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

Options

A. `13` cm
B. `14` cm
C. `15` cm
D. `16` cm

Answer

A

Explanation

Let `AB` and `CD` be two chords of a circle `C(O,r)` such that `AB ||CD`. Also , `AB` `=10cm` and `CD=24cm`.
Draw `OL` `_|_` AB and OM `_|_ CD`

Join `OA` and `OC` .
Then , `OA` `=` `OC` `=`r cm
Since OL `_|_` AB and OM `_|_` CD and `AB ||CD` , the points L,O,M are collinear
`:. LM =17cm`
Let Ol `=` X cm . Then, OM `=(17-X)cm`.
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have
`AL =(1)/(2)AB -((1)/(2)xx10)cm=5cm`, and `CM=(1)/(2)CD=((1)/(2)xx24)cm=12cm`.
From the right `Delta `OLA, we have
`OA^(2)=OL^(2)+AL^(2)impliesr^(2)=x^(2)+(5)^(2)`. ....(i)
From the right `Delta OMC,` we have
`OC^(2)=OM^(2)+CM^(2)implies r^(2)= (17-x)^(2)=(12)^(2)`. ....(ii)
From (i) and (ii) , we get
`x^(2)+(5)^(2)=(17-x)^(2)+(12)^(2)`
`hArrx^(2)+25=x^(2)-34x+433`
`hArr 34x=408hArrx=12`
Putting `x=12` in (i) , we get `r^(2)=169 rArr r=13`
Hence, the radius of the circle is 13 cm.