AB and CD are two chords such that AB = 10 cm , CD = 24 cm and AB // CD The distance between the chords is 17 cm . Find the radius of the circle.

Explanation

In the adjoining figure, O is the centre of the circle.
Here, `AB=10 cm`
`CD=24 cm`
OM and ON are perpendicular from centre `O` to AB and CD respectively.
`because AB||CD`
`therefore` MON is a transversal
Now, `AM=(1)/(2) AB=(1)/(2)xx10=5 cm`
`CN=(1)/(2)CD=(1)/(2) xx24=12 cm`
Let `OM=x`
`therefore ON=17-x`
In `Delta OMA`,
`OA^2=OM^2+AM^2` (where `OA = r` is the radius of circle)
`rArrr^2=x^2+5^2` ........(1)
`therefore` In `Delta OCN`,
`OC^2=ON^2+CN^2`
`rArre^2=(17-x)^2+12^2` .........(2)
From eqs. (1) and (2).
`x^2+5^2=(17-x)^2+12`
`rArrx^2+25=289+x^2-34x+144`
`rArr34x=289+144-25`
`=408`
`rArrx=(408)/(34)=12`
Put `x=12` in eq. (1)
`r^2=x^2+5^2`
`=12^2+5^2`
`=144+25`
`=169`
`rArrr=sqrt(169)=13cm`
`therefore` Radius of circle =13cm.