In triangle $ABC$, $AB = 7$, $AC = 17$, and the length of median $AM$ is $12$. Find the area of triangle $ABC$.
Well, unfortunately, this triangle can't ever exist.
Using the Law of Cosines, we have
AC2=CM2+AM2−2(AM∗MC)cos(AMC)AB2=BM2+AM2−2(AM∗BM)(−cos(AMC)172=x2+122−2(AM∗x)cos(AMC)72=x2+122+2(AM∗x)cos(AMC)
Adding these two equations together, we get
338=2x2+28850=2x225=x25=x
Alright.
Now, we have BM+CM=5+5=10=BC.
However, according to the Triangle Inequality Theorem, we have
AB+BC>ACAB+BC>17
However, we have
AB+BC=7+10=17
Meaning this is an invalid triangle.
Thanks! :)
Well, unfortunately, this triangle can't ever exist.
Using the Law of Cosines, we have
AC2=CM2+AM2−2(AM∗MC)cos(AMC)AB2=BM2+AM2−2(AM∗BM)(−cos(AMC)172=x2+122−2(AM∗x)cos(AMC)72=x2+122+2(AM∗x)cos(AMC)
Adding these two equations together, we get
338=2x2+28850=2x225=x25=x
Alright.
Now, we have BM+CM=5+5=10=BC.
However, according to the Triangle Inequality Theorem, we have
AB+BC>ACAB+BC>17
However, we have
AB+BC=7+10=17
Meaning this is an invalid triangle.
Thanks! :)