1953 AMC 12 Problem 37

Question

The base of an isosceles triangle is

6

inches and one of the equal sides is

12

inches. The radius of the circle through the vertices of the triangle is:

\dfrac{7\sqrt{15}}5

4\sqrt3

3\sqrt5

6\sqrt3

none of these

Drop the altitude to split the base into two segments of length

3

Find the area, then use

R=\frac{abc}{4K}

The altitude is

\sqrt{12^2-3^2}=3\sqrt{15},

so the area is

K=\tfrac12(6)(3\sqrt{15})=9\sqrt{15}.

Hence the circumradius is

R=\frac{(12)(12)(6)}{4(9\sqrt{15})} =\frac{8\sqrt{15}}5.

This value is not listed.

Thus, the correct answer is E.

Problem 37 in other years