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==
Link to Overleaf: [[https://www.overleaf.com/read/rmkbjhcyynbv]]
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<asy>
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draw((0,0) -- (9/5,12/5) -- (5,0) -- cycle);
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draw((9/5,12/5) -- (9/5,0));
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</asy>
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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:
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<cmath>\frac{c*a}{2}=A</cmath>
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<cmath>c*a=2A</cmath>
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<cmath>a=\textbf{(B)} \frac{2A}{c}</cmath>
  
 
==See Also==
 
==See Also==

Latest revision as of 22:14, 15 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

[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
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


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