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.

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Solution:

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

Now A+C=90o

Angles A and C are less than 90o

Now,    B>A

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again, B>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 are extended to points P and Q  respectively. Also, PBC<QCB . Show that AC > AB.

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Solution:

ABC+PBC=180o(Linearpair) ABC=180oPBC

Similarly, ACB=180oQCB

It is given that PBC<QCB

180oQCB<180oPBC

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

AB<AC AC>AB

Hence proved.

 

Question 3:

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

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Solution:

B<A (Given)

BO > AO …(i)

(Side opposite to greater angle is longer)

C<D (Given)

CO > DO …(ii)

(Same reason)

Adding (i) and (ii)

BO + CO > AO + DO

BC > AD

AD < BC

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 .

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Solution:

Join AC.

Mark the angles as shown in the figure..

In ABC ,

BC > AB (AB is the shortest side)

2>4 …(i)

[Angle opposite to longer side is greater]

In ADC ,

CD > AD (CD is the longest side)

1>3 …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

2+1>4+3 A>C

Similarly, by joining BD, we can prove that

B>D

 

Question 5:

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

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Solution:

PR > PQ

PQR>PRQ …(i)

[Angle opposite to longer side is greater]

QPS>RPS (PS bisects QPR) …(ii)

In PQS , PQS+QPS+PSQ=180o

PSQ=180o(PQS+QPS) …(iii)

Similarly in PRS , PSR=180o(PRS+RPS)

PSR=180o(PSR+QPS) [from (ii) … (iv)

From (i), we know that PQS<PSR

So from (iii) and (iv), PSQ<PSR

PSR>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.            

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Solution:

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

OPi

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

OP < OS

In OPQ , P=90o

Therefore, Q is an acute angle (i.e, Q<90o )

Q<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

 

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