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

BRAlNBOLT Jun 2, 2024

#1**+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! :)

NotThatSmart Jun 2, 2024

#1**+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