Question 1364

 As the diagram below,chord AB = 48 ft,center O, 
 OA (radius) = 50/2 = 25,M is the mid point of AB. So, AM = 24.
 
   A    M     B
     ---------
         |
         |
         O

 Since AM is perpendicular to AB, we have AO^2 = AM^2 + OM^2,
 so OM^2 = AO^2 - AM^2 = 25^2 - 24^2 = 625 - 576 =49

 Hence,the distance from the center of the circle to the chord is 7 ft 
 ...Anwer