Question

please answer friends

Answer 1

Answer

Let center of circle be O and the line through O touches AB at M and CD at N.

Now, as we know line from center to chord intersect it at 90° and bisect the chord

MN= 7 cm (given)

Let MO be x cm

Therefore NO=7-x cm

Now,AM=5 cm an CN=12 cm

IN rt Triangle AMO

AO^2=AM^2+OM^2

AO^2=5^2+x^2

AO^2=25+x^2

In rt triangle CNO

CO^2=ON^2+CN^2

25+x^2=12^2+(7-x)^2      [AO=CO=radius}

25+x^2=144+49+x^2-14x

25=193-14x

14x=168

x=12

Putting value of x in eq 1

AO^2=25+12^2

AO^2=169

AO=13cm =Radius

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