1950 AHSME Problems/Problem 46

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Problem

In triangle $ABC$, $AB=12$, $AC=7$, and $BC=10$. If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:

$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The median is unchanged} \qquad\\ \textbf{(E)}\ \text{The area of the triangle is 0}$

Solution

If you double sides $AB$ and $AC$, they become $24$ and $14$ respectively. If $BC$ remains $10$, then this triangle has area $0$ because ${14} + {10} = {24}$, so two sides overlap the third side. Therefore the answer is $\boxed{\textbf{(E)}\ The area of the triangle is 0}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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