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Difference between revisions of "1956 AHSME Problems/Problem 38"

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== Solution==
 
== Solution==
  
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The area of this triangle can be expressed in two ways; the first being <math>\frac{ab}{2}</math> (as this is a right triangle), and the second being <math>\frac{cx}{2}</math>. But by the Pythagorean Theorem, <math>c=\sqrt{a^2+b^2}</math>. Thus a second way of finding the area of the triangle is <math>\frac{x\sqrt{a^2+b^2}}{2}</math>. By setting them equal to each other we get <math>x=\frac{ab}{\sqrt{a^2+b^2}}</math>, and we can observe that the correct answer was <math>\boxed{\textbf{(D) \ }}</math>.
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The area of this triangle can be expressed in two ways; the first being <math>\frac{ab}{2}</math> (as this is a right triangle), and the second being <math>\frac{cx}{2}</math>. But by the [[Pythagorean Theorem]], <math>c=\sqrt{a^2+b^2}</math>. Thus a second way of finding the area of the triangle is <math>\frac{x\sqrt{a^2+b^2}}{2}</math>. By setting them equal to each other we get <math>x=\frac{ab}{\sqrt{a^2+b^2}}</math>, and we can observe that the correct answer is <math>\boxed{\textbf{(D) \ }}</math>.
  
 
~anduran
 
~anduran
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== See Also ==
 
== See Also ==
  
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{{AHSME box|year=1956|num-b=17|num-a=19}}
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{{AHSME box|year=1956|num-b=37|num-a=39}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 22:35, 2 January 2024

Problem 38

In a right triangle with sides $a$ and $b$, and hypotenuse $c$, the altitude drawn on the hypotenuse is $x$. Then:

$\textbf{(A)}\ ab = x^2 \qquad \textbf{(B)}\ \frac {1}{a} + \frac {1}{b} = \frac {1}{x} \qquad \textbf{(C)}\ a^2 + b^2 = 2x^2 \\ \textbf{(D)}\ \frac {1}{x^2} = \frac {1}{a^2} + \frac {1}{b^2} \qquad \textbf{(E)}\ \frac {1}{x} = \frac {b}{a}$

Solution

The area of this triangle can be expressed in two ways; the first being $\frac{ab}{2}$ (as this is a right triangle), and the second being $\frac{cx}{2}$. But by the Pythagorean Theorem, $c=\sqrt{a^2+b^2}$. Thus a second way of finding the area of the triangle is $\frac{x\sqrt{a^2+b^2}}{2}$. By setting them equal to each other we get $x=\frac{ab}{\sqrt{a^2+b^2}}$, and we can observe that the correct answer is $\boxed{\textbf{(D) \ }}$.

~anduran

See Also

1956 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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
All AHSME Problems and Solutions

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