Ab and cd are two chords of a circle

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AB and CD are two chords of a circle such that AB= 10cm CD=24cm and AB parallel CD the distance between AB and CD is 17cm. Find the radius of the circle if the chords are on opposite sides of the centre

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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)square + (5)square = (x)square + (12)square
on solving we get x=5
r square=x square+12 square
r square=5 square+12 square
r square= 25+144

r = \sqrt{169 = 13}
r = 13cm
radius of the circle is 13cm

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