Bài tập 2.33 trang 101 SBT Hình học 10

a) \(m_a^2 = \frac{{{b^2} + {c^2}}}{2} - \frac{{{a^2}}}{4} = \frac{{{{18}^2} + {{16}^2}}}{2} - \frac{{{{26}^2}}}{4} = \frac{{484}}{4} \Rightarrow {m_a} = \frac{{22}}{2} = 11\)

b) \(\left\{ \begin{array}{l}
m_a^2 = \frac{{{b^2} + {c^2}}}{2} - \frac{{{a^2}}}{4}\\
m_b^2 = \frac{{{c^2} + {a^2}}}{2} - \frac{{{b^2}}}{4}\\
m_c^2 = \frac{{{a^2} + {b^2}}}{2} - \frac{{{c^2}}}{4}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m_a^2 = 2\left( {{b^2} + {c^2}} \right) - {a^2}\\
m_b^2 = 2\left( {{a^2} + {c^2}} \right) - {b^2}\\
m_c^2 = 2\left( {{a^2} + {b^2}} \right) - {c^2}
\end{array} \right.\)

Suy ra \(4\left( {m_a^2 + m_b^2 + m_c^2} \right) = 3\left( {{a^2} + {b^2} + {c^2}} \right)\)

-- Mod Toán 10 HỌC247