triangles _ exercise 7.4
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Exercise 7.4TRANSCRIPT
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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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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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(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
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
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
⇒
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