Ncert Solutions For Class 9 Maths Ex 7.2

Ncert Solutions For Class 9 Maths Chapter 7 Ex 7.2

Question 1:

In an isosceles triangle XYZ with XY=XZ ,the bisector of B and C

Intersect each other at O .join A at O .Show that:

(1)OY=OZ    (2)XO bisects A

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Solution:

12XYZ=12XZY ZYO=XZY

[OY and OZ are bisector of

Y and Z  respectively]

OBY=OZ [ sides opposite to equal angles are equal]

Again,     12XYZ=12XZY

XYO=XYO               [ OY and OZ are bisector  of Y and  Z respectively   ]

In ΔXYO=ΔXYO , we have

XY=XZ  [Given]

OY=OZ  [proved XYove]

XYO=XZO         [proved above]

ΔXYO=ΔXZO          [SAS congruence]

ΔXYO= ZXO   [CPCT]

XO bisects X             proved    

 

Question 2:

In ΔXYZ   ,XO is the perpendicular bisector of YZ .Prove that ΔXYZ is an  isosceles triangle in which  XY=XZ

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 Solution:

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In ΔXYO=ΔXZO , we have

XOZ=XOZ    [] each =90°]

YO=ZO [ XO bisects YZ]

XO=XO [common]

ΔXYO=ΔXZO   [SAS]

XY=XZ    [CPCT]

Hence, ΔXYZ is an isosceles triangle .Proved

 

Question 3:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and

AB respectively (see figure).Show that these altitudes are equal.

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Solution:

In ΔABC,

AB=AC   [Given]

B=C     [angles opposite to equal sides of a triangle are equal]

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Now in right triangles BFC and CEB ,

BFC=CEB                  [Each =90° ]

FBC=ECB                 [proved above]

BC=BC

ΔBFC=ΔCEB            [AAS]

Hence, BE=CF [CPCT] proved

 

Question 4:

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see.fig) show that

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(1) ΔABEΔACF

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(2) AB=AC   i,e ABC is an isosceles triangle.

Solution(1) in ΔABE and ACF , we have

BE=CF         [Given]

BAE=CAF  [common]

BEA=CFA    [ Each =90°]

So, ΔABE=ACF        [AAS]proved

(2)  also, AB=AC                 [ CPCT]

i,e., ABC is an isosceles triangle Proved.

 

Question 5:

ABC and DBC are two isosceles triangle on the same base BC .show that

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ABD=ACD

Solution. In isosceles ΔABC, We have

AB= AC

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ABC=ACB   ———-(i)

[Angles opposite to equal sides are equal]

Now , in isosceles ΔDCB, We have

BD=CD

DBC=DCB —————–(ii)

[Angle opposite to equal sides are equal]

Adding (i) and (ii)  , we have

ABC+DBC=ACB+DCB

ABD=ACD .proved

 

Question 6:

ΔABC is an isosceles triangle in which AB=AC side BA is produced to D such that AD =AB  .show that  BCD is a right angle.

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Solution:

AB=AC

ACB=ABC—————-(1)

[Angles opposite to equal sides are equal]

AB=AD          [Given]

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AD=AC             [AB=AC]

ACD=ADC ——————(ii)

[Angles opposite to equal sides are equal]

Adding (i) and (ii)

ACB+ACD=ABC+ADC BCD=ABC+ADC

Now in ΔBCD,Wehave

BCD+DBC+BDC== 180  [Angle sum property of a triangle]

BCD+BCD=180°

2BCD=180 2BCD=180

 

Question 7:

ABC is a right angled triangle in which A=90 and AB=AC.Find  

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BandC

Solution.  In ΔABC, We have

A=90

AB =AC

We know that angles opposite to equal sides of an isosceles triangle are equal

So, B=C

Since A=90 , therefore sum of remaining two angles =90

Hence B=C=45

 

Question 8:

Show that the angles of an equilateral triangle are 60 each.

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Solution:

As ΔABC  is an equilateral

So, AB=BC=AC

Now, AB=AC

ACB=ABC——-(ii)[ angles sum property of a triangle]

Again BC=AC

BAC=ABC—-(ii)    [same reason]

Now in ΔABC

ABC+ACB+BAC=180  [angle sum property of a triangle are equal]

ABC+ACB+BAC=180=180 [from (i) (ii)]

3ABC=180

ABC =180/3=60°

Also from (i) and (ii)

ACB=60 and BAC

Hence each angle of an equilateral triangle is 60°  Proved