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In triangle $ABC$, $AB = 7$, $AC = 17$, and the length of median $AM$ is $12$. Find the area of triangle $ABC$.

 Apr 22, 2024
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                             A

                       

                     7        12           17

 

 

             B                    M                      C

 

Law of Cosines

 

12^2 =  7^2  + (BC/2)^2  - 2 (7 * BC /2) cos ( ABC)

17^2  = 7^2 + BC^2 - 2 (7 * BC) cos (ABC)

 

Equate cosines and simplify

 

[144 - 49 - BC^2/4] / [ 7 BC ]   =   [289 - 49 - BC^2 ] / [  14 BC]

 

[ 95  - BC^2 / 4 ]  / 7  =  [ 240 - BC^2 ] / 14

 

[95 - BC^2/4 ]  = [ 240 - BC^2 ] / 2

 

95 -BC^2 / 4 = 120 - BC^2/2

 

120 - 95  = BC^2 / 4

 

25 = BC^2  /4

 

BC^2 = 25 * 4

 

BC = 5 * 2  = 10

 

Impossible  because  of the trianle inequality,  AB + BC   > AC

 

But

 

AB + BC  = 

 

7 + 10  not greater than  17

 

cool cool cool

 Apr 22, 2024

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