Difference between revisions of "1959 AHSME Problems/Problem 7"
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| − | ==Solution== | + | == Problem == |
| − | 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{D} | + | 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> | ||
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| + | == 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>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1959|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 11:01, 21 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 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
