If the length of a chord of a circle is 48 cm and the distance of this chord from the centre is 7 cm , then find the length of the chord , the distance of which form the centre is 20 cm.

Explanation

Let AB and PQ are two chords of the circle with centre at O , OC and OD are two perpendicular draw form O to AB and PQ respectively.
As per question , OC = 7 cm .
AB = 48 cm , OD = 20 cm , PQ = ?
Now AB = 48 cm , `therefore BC = (1)/(2) xx 48 cm = 24 cm`
Then , form the right angled triangle OBC we get ,
`OB^(2) = BC^(2) + OC^(2) or OB^(2) = (24)^(2) + (7)^(2)`
or `OB^(2) = 576 + 49 or OB^(2) = 625 or , OB = sqrt(625) = 25`
` therefore OQ = 25 cm [ because OB = OQ ` (radii of same circle )]
Again , in right -angled triangle `OQ , OQ^(2) = OD^(2) + DQ^(2)`
` or (25)^(2) = (20)^(2) + DQ^(2) or 625 = 400 + DQ^(2)`
` therefore " " PQ = 2DQ = 2 xx 15 cm = 30 cm `
Hence lenght of the chord ` Pq = 30 cm `