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

(Created page with "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 trian...")
 
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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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==Video Solution==
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https://www.youtube.com/watch?v=l4lAvs2P_YA&t=97s
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~MathProblemSolvingSkills.com
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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 15:49, 15 July 2023

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

Video Solution

https://www.youtube.com/watch?v=l4lAvs2P_YA&t=97s

~MathProblemSolvingSkills.com

See Also

1950 AHSC (Problems • Answer Key • Resources)
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

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