$20 \cdot 42$ A circle passes through the three vertices of a triangle with side lengths $7 \frac{1}{2}, 10, 12 \frac{1}{2}$. Then the radius of this circle is 
(A) $\frac{15}{4}$.
(B) 5 .
(C) $\frac{25}{4}$.
(D) $\frac{35}{4}$.
(E) $\frac{15 \sqrt{2}}{2}$.
(19th American High School Mathematics Examination, 1968)
［Solution］Assuming and noting that
$$
\left(12 \frac{1}{2}\right)^{2}=10^{2}+\left(7 \frac{1}{2}\right)^{2} \text {, }
$$

we know that this triangle is a right triangle, so half of the hypotenuse is the radius of the circumcircle of the triangle. Therefore, the answer is $(C)$.