1956 AHSME Problems/Problem 38: Difference between revisions

Revision as of 13:11, 1 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 was $\boxed{\textbf{(D) \ }}$ .

~anduran

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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 == See Also ==
-{{AHSME box|year=1956|num-b=17|num-a=19}}
+{{AHSME box|year=1956|num-b=37|num-a=39}}
 {{MAA Notice}}
 [[Category:AHSME]][[Category:AHSME Problems]]

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1956 AHSME Problems/Problem 38: Difference between revisions

Revision as of 13:11, 1 January 2024

Problem 38

Solution

See Also