The length of each of two parallel chords AB and CD is 16 cm. If the radius of the circle be 10 cm, the distance between the two chords is .

Options

A. 12 cm
B. 16cm
C. 20 cm
D. 5 cm

Answer

A

Explanation

Let AB and CD be two parallel chords of the circle with centre at O. OP and OQ are two perpendicluar form O to AB and CD respectively.
`because AB || CD, therefore , OP ` and OQ are in the same straight line .
`therefore ` the distance between the two chords `= OP +OQ....... (i)`
As per question,
OA = OC = = radius of circle = 10 cm
Now ` OP bot AB therefore P` is the mid -point of AB . Similarly , Q is the mid -point of CD.
`therefore " " AP = (1)/(2) AB = (1)/(2) AB = (1)/(2) xx 16 cm = 8cm`,
Again `CQ = (1)/(2) CD = (1)/(2) xx 16 cm = 8 cm`
Now , in right - angled triangle OAP,
`OA^(2) = OP^(2) + AP^(2)`
or `(10)^(2) = OP^(2) + AP^(2)`
or ` 100= OP^(2) + 64`
or `OP^(2) = 100- 64 or OP^(2) = 36 or , OP = sqrt(36) = 6`
`therefore` from (1) we get , distance between two chords ` = OP + OQ = 6 cm + 6 cm = 12 cm`
`therefore ` The required distance = 12 cm `rArr` (a) is correct.