1950 AHSME Problems/Problem 46: Difference between revisions
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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) | 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 22:05, 25 November 2011
If you double sides 
and 
C, they become 
and 
respectively. If 
remains 
, then this triangle has area 
because 
, so two sides overlap the third side. Therefore the answer is (E)
