1959 AHSME Problems/Problem 43

Problem

The sides of a triangle are $25,39$ , and $40$ . The diameter of the circumscribed circle is: $\textbf{(A)}\ \frac{133}{3}\qquad\textbf{(B)}\ \frac{125}{3}\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 40$

Solution

The semiperimeter $s$ of the triangle is $\frac{25+39+40}{2}=52$ . Therefore, by Heron's Formula, we can find the area of the triangle as follows: $\begin{aligned} A & = \sqrt{52 (52 - 25) (52 - 39) (52 - 40)} \\ = \sqrt{52 (27) (13) (12)} \\ = \sqrt{13 * 4 (9 * 3) (13) (4 * 3)} \\ = 13 * 4 * 3 * 3 \\ = 468 \end{aligned}$ Also, we know that the area of the triangle is $\frac{abc}{4R}$ , where $a$ , $b$ , and $c$ are the sides of the triangle and $R$ is the triangle's circumradius. Thus, we can equate this expression for the area with $468$ to solve for $R$ : $\begin{aligned} \frac{(25) (39) (40)}{4 R} & = 468 = 39 * 12 \\ \frac{25 * 40}{12} & = 4 R \\ R & = \frac{250}{12} = \frac{125}{6} \end{aligned}$ Because the problem asks for the diameter of the circumcircle, our answer is $2R=\boxed{\textbf{(B) }\frac{125}{3}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1959 AHSME Problems/Problem 43

Problem

Solution

See also