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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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)

Revision as of 23:05, 25 November 2011

If you double sides $AB$ and $A$C, 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 (E)

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