Difference between revisions of "1959 AHSME Problems/Problem 7"
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| − | ==Solution== | + | == Problem == |
| + | The sides of a right triangle are <math>a</math>, <math>a+d</math>, and <math>a+2d</math>, with <math>a</math> and <math>d</math> both positive. The ratio of <math>a</math> to <math>d</math> is: | ||
| + | <math>\textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 1:4 \qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 3:1\qquad\textbf{(E)}\ 3:4 </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
If we let <math>a=3</math> and <math>d=1</math>, then we will get a <math>3</math>-<math>4</math>-<math>5</math> triangle, which is a right triangle. So, the answer is <math>\boxed{\textbf{(D)} \ 3:1}</math>. | If we let <math>a=3</math> and <math>d=1</math>, then we will get a <math>3</math>-<math>4</math>-<math>5</math> triangle, which is a right triangle. So, the answer is <math>\boxed{\textbf{(D)} \ 3:1}</math>. | ||
Revision as of 13:56, 16 July 2024
Problem
The sides of a right triangle are 
, 
, and 
, with and 
both positive. The ratio of to is:
Solution
If we let 
and 
, then we will get a 
-
-
triangle, which is a right triangle. So, the answer is 
.
