1950 AHSME Problems/Problem 46

Revision as of 12:30, 5 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\textbf{(E)}\ \text{The area of the triangle is 0}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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

Invalid username
Login to AoPS