1959 AHSME Problems/Problem 7

Problem

The sides of a right triangle are $a$ , $a+d$ , and $a+2d$ , with $a$ and $d$ both positive. The ratio of $a$ to $d$ is: $\textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 1:4 \qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 3:1\qquad\textbf{(E)}\ 3:4$

Solution

If we let $a=3$ and $d=1$ , then we will get a $3$ - $4$ - $5$ triangle, which is a right triangle. So, the answer is $\boxed{\textbf{(D)} \ 3:1}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1959 AHSME Problems/Problem 7

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