Question

AB and CD are two chords of a circle such
that AB = 10 cm, CD = 24 cm and AB
||CD. The distance between AB and CD is
17 cm. Then, the radius of the circle is
equal to​

Answer 1

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Since op perpendicular to ab and oq perpendicular to cd nd ab parallel to cd

poq is a st. line

ab=10cm,cd=24cm nd pq is 17cm

ap=bp=half ab = 5cm and cq=dq=half cd =12cm

if oq=x cm ,then op= (17-x)cm

join oa nd oc

oa=oc=r(radius)

now in right angled triangle oap,

oa square=op square+ap square

=r square=(17-x)square+ 5square_ist equation

in right angled triangle ocq,

oc square=oq square+cq square

=r square=x square+12square_2nd equation

frm ist nd 2nd equations ,we get :

(17-x)² +(5)²=x²+(12)²

on solving we get x=5

r square=x square+12 square

r square=5 square+12 square

r square= 25+144

r=√169

r=13cm.

RADIUS OF CIRLCE =13cm.

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Answer 2

Let OL = x. Then OM = (17−x).

Also LB=5 cm, MD=12 cm,

Then x²+25=r²

and (17−x)²+144=r²

Solving these we get

x=12 cm and r=13cm

Thus, diameter = 2×13=26 cm.