Difference between revisions of "1956 AHSME Problems/Problem 39"
Coolmath34 (talk | contribs) (Created page with "== Problem 39== The hypotenuse <math>c</math> and one arm <math>a</math> of a right triangle are consecutive integers. The square of the second arm is: <math> \textbf{(A)}...") |
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Squaring the middle length gets <math>2a + 1 = a + c,</math> so the answer is <math>\boxed{\textbf{(C)}}.</math> | Squaring the middle length gets <math>2a + 1 = a + c,</math> so the answer is <math>\boxed{\textbf{(C)}}.</math> | ||
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-coolmath34 | -coolmath34 | ||
Latest revision as of 21:59, 12 February 2021
Problem 39
The hypotenuse 
and one arm 
of a right triangle are consecutive integers. The square of the second arm is:
Solution
The sides of the triangle are 
and 
We know that the hypotenuse and one leg are consecutive integers, so we can rewrite the side lengths as 
and 
Squaring the middle length gets 
so the answer is 
-coolmath34
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 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. 
