# 1958 AHSME Problems/Problem 36

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## 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

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