Question

Answer 1

Có \(S_{ABC}=\dfrac{1}{2}AB.AC.sinA=\dfrac{1}{2}.2a.3a.sin60^0=\dfrac{3\sqrt{3}a^2}{2}\)

Do góc A nhọn => \(cosA>0\) => \(cosA=\sqrt{1-sin^2A}=\dfrac{1}{2}\)

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}\)\(=\sqrt{\left(2a\right)^2+\left(3a\right)^2-2.2a.3a.\dfrac{1}{2}}\)\(=a\sqrt{7}\)

Có \(S_{ABC}=\dfrac{1}{2}BC.h_a=\dfrac{3\sqrt{3}a^2}{2}\)

\(\Leftrightarrow\dfrac{1}{2}.a\sqrt{7}.h_a=\dfrac{3\sqrt{3}a^2}{2}\) \(\Leftrightarrow h_a=\dfrac{3a\sqrt{21}}{7}\)

Ý D