ta có : \(S_{\Delta ABC}=\dfrac{abc}{4R}\Leftrightarrow10\sqrt{3}=\dfrac{7.8.5}{4R}\Leftrightarrow R=\dfrac{7.8.5}{4.10\sqrt{3}}=\dfrac{7\sqrt{3}}{3}\)
(3) ta có : \(p=\dfrac{7+8+5}{2}=10\)
ta có : \(S_{\Delta ABC}=p.r\Leftrightarrow10\sqrt{3}=10.r\Leftrightarrow r=\dfrac{10\sqrt{3}}{10}=\sqrt{3}\)
(4) ta có : \(S_{\Delta ABC}=\dfrac{1}{2}a.h_a\Leftrightarrow10\sqrt{3}=\dfrac{1}{2}.7.h_a\Leftrightarrow h_a=10\sqrt{3}.\dfrac{2}{7}=\dfrac{20\sqrt{3}}{7}\)
(5) ta có : \(\left(m_a\right)^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{8^2+5^2}{2}-\dfrac{7^2}{4}=\dfrac{129}{4}\)