12:
\(p=\dfrac{12+13+15}{2}=20\)
\(S=\sqrt{20\cdot\left(20-12\right)\left(20-13\right)\left(20-15\right)}\)
\(=\sqrt{20\cdot5\cdot8\cdot7}=20\sqrt{14}\)
Xét ΔABC có \(cosA=\dfrac{13^2+15^2-12^2}{2\cdot13\cdot15}=\dfrac{25}{39}\)
=>\(\widehat{A}\simeq50^0\)
Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)
=>\(2R=\dfrac{12}{sin50}\simeq15,67\)
=>R=7,835
S ABC=p*r
=>r*20=20*căn 14
=>r=căn 14
\(AH=2\cdot\dfrac{S_{ABC}}{BC}=\dfrac{2\cdot20\sqrt{14}}{12}=\dfrac{10}{3}\cdot\sqrt{14}\)
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