Art of Problem Solving

1955 AHSME Problems/Problem 21: Difference between revisions

Angrybird029 (talk | contribs)
Created page with "== Problem 21== Represent the hypotenuse of a right triangle by <math>c</math> and the area by <math>A</math>. The altitude on the hypotenuse is: <math> \textbf{(A)}\ \frac..."
 
Angrybird029 (talk | contribs)
Line 5: Line 5:
<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==
Link to Overleaf: [[https://www.overleaf.com/read/rmkbjhcyynbv|https://www.overleaf.com/read/rmkbjhcyynbv]]
Link to Overleaf: [[https://www.overleaf.com/read/rmkbjhcyynbv]]
 
==See Also==
==See Also==
In order to return to the problem set, click [[1955 AHSME Problems|here]]
In order to return to the problem set, click [[1955 AHSME Problems|here]]


{{MAA Notice}}
{{MAA Notice}}

Revision as of 22:40, 14 August 2020

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

Link to Overleaf: [[1]]

See Also

In order to return to the problem set, click here

These problems are copyrighted © by the Mathematical Association of America.