Question 1157990
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For each case, let P be the point on diameter AB such that CP is perpendicular to AB.  Then in each case the length of CP is what we need to find.<br>
(a) AC = CB<br>
Triangle ABC is isosceles; CP is a radius of the circle -- length 10.<br>
(b) AC = 10<br>
AC is half of AB, so triangles ABC, APC, and BPC are all 30-60-90 right triangles.  AC=10 means CP = 5*sqrt(3).<br>
(c) AC = 12<br>
Triangle ABC has side lengths 12, 16, and 20 and so is similar to a 3-4-5 right triangle.  Triangles APC and BPC are also similar to a 3-4-5 right triangle.  CP is the long leg in the triangle with hypotenuse 12 -- length 4/5 of 12, which is 48/5, or 9.6.<br>