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# Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.4

• Last Updated : 22 Sep, 2021

### Question 1. Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution:

Given: Right angle triangle intersect âˆ B=90Â°

To show: AC>AB and AC>BC

Solution:âˆ A+âˆ B+âˆ C=180Â°           —————-[angle sum property]

âˆ A+90Â°+âˆ C=180Â°

âˆ A+âˆ C=180Â°=90Â°

âˆ A+âˆ C=90Â°

âˆ´âˆ A<90Â°         and  âˆ C<90Â°

BC<AC                         AB<AC                   ———-[sides opposite to longer angle is greater]

âˆ´Hypotenuse  is the longest side.

### Question 2. In Fig. 7.48, sides AB and AC of Î”ABC are extended to points P and Q respectively. Also, âˆ PBC < âˆ QCB. Show that AC > AB.

Solution:

Given : âˆ PBC < âˆ QCB       LET this be  [âˆ 1 < âˆ 2]

To Show : AD < BC

Solution: âˆ 1 < âˆ 2                       ——–[given]

-âˆ 1 > -âˆ 2

180-âˆ 1>180-âˆ 2

âˆ 3>âˆ 4                                  ———[linear pair]

In âˆ†ABC,

âˆ 3>âˆ 4

AC>AB

### Question 3. In Fig. 7.49, âˆ B < âˆ A and âˆ C < âˆ D. Show that AD < BC.

Solution:

Given: âˆ B < âˆ A and âˆ C < âˆ D

Solution: In  âˆ†BOA

âˆ B < âˆ A

âˆ´AO<BO————-1

In âˆ†COD

âˆ C < âˆ D

âˆ´OD<OC————-2

AO+OD+<BO+OC

### Question 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that âˆ A > âˆ C and âˆ B > âˆ D.

Solution:

Given: AB is smaller side

CD is longest side

To show: âˆ A > âˆ C and âˆ B > âˆ D.

Solution : In âˆ†ABC

BC>AB

âˆ 1>âˆ 2              ———-[angle opposite to greater side is greatest]-1

In âˆ†ABC

âˆ 3>âˆ 4             ———[ angle opposite to greater side is greatest]-2

âˆ 1+âˆ 2+âˆ 3+âˆ 4

âˆ A>âˆ C

ii) In âˆ†ABD

âˆ 5>âˆ 6             ——————-[ angle opposite to greater side is greatest]-3

In âˆ†BCD

CD>BC

âˆ 7>âˆ 8             ——————-[ angle opposite to greater side is greatest]-4

âˆ 5+âˆ 6+âˆ 7+âˆ 8

âˆ B > âˆ D

### Question 5. In Fig 7.51, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR > âˆ PSQ.

Solution:

Given: PR>PQ

PS is angle bisects âˆ QPR

To show: âˆ PSR > âˆ PSQ

Solution: PR>PQ

âˆ´âˆ 3+âˆ 4          ————[angle opposite to greater side is larger]

âˆ 3+âˆ 1+x=180Â° ————-[angle sum property of âˆ†]

âˆ 3=180Â°-âˆ 1-x ————1

In âˆ†PSR

4+âˆ 2+x=180Â° ————-[angle sum property of âˆ†]

âˆ 4=180Â°-âˆ 2-x ————2

Because âˆ 3>âˆ 4

180Â°-âˆ 1-x >180Â°-âˆ 2-x

-âˆ 1>-âˆ 2

âˆ 1<âˆ 2

âˆ PSQ<âˆ PSR  OR âˆ PSR>âˆ PSQ

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

Solution:

Given: Let P be any point not lying on a line  L.

PM âŠ¥ L

Now, âˆ  is any point another than M lying on line=L

In âˆ†PMN

âˆ M90Â°                  ———-[ PM âŠ¥ L]

âˆ L<90Â°                 ——-[âˆ´âˆ M90Â° âˆ L<90Â°  âˆ L<90Â°]

âˆ L<âˆ M

AM<PL                ———[side opposite to greater is greater ]

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