1955 AHSME Problems/Problem 21

Problem 21

Represent the hypotenuse of a right triangle by $c$ and the area by $A$. The altitude on the hypotenuse is:

$\textbf{(A)}\ \frac{A}{c}\qquad\textbf{(B)}\ \frac{2A}{c}\qquad\textbf{(C)}\ \frac{A}{2c}\qquad\textbf{(D)}\ \frac{A^2}{c}\qquad\textbf{(E)}\ \frac{A}{c^2}$

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 $A$, and the formula for the area of a triangle is $\frac{bh}{2}=A$, we can replace $b$ (the base) and $h$ (the height) with $c$ (the hypotenuse) and $a$ (the altitude), we can rearrrange as follows: \[\frac{c*a}{2}=A\] \[c*a=2A\] \[a=\textbf{(B)} \frac{2A}{c}\]

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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