# Difference between revisions of "1968 AHSME Problems/Problem 12"

## Problem

A circle passes through the vertices of a triangle with side-lengths $7\tfrac{1}{2},10,12\tfrac{1}{2}.$ The radius of the circle is:

$\text{(A) } \frac{15}{4}\quad \text{(B) } 5\quad \text{(C) } \frac{25}{4}\quad \text{(D) } \frac{35}{4}\quad \text{(E) } \frac{15\sqrt{2}}{2}$

## Solution

The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle. $(7\frac{1}{2})^{2}+10^{2}=(12\frac{1}{2})^{2}$, so the triangle is a right triangle.The radius of a circumcircle of a right triangle is half the hypotenuse. $\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}\implies$ $\fbox{C}$

## See also

 1968 AHSME (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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