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

 Jan 13, 2024
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                                A

                          7                 17

                   B                  M                 C

 

We can show that such a triangle  cannot exist

 

angles AMB and AMC are supplemental....so cos (AMB)  =  -cos (AMC) 

 

Since AM is a median,  then BM = CM  = x

 

Using the Law of Cosines

 

AC^2 = CM^2 + AM^2  - 2 (AM * MC)cos (AMC)

AB^2  = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC)

 

17^2  = x^2 + 12^2  - 2 (AM * x) cos (AMC)

7^2  = x^2 +  12^2  + 2(AM * x) cos (AMC)         add these

 

338 = 2x^2  + 288

 

50  = 2x^2

 

25  = x^2

 

5  = x

 

Then BM  + CM =  5 + 5   =  10  = BC

 

But, by the Triangle Inequality,   AB + BC    must be > AC =  17

 

However  AB + BC =   7 + 10  = 17

 

Which means  that no  such triangle  can exist

 

cool cool cool

 Jan 13, 2024

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