Question

1. Two chords AB and CD of lengths 5 cm and 11 cm respectively, of a circle are parallel
The distance between AB and CD is 3 cm. The radius of the circle is​

Answer 1

ANSWER:

Let OM=X

ON=3-x

Then in ∆ONC,M∆ONC=90

OC^2=ON^2+CN^2

=(3-x)^2+(5.5)^2

=9-6x + x^2 + 30.25

=x^2 - 6x + 39.25_(1)

∆OMA,M∆OMA=90

AO^2=OM^2+MA^2

=x^2+(2.5)^2

=x^2+6.25_(2)

compare EQ(1) with EQ(2)

x^2+6.25=x^2-6x+39.25

(-6x )-625/100+3925/100=0

(-600x-625+3925)/100=0

(-600x-625-3925)=0

3300=600x

X=3300/600

X=5.5

put X value in EQ(2)

AO^2=(5.5)^2+(2.5)^2

=30.25+6.25

=36.50

=√36.50=6.04

so,radius of circle is 6cm