Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1956 AHSME Problems/Problem 38: Difference between revisions

Anduran (talk | contribs)
editor
52 edits
Created page with "== Problem 38== In a right triangle with sides <math>a</math> and <math>b</math>, and hypotenuse <math>c</math>, the altitude drawn on the hypotenuse is <math>x</math>. Then..."
 
Anduran (talk | contribs)
editor
52 edits
 
(3 intermediate revisions by the same user not shown)
Line 11: Line 11:
== Solution==
== Solution==


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>.
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
== See Also ==
{{AHSME box|year=1956|num-b=37|num-a=39}}
{{MAA Notice}}
[[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

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

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1956_AHSME_Problems/Problem_38&oldid=209655"