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Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.4
  • Last Updated : 03 Jan, 2021

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

Solution:

Given: Right angle triangle intersect ∠B=90°

To show: AC>AB and AC>BC

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

∠A+90°+∠C=180°        



∠A+∠C=180°=90°  

∠A+∠C=90°  

∴∠A<90°         and  ∠C<90°  

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

∴Hypotenuse  is the longest side.

Question 2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:



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

To Show : AD < BC

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

-∠1 > -∠2

180-∠1>180-∠2                        

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

In ∆ABC,

∠3>∠4  

AC>AB

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

Solution:

Given: ∠B < ∠A and ∠C < ∠D

To show: AD<BC

Solution: In  ∆BOA

∠B < ∠A

∴AO<BO————-1

In ∆COD

∠C < ∠D

∴OD<OC————-2

Adding 1 and 2  

AO+OD+<BO+OC

AD<BC

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

Solution:

Given: AB is smaller side

CD is longest side

To show: ∠A > ∠C and ∠B > ∠D.

Solution : In ∆ABC

BC>AB  

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

In ∆ABC

CD>AD

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

Adding 1 and 2

∠1+∠2+∠3+∠4

∠A>∠C

ii) In ∆ABD

AD>AB

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

In ∆BCD

CD>BC

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

Adding  3 and 4

∠5+∠6+∠7+∠8

∠B > ∠D

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

Solution:

Given: PR>PQ

PS is angle bisects ∠QPR

To show: ∠PSR > ∠PSQ

Solution: PR>PQ

∴∠3+∠4          ————[angle opposite to grater side is larger]

∠3+∠1+x=180° ————-[angle sum property of ∆]  

∠3=180°-∠1-x ————1

In ∆PSR

4+∠2+x=180° ————-[angle sum property of ∆]  

∠4=180°-∠2-x ————2

Because ∠3>∠4

180°-∠1-x >180°-∠2-x

-∠1>-∠2  

∠1<∠2  

∠PSQ<∠PSR  OR ∠PSR>∠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 ⊥ L

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

In ∆PMN

∠M90°                  ———-[ PM ⊥ L]

∠L<90°                 ——-[∴∠M90° ∠L<90°  ∠L<90°]  

∠L<∠M  

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

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