



 
Hi Lorraine. The distance from the center to a chord would meet the chord in at a right angle right at the midpoint of the chord. If you draw this, and also draw the radii involved, you can see two distinct right triangles: So let a = the distance to the 10 inch chord, then 2a is the distance to the 8 inch chord, according to the description. Now take a look at what pythagoras has to say: r^{2} = 5^{2} + a^{2} and r^{2} = 4^{2} + (2a)^{2} Can you solve this system of simultaneous equations for r? Cheers,  


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