A(1;3); B(-2;-2); C(3;-2)
\(AB=\sqrt{\left(-2-1\right)^2+\left(-2-3\right)^2}=\sqrt{34}\)
\(AC=\sqrt{\left(3-1\right)^2+\left(-2-3\right)^2}=\sqrt{29}\)
\(BC=\sqrt{\left(3+2\right)^2+\left(-2+2\right)^2}=5\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{34+29-25}{2\cdot\sqrt{34}\cdot\sqrt{29}}=\dfrac{19}{\sqrt{986}}\)
=>\(sinBAC=\sqrt{1-\dfrac{19^2}{986}}=\dfrac{25}{\sqrt{986}}\)
Diện tích tam giác BAC là:
\(S_{BAC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\sqrt{34}\cdot\sqrt{29}\cdot\dfrac{25}{\sqrt{986}}=\dfrac{25}{2}\)
Chu vi tam giác ABC là:
\(C_{ABC}=AB+AC+BC=\sqrt{34}+\sqrt{29}+5\)
