Difference between revisions of "1958 AHSME Problems/Problem 36"

Revision as of 20:21, 27 January 2023

Problem

The sides of a triangle are $30$ , $70$ , and $80$ units. If an altitude is dropped upon the side of length $80$ , the larger segment cut off on this side is:

$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

Solution

Let the shorter segment be $x$ and the altitude be $y$ . The larger segment is then $80-x$ . By the Pythagorean Theorem , $30^2+y^2=x^2 \qquad(1)$ and $(80-x)^2=70^2+y^2 \qquad(2)$ Adding $(1)$ and $(2)$ and simplifying gives $x=15$ . Therefore, the answer is $80-15=\boxed{\textbf{(D)}~65}$

~megaboy6679

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 35	Followed by Problem 37
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 == Solution ==
-Let the shorter segment be x. The larger segment is therefore 80-x. Let the altitude be y. By the [[Pythagorean Theorem]]
+Let the shorter segment be <math>x</math> and the altitude be <math>y</math>. The larger segment is then <math>80-x</math>.  By the [[Pythagorean Theorem]]
 , <cmath>30^2+y^2=x^2 \qquad(1)</cmath>
 and <cmath>(80-x)^2=70^2+y^2 \qquad(2)</cmath>

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Difference between revisions of "1958 AHSME Problems/Problem 36"

Revision as of 20:21, 27 January 2023

Problem

Solution

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