Question

Answer 1

4:

b: A(2;3); B(-2;0); C(4;3)

\(AB=\sqrt{\left(-2-2\right)^2+\left(0-3\right)^2}=5\)

\(AC=\sqrt{\left(4-2\right)^2+\left(3-3\right)^2}=2\)

\(BC=\sqrt{\left(4+2\right)^2+\left(3-0\right)^2}=\sqrt{6^2+3^2}=3\sqrt5\)

Chu vi tam giác ABC là:

AB+AC+BC

\(=5+2+3\sqrt5=7+3\sqrt5\)

Xét ΔABC có \(cosA=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{5^2+2^2-45}{2\cdot5\cdot2}=\frac{25+4-45}{20}=\frac{29-45}{20}=-\frac45\)

=>\(\sin BAC=\sqrt{1-\left(-\frac45\right)^2}=\frac35\)

Diện tích tam giác ABC là:

\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\)

\(=\frac12\cdot5\cdot2\cdot\frac35=3\)