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== | ||
− | + | {{AHSME 50p box|year=1955|num-b=20|num-a=22}} | |
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
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
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:
See Also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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