Difference between revisions of "1950 AHSME Problems/Problem 46"

 
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If you double sides <math>AB</math> and <math>A</math>C, they become <math>24</math> and <math>14</math> respectively. If <math>BC</math> remains <math>10</math>, then this triangle has area <math>0</math> because <math>{14} + {10} = {24}</math>, so two sides overlap the third side. Therefore the answer is (E)
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==Problem==
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In triangle <math>ABC</math>, <math>AB=12</math>, <math>AC=7</math>, and <math>BC=10</math>. If sides <math>AB</math> and <math>AC</math> are doubled while <math>BC</math> remains the same, then:
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<math>\textbf{(A)}\ \text{The area is doubled} \qquad\\
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\textbf{(B)}\ \text{The altitude is doubled} \qquad\\
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\textbf{(C)}\ \text{The area is four times the original area} \qquad\\
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\textbf{(D)}\ \text{The median is unchanged} \qquad\\
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\textbf{(E)}\ \text{The area of the triangle is 0}</math>
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==Solution==
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If you double sides <math>AB</math> and <math>AC</math>, they become <math>24</math> and <math>14</math> respectively. If <math>BC</math> remains <math>10</math>, then this triangle has area <math>0</math> because <math>{14} + {10} = {24}</math>, so two sides overlap the third side. Therefore the answer is <math>\boxed{\textbf{(E)}\ \text{The area of the triangle is 0}}</math>.
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==See Also==
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{{AHSME 50p box|year=1950|num-b=45|num-a=47}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:30, 5 July 2013

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
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All AHSME Problems and Solutions

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