1955 AHSME Problems/Problem 21: Difference between revisions
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==See Also== | ==See Also== | ||
{{AHSME 50p box|year=1955|num-b=20|num-a=22}} | |||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:43, 14 August 2020
Problem 21
Represent the hypotenuse of a right triangle by 
and the area by 
. The altitude on the hypotenuse is:
Solution
Link to Overleaf: [[1]]
See Also
| 1955 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
