Question

Cho tam giac ABC co BC=3 hai trung tuyen BM=4,CN=2 .tính AB

 

Answer 1

Theo công thức trung tuyến:

\(\left\{{}\begin{matrix}\dfrac{2AB^2-AC^2}{4}=BM^2-\dfrac{BC^2}{2}\\\dfrac{2AC^2-AB^2}{4}=CN^2-\dfrac{BC^2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2AB^2-AC^2=46\\2AC^2-AB^2=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4AB^2-2AC^2=92\\2AC^2-AB^2=-2\end{matrix}\right.\)

\(\Rightarrow3AB^2=90\)

\(\Rightarrow AB=\sqrt{30}\)