Math, asked by ridhisingh794, 2 months ago


In the given diagram 'O' is the centre of the circle and AB is parallel to CD
AB = 24 cm and distance between the chords AB and CD is 17 cm. If the radius
of the circle is 13 cm, find the length of the chord CD.

Attachments:

Answers

Answered by prabhas24480
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 :

on solving we get x=5

r square=x square+12 square

r square=5 square+12 square

r square= 25+144

r = 13cm

radius  \:  \: of  \:  \: the \:  \:  circle  \:  \: is  \:  \: 13cm.

________________________________________

Answered by UniqueBabe
21

Answer:

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 :

on solving we get x=5

r square=x square+12 square

r square=5 square+12 square

r square= 25+144

r = 13cmr=13cm

radiusofthecircleis13cm.

________________________________________

Similar questions