Difference between revisions of "1955 AHSME Problems/Problem 21"
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<math> \textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2} </math> | <math> \textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2} </math> | ||
==Solution== | ==Solution== | ||
| − | + | <asy> | |
| + | draw((0,0) -- (9/5,12/5) -- (5,0) -- cycle); | ||
| + | draw((9/5,12/5) -- (9/5,0)); | ||
| + | </asy> | ||
| + | Given that the area of the triangle is <math>A</math>, and the formula for the area of a triangle is <math>\frac{bh}{2}=A</math>, we can replace <math>b</math> (the base) and <math>h</math> (the height) with <math>c</math> (the hypotenuse) and <math>a</math> (the altitude), we can rearrrange as follows: | ||
| + | <cmath>\frac{c*a}{2}=A</cmath> | ||
| + | <cmath>c*a=2A</cmath> | ||
| + | <cmath>a=\textbf{(B)} \frac{2A}{c}</cmath> | ||
==See Also== | ==See Also== | ||
Latest revision as of 21:14, 15 August 2020
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 | |
| 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. 
