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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+1953 
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Well, unfortunately, this triangle can't ever exist. 

 

Using the Law of Cosines, we have

AC2=CM2+AM22(AMMC)cos(AMC)AB2=BM2+AM22(AMBM)(cos(AMC)172=x2+1222(AMx)cos(AMC)72=x2+122+2(AMx)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! :)

 Jun 2, 2024
 #1
avatar+1953 
+1
Best Answer

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

 

Using the Law of Cosines, we have

AC2=CM2+AM22(AMMC)cos(AMC)AB2=BM2+AM22(AMBM)(cos(AMC)172=x2+1222(AMx)cos(AMC)72=x2+122+2(AMx)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! :)

NotThatSmart Jun 2, 2024

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