Question

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

Answer 1

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

Answer 2

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(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=

169=13

r = 13cm

radius of the circle is 13cm