Difference between revisions of "1950 AHSME Problems/Problem 46"
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==Solution== | ==Solution== | ||
− | 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)}\ The area of the triangle is 0}</math>. | + | 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>. |
==See Also== | ==See Also== |
Revision as of 08:42, 29 April 2012
Problem
In triangle , , , and . If sides and are doubled while remains the same, then:
Solution
If you double sides and , they become and respectively. If remains , then this triangle has area because , so two sides overlap the third side. Therefore the answer is .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 45 |
Followed by Problem 47 | |
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