1959 AHSME Problems/Problem 7: Difference between revisions
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==Solution== | ==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 04:56, 24 September 2018
Solution
If we let 
and 
, then we will get a 
-
-
triangle, which is a right triangle. So, the answer is 
.
