Ncert Solutions For Class 9 Maths Ex 10.2

Ncert Solutions For Class 9 Maths Chapter 10 Ex 10.2

Long answer questions

1.If from an external point A of a circle with centre X,two tangents AC and AD are drawn such that DAC=120° , prove that AC+AD=AX i.e.,AX=2AC.

 

Sol. Join XC and XD.

since,                XCAC

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∴ [tangent to any circle is perpendicular to its radius at point of contact]

 

∴                XCA=90°

In ΔXCB and ΔXDB,we have

CA=DA

[tangents from an external point]

XA=XA                       [common side]

XC=XD                     [radii of a circle]

∴   By SSS  congruency , we have

ΔXCAΔXDA

   XAC=XAD

=12CAD=12×120=60

In  ΔXCA ,XCA=90°

∴  ACAX=cos60

 

BCBX=12

       AX=2AC

Also, AX=AC + AC

    AX=AC + AD     [Since,  AC=AD]

 

2.If x,y,z are the sides of a right triangle where c is hypotenuse,prove that the radius r of the circle which touches the sides of the triangle is given by

 

                                            r  = x+yz2

 

Sol.   Let the circle touches the sides YZ,ZX,XY of the right triangle XYZ at D,E,F respectively,where YZ=x,ZX=y and XY=z.Then XE = XF and YD = YF.

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Also,           ZE=ZD=r

I.e.,               y-r=XF,

x-r=yF

Or                 XY=z=XF+YF

    Z=y-r+x-r

    r  = x+yz2

 

3.In the given figure,from an external poiny P,a tangent PT and a line segment PXY is drawn to a circle with the centre O.ON is perpendicular on the chord XY. Prove that:

                            (i)PX.PY= PY2XN2

                            (ii) PN2XN2= OP2OT2

                            (iii) PX.PY=PT2

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Sol.                   (i) PX.PY=(PN-XN)(PN+YN)

=(PN-XN)(PN+XN)

since, XN=YN,as ONChord XY]

= PN2XN2

(ii) PN2XN2=(OP2ON2)- XN2

Since,in rt. ONP OP2= ON2+ PN2]

=OP2ON2XN2

=OP2-(ON2)+ XN2)

=OP2OX2

=OP2OT2

Since,OX=OT=r]

(iii)From (i) and(ii),we have

PX.PY= OP2OT2

PX.PY=PT2

∴ In rt. OTP; OP2= OT2+ PT2

 

4.XY is a diameter and XZ is a chord with centre O such that YXZ= 30.The tangent at Z intersects XY extended at a point D. Prove that YZ=YD.

Sol. Here, XOY is a diameter of the circle, such that YXZ=30 and ZD be the tangent at Z.

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XZY=90in a semi-circle]

Also,          ZYD=YXZ+XZY

[ext. of a is equal to sum of interior opp.angles]

=30+ 90

=120

YZD=YXZ=30

In ,by angles sum property,we have

YDZ+ZYD+YZD=180

YDZ+120+30=180

YDZ=18012030

=30

\Rightarrow                 YDZ=YDZ

=30

Hence,                           YD=YZ

[sides opp.to equal angles]

 

5.Two circles with centres O and O of radii 3 cm and 4cm respectively intersect at two points P and Q such that OP are tangents to the two circles.Find the length of the common chord PQ.

 

Sol.       Clearly, OPO= 90

OO= 32+42

=9+16=25

=5cm

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Let                       RO=x

⇒             RO=5-x

Now,                    PR2= PO2RO2

=9-x2    ……..(1)

Also,                    PR2= PO2RO2

=16-(5x)2……….(2)

From (1) and (2),we have

9-x2=16-(25+x2-10x)

9-x2=16-25+x2+10x)

10x=18

X=1.8cm

From (1),we have

PR2=9-1.82

=9-3.24

=5.76

PR=5.76

=2.4cm

Hence, the required length of the common chord is 2 x PR i.e., 2×2.4 i.e., 4.8cm.