Do tam giác vuông cân tại A \(\Rightarrow\left\{{}\begin{matrix}BA=\dfrac{BC}{\sqrt{2}}=a\\BM=\dfrac{1}{2}BC=\dfrac{a\sqrt{2}}{2}\\\left(\overrightarrow{BA};\overrightarrow{BM}\right)=45^0\end{matrix}\right.\)
\(\left|\overrightarrow{BA}+\overrightarrow{BM}\right|^2=BA^2+BM^2+2\overrightarrow{BA}.\overrightarrow{BM}=BA^2+BM^2+2BA.BM.cos45^0\)
\(=a^2+\dfrac{a^2}{2}+a^2=\dfrac{5a^2}{2}\)
\(\Rightarrow\left|\overrightarrow{BA}+\overrightarrow{BM}\right|=\dfrac{a\sqrt{10}}{2}\)
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