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) | + | 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 
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)
