Difference between revisions of "1956 AHSME Problems/Problem 38"
(→Solution) |
(→See Also) |
||
| Line 17: | Line 17: | ||
== See Also == | == See Also == | ||
| − | {{AHSME box|year=1956|num-b= | + | {{AHSME box|year=1956|num-b=37|num-a=39}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:AHSME]][[Category:AHSME Problems]] | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Revision as of 13:11, 1 January 2024
Problem 38
In a right triangle with sides 
and 
, and hypotenuse 
, the altitude drawn on the hypotenuse is 
. Then:
Solution
The area of this triangle can be expressed in two ways; the first being 
(as this is a right triangle), and the second being 
. But by the Pythagorean Theorem, 
. Thus a second way of finding the area of the triangle is 
. By setting them equal to each other we get 
, and we can observe that the correct answer was 
.
~anduran
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 37 |
Followed by Problem 39 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
