(i)

[Angle opposite to longer side is greater]

QPS>RPS (PS bisects QPR)$\angle QPS > \angle RPS\ (PS\ bisects\ \angle QPR)$ …(ii)

In PQS$\triangle PQS$ , PQS+QPS+PSQ=180o$\angle PQS + \angle QPS + \angle PSQ = 180^{o}$

PSQ=180o(PQS+QPS)$\Rightarrow \angle PSQ = 180^{o} – (\angle PQS + \angle QPS)$ …(iii)

Similarly in PRS$\triangle PRS$ , PSR=180o(PRS+RPS)$\triangle PSR = 180^{o} – (\angle PRS + \angle RPS)$

PSR=180o(PSR+QPS)$\Rightarrow \angle PSR = 180^{o} – (\angle PSR + \angle QPS)$ [from (ii) … (iv)

From (i), we know that PQS<PSR$\angle PQS < \angle PSR$

So from (iii) and (iv), PSQ<PSR$\angle PSQ < \angle PSR$

PSR>PSQ$\Rightarrow \angle PSR > \angle PSQ$

Hence proved.

Question 6:

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

Solution:

We have a line I and O is the point not on I

OPi$OP \perp i$

We have to prove that OP < OQ, OP < OR and OP < OS.

OP < OS

In OPQ$\triangle OPQ$ , P=90o$\angle P = 90^{o}$

Therefore, Q$\angle Q$ is an acute angle (i.e, Q<90o$\angle Q < 90^{o}$ )

Q<P$\angle Q < \angle P$

Hence, OP < OQ (Side opposite to greater angle is longer)

Similarly, we can prove that OP is shorter than OR, OS, etc.

Hence proved