1952 AHSME Problems/Problem 15

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Problem

The sides of a triangle are in the ratio $6:8:9$. Then:

$\textbf{(A) \ }\text{the triangle is obtuse}$

$\textbf{(B) \ }\text{the angles are in the ratio }6:8:9$

$\textbf{(C) \ }\text{the triangle is acute}$

$\textbf{(D) \ }\text{the angle opposite the largest side is double the angle opposite the smallest side}$

$\textbf{(E) \ }\text{none of these}$

Solution

In a triangle with sides $a$, $b$, and $c$, where $a\le b\le c$, $\bigtriangleup ABC$ is acute if $a^2+b^2>c^2$, right if $a^2+b^2=c^2$, and obtuse if $a^2+b^2<c^2$. If $c$ were equal to $10$, $\bigtriangleup ABC$ would be right, as a multiple of the Pythagorean triple $(3,4,5)$. Because $c<10$, $\boxed{\textbf{(C)}\ \text{the triangle is acute}}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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