Đáp án:
$\begin{array}{l}
a)A\left( {1;2} \right);B\left( {3;4} \right);C\left( {7;0} \right)\\
\Rightarrow G\left( {x;y} \right)\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{1 + 3 + 7}}{3} = \dfrac{{11}}{3}\\
y = \dfrac{{2 + 4 + 0}}{3} = 2
\end{array} \right.\\
\Rightarrow G\left( {\dfrac{{11}}{3};2} \right)\\
b)AB = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {4 - 2} \right)}^2}} = 2\sqrt 2 \\
AC = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( {0 - 2} \right)}^2}} = 2\sqrt {10} \\
BC = \sqrt {{{\left( {7 - 3} \right)}^2} + {{\left( {0 - 4} \right)}^2}} = 4\sqrt 2 \\
\Rightarrow A{B^2} + B{C^2} = 8 + 32 = 40 = A{C^2}\\
\Rightarrow \left\{ \begin{array}{l}
Chu\,vi:C = 6\sqrt 2 + 2\sqrt {10} \\
{S_{ABC}} = \dfrac{1}{2}.AB.BC = \dfrac{1}{2}.2\sqrt 2 .4\sqrt 2 = 8
\end{array} \right.
\end{array}$
