A(2;3); B(-2;0); C(4;3)
\(AB=\sqrt{\left(-2-2\right)^2+\left(0-3\right)^2}=\sqrt{16+9}=5\)
\(AC=\sqrt{\left(4-2\right)^2+\left(3-3\right)^2}=\sqrt{4}=2\)
\(BC=\sqrt{\left(4+2\right)^2+\left(3-0\right)^2}=\sqrt{36+9}=3\sqrt{5}\)
Chu vi tam giác ABC là:
\(C_{ABC}=AB+AC+BC=7+3\sqrt{5}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{25+4-45}{2\cdot5\cdot2}=\dfrac{-4}{5}\)
=>\(sinBAC=\sqrt{1-\left(-\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)
Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{5}\cdot5\cdot2=3\)
Đúng 0
Bình luận (0)