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

 Jun 2, 2024

Best Answer 

 #1
avatar+806 
+1

Well, unfortunately, this triangle can't ever exist. 

 

Using the Law of Cosines, we have

\(AC^2 = CM^2 + AM^2 - 2 (AM * MC)cos (AMC) \\ AB^2 = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC) \\ 17^2 = x^2 + 12^2 - 2 (AM * x) cos (AMC) \\ 7^2 = x^2 + 12^2 + 2(AM * x) cos (AMC)\)

 

Adding these two equations together, we get 

\(338 = 2x^2 + 288 \\ 50 = 2x^2 \\ 25 = x^2 \\ 5 = x \\\)

 

Alright. 

 

Now, we have \( BM + CM = 5 + 5 = 10 = BC\)

 

However, according to the Triangle Inequality Theorem, we have

\(AB+BC>AC \\ AB +BC > 17\)

 

However, we have

\( AB + BC = 7 + 10 = 17\)

 

Meaning this is an invalid triangle. 

 

Thanks! :)

 Jun 2, 2024
 #1
avatar+806 
+1
Best Answer

Well, unfortunately, this triangle can't ever exist. 

 

Using the Law of Cosines, we have

\(AC^2 = CM^2 + AM^2 - 2 (AM * MC)cos (AMC) \\ AB^2 = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC) \\ 17^2 = x^2 + 12^2 - 2 (AM * x) cos (AMC) \\ 7^2 = x^2 + 12^2 + 2(AM * x) cos (AMC)\)

 

Adding these two equations together, we get 

\(338 = 2x^2 + 288 \\ 50 = 2x^2 \\ 25 = x^2 \\ 5 = x \\\)

 

Alright. 

 

Now, we have \( BM + CM = 5 + 5 = 10 = BC\)

 

However, according to the Triangle Inequality Theorem, we have

\(AB+BC>AC \\ AB +BC > 17\)

 

However, we have

\( AB + BC = 7 + 10 = 17\)

 

Meaning this is an invalid triangle. 

 

Thanks! :)

NotThatSmart Jun 2, 2024

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