A circle passes through the vertices of a triangle with side-lengths 712, 10, 1212. The radius of the circle is:

1968 AMC 12 Problem 12

Attempt Problem 12 of the 1968 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1968 AMC 12 solutions, or check the answer key.

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12.

A circle passes through the vertices of a triangle with side-lengths

7\dfrac12,

10,

12\dfrac12.

The radius of the circle is:

\dfrac{15}{4}

5

\dfrac{25}{4}

\dfrac{35}{4}

\dfrac{15\sqrt2}{2}

The side lengths are

\frac{5}{2}

times a familiar Pythagorean triple

A right triangle’s hypotenuse is the circumcircle’s diameter

The sides are

\frac52(3,4,5),

so the triangle is right with hypotenuse

\frac{25}{2}.

The hypotenuse is the circumdiameter, making the radius

\frac{25}{4}.

Therefore, the correct answer is C.