1955 AHSME Problems/Problem 21
Problem 21
Represent the hypotenuse of a right triangle by 
and the area by 
. The altitude on the hypotenuse is:
Solution
![[asy] draw((0,0) -- (9/5,12/5) -- (5,0) -- cycle); draw((9/5,12/5) -- (9/5,0)); [/asy]](http://latex.artofproblemsolving.com/d/0/8/d08a8e664225cc94966e82429d6b45696d37db42.png)
Given that the area of the triangle is , and the formula for the area of a triangle is 
, we can replace 
(the base) and 
(the height) with (the hypotenuse) and 
(the altitude), we can rearrrange as follows:
![\[\frac{c*a}{2}=A\]](http://latex.artofproblemsolving.com/e/3/0/e30a27253f07c0d7faa3baa36486a62f705cb626.png)
![\[c*a=2A\]](http://latex.artofproblemsolving.com/8/8/0/88015e6b3e0136549f0b5c4a5e2bc8d8662c312b.png)
![\[a=\textbf{(B)} \frac{2A}{c}\]](http://latex.artofproblemsolving.com/1/b/a/1ba96cdc040dd496e43ec94b1d0e6e08c65d66e4.png)
See Also
| 1955 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
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| All AHSME Problems and Solutions | ||
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