The two parallel chords AB and CD with length of 10 cm and 24 cm in a circle situated on the opposite side of the centre. If the distance between two chards AB and CD is 17 cm , the calculate the length of the radiuys of the circle

Explanation

Let the length of the chord AB is 10 cm of the circle with centre at O and that of CD is 24 cm . PQ is a straight line throught O , which is perpendicular to both AB and CD.
`therefore OP bot AB rArr P` is the mid -point of AB .
`therefore AP = (1)/(2) AB =(1)/(2) xx 10 vm = 5 cm`
Again , `OQ bot CD rArr Q ` is the mid - point of CD .
`therefor e DQ = (1)/(2) CD = (1)/(2) xx 24 cm = 12 cm `
As per question PQ = 17 cm `therefore ` if OP= x cm then OQ = (17 - x) cm .
Now form the right - angled triangle OAP we get .
` OA^(2) = AP^(2) + P^(2)`
or ` OA^(2) = (5)^(2) + (x)^(2) or , OA^(2) = 25 + x^(2) ............(1)`
Again form OQD right - angled triangle we get ,
` OD^(2) = DQ^(2) + OQ^(2)`
or `OD^(2) = (12)^(2) + (17-x)^(2)`
or ` OD^(2) = 14 + (17 -x)^(2)`
or `OD^(2) = 144 + (17 -x)^(2)`
or ` OA^(2) = 144 + (17 - x)^(2) ..............(2)` [ ` because OD = OA` = radii of same circle ]
Then form (1) and (2) we get ,
` 144 + (17 -x)^(2) = 25+ x^(2)`
or ` (17 -x)^(2) - x^(2) = 25 - 144`
or `(17 - x + x) (17 - x -x) = - 119`
or ` 17 (17 - 2x) =- 119`
` or , 17 - 2x = - 7 or , 2x = 24 or , x = 12`
from (1) we get ` OA^(2) = 25 + (12)^(2) = 25 + 144 = 169`
`therefore OA = sqrt(169) = 13`
Hence the required radius of the circle = 13 cm .