Đáp án:
$\begin{array}{l}
c)M\left( {2;1} \right);A\left( {0;5} \right)\\
\Rightarrow OM = \sqrt {{2^2} + {1^2}} = \sqrt 5 \\
OA = \sqrt {{0^2} + {5^2}} = 5\\
AM = \sqrt {\left( {{2^2}} \right) + {{\left( {5 - 1} \right)}^2}} = 2\sqrt 5 \\
\Rightarrow O{M^2} + A{M^2} = O{A^2}\\
\Rightarrow \Delta MOA \bot M\\
\Rightarrow \left\{ \begin{array}{l}
Chu\,vi = OA + OM + AM = 5 + 3\sqrt 5 \left( {cm} \right)\\
{S_{OAM}} = \dfrac{1}{2}.OM.AM = \dfrac{1}{2}.\sqrt 5 .2\sqrt 5 = 5\left( {c{m^2}} \right)
\end{array} \right.
\end{array}$
