+153 In triangle $ABC$, $AB = 7$, $AC = 17$, and the length of median $AM$ is $12$. Find the area of triangle $ABC$.
+1365 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! :)
+1365 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! :)
