triangles _ exercise 7.4

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Exercise 7.4

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Page 1: Triangles _ Exercise 7.4

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Triangles : Exercise 7.4 (Mathematics NCERT Class9th)

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

 

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Page 2: Triangles _ Exercise 7.4

Let ABC be a right angled   in which   90ºBut   180º       [By angle­sum property]

(http://dronstudy.com/wp­content/uploads/2013/12/28.png)  90º +   180º   90º   are acute angles   AC > AB and AC > BC [Since side opp. to greater angle is larger]

Hence , in a right triangle , the hypotenuse is the  longest side.

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Q.2     In figure, sides AB and AC of   ABC are extended to points P and Qrespectively. Also  . Show that AC > AB.

(http://dronstudy.com/wp­content/uploads/2013/12/241.png)Sol.

Since      (given)   (Both sides multiply by ­ ) 180º   (Adding 180º on both sides)

[Since   and    as well as,   and   are linear pair]  AC > AB [Since side opp. to greater angle is larger]

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Q.3     In figure ,   Show that AD < BC.

Δ ∠ABC =∠ABC + ∠BCA + ∠CAB =

⇒ ∠BCA + ∠CAB =⇒ ∠BCA + ∠CAB =⇒ ∠BCA and ∠CAB

⇒ ∠BCA < and ∠CAB <90o 90o

⇒ ∠BCA < ∠ABC and ∠CAB < ∠ABC

Δ∠PBC < ∠QCB

∠PBC < ∠QCB

⇒ −∠PBC > −∠QCB

⇒ −∠PBC > − ∠QCB180o

∠PBC ∠ABC ∠QCB ∠ACB

⇒ ∠ABC > ∠ACB

∠B < ∠A and ∠C < ∠D.

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Page 3: Triangles _ Exercise 7.4

(http://dronstudy.com/wp­content/uploads/2013/12/27.png)Sol.

Since    (Given)Therefore AO < BO ..................(1)and OD < OC.......................(2)[Since side opp. to greater angle is larger]Adding  (1) and (2) , we getAO + OD < BO + OC  AD < BC.

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Q.4     AB and CD are respectively the smallest and longest sides of a quadrilateralABCD (see figure). Show that 

(http://dronstudy.com/wp­content/uploads/2013/12/29.png)Sol.

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. 

(http://dronstudy.com/wp­content/uploads/2013/12/11.jpg)Therefore , in   ABC, we have BC > AB        ... (1) 

[Since angle opp. to longer side is greater]

∠B < ∠A and ∠C < ∠D

∠A > ∠C and ∠B > ∠D

Δ

⇒ ∠8 > ∠3

Page 4: Triangles _ Exercise 7.4

Since CD is the longest side of quadrilateral ABCD. Now , in    ACD, we have CD > AD          ... (2) 

[Since angle opp. to longer side is greater]Adding (1) and (2), we get 

      Again , in   ABD , we have AD > AB [Since AB is the shortest side]       ... (3) 

In   BCD, we have CD > BC [Since CD is the longest side]         ... (4) 

Adding (3) and (4), we get  

 Thus , 

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Q.5      In figure PR > PQ and PS bisects  . Prove that 

(http://dronstudy.com/wp­content/uploads/2013/12/311.png)Sol.

In   PQR ,  we have PR > PQ [Given]         [Since angle opp. to larger  side is greater]

(http://dronstudy.com/wp­content/uploads/2013/12/331.png)       [Adding   on both sides]

      [Since PS is the bisector of   since  ]Now , in   PQS and PSR ,we have 

 180ºand    180º  

and  

Δ

⇒ ∠7 > ∠4

∠8 + ∠7 > ∠3 + ∠4⇒ ∠A > ∠C

Δ

⇒ ∠1 > ∠6Δ

⇒ ∠2 > ∠5

∠1 + ∠2 > ∠5 + ∠6⇒ ∠B > ∠D

∠A > ∠C and ∠B > ∠D

∠QPR ∠PSR > ∠PSQ

Δ

⇒ ∠PQR > ∠PRQ

⇒∠PQR + ∠1 > ∠PRQ + ∠1 ∠1⇒ ∠PQR + ∠1 > ∠PRQ + ∠2 ∠P ∠1 = ∠2

Δs

∠PQS + ∠1 + ∠PSQ =∠PRS + ∠2 + ∠PSR =

⇒ ∠PQS + ∠1 = − ∠PSQ180o

∠PRS + ∠2 = − ∠PSR180o

Page 5: Triangles _ Exercise 7.4

Therefore 180º –                    [From (1)]      i.e. 

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Q.6        Show  that  of  all  line  segments  drawn  from  a  given  point  not  on  it,  theperpendicular line segment is the shortest. Sol.

Let P be any point not on the straight line l. PM   l and N is any point on l other than M. In   PMN , we have 

 

(http://dronstudy.com/wp­content/uploads/2013/12/321.png)[Since       

          PM <  PN                   [Since side opp. to greater angle is larger]

Hence, PM is the shortest of all line segments from P to AB.

∠PSQ > − ∠PSR180o

⇒ −∠PSQ > −∠PSR

⇒ ∠PSQ < ∠PSR ∠PSR > ∠PSQ

⊥Δ

∠M = 90o

⇒ ∠N < 90o

∠M = 90o ⇒∠MPN + ∠PNM = 90o ⇒ ∠P + ∠N = ⇒ ∠N < ]90o 90o

⇒ ∠N < ∠M

Page 6: Triangles _ Exercise 7.4

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