Question

M (2; 1)
N (6; 3)
P (-4; -2)
AΔ=?

Answer 1

[tex]MN= \sqrt{(6-2)^2+(3-1)^2} =\ \textgreater \ MN= \sqrt{16+4} =\ \textgreater \ MN=2 \sqrt{5} \\ NP= \sqrt{(-4-6)^2+(-2-3)^2} =\ \textgreater \ NP= \sqrt{100+25} =\ \textgreater \ NP=5 \sqrt{5} \\ MP= \sqrt{(-4-2)^2+(-2-1)^2} =\ \textgreater \ MP= \sqrt{36+9} =\ \textgreater \ 3 \sqrt{5} \\ P=2 \sqrt{5} +5 \sqrt{5} +3 \sqrt{5} \\ P=10 \sqrt{5} =\ \textgreater \ p=5 \sqrt{5} \\ A=\sqrt{p(p-a)(p-b)(p-c)} \\\\ A= \sqrt{5 \sqrt{5}(5 \sqrt{5} -2 \sqrt{5})(5 \sqrt{5} -5 \sqrt{5} )(5 \sqrt{5}-3 \sqrt{5} }) } \\ A= \sqrt{5 \sqrt{5} (3 \sqrt{5}-0-2 \sqrt{5} )} \\ A= \sqrt{5 \sqrt{5}* \sqrt{5} } \\ A=\sqrt{5*5}\\ A=\sqrt{25}\\ A=5cm^2[/tex]