



 
Hi Tricia, I drew the base circle, the specific diameter and one chord BD that is perpendicular to the diameter and intersects the diameter at A. C is the centre of the circle and I let the distance CA be x centimeters.
The triangle ABC is a right triangle so using Pythagoras theorem the distance from A to B is √(r^{2}  x^{2}). Thus the length of the chord BD is 2 √(r^{2}  x^{2}). Does this help?  


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