Question

Cho tam giác ABC đều cạnh a, có AH là đường trung tuyến. Tính|vecto AC+ vecto AH|

Answer 1

\(\widehat{HAC}=30^0\) ; \(AH=\frac{a\sqrt{3}}{2}\)

\(x=\left|\overrightarrow{AH}+\overrightarrow{AC}\right|\Rightarrow x^2=AH^2+AC^2+2\overrightarrow{AH}.\overrightarrow{AC}\)

\(\Rightarrow x^2=AH^2+AC^2+2AH.AC.cos\widehat{HAC}\)

\(\Rightarrow x^2=\left(\frac{a\sqrt{3}}{2}\right)^2+a^2+2.\frac{a\sqrt{3}}{2}.a.cos30^0\)

\(\Rightarrow x^2=\frac{13a^2}{4}x\Rightarrow x=\frac{a\sqrt{13}}{2}\)