oblicz pole prostokąta o wierzchołkach : A= ( 2, 0) B=(1,-3) C=(-5,-1) D= (-4,2) pilne proszę o pomoc DAJE NAJ :)

[latex]\|AB|= sqrt{(1-2)^2+(-3-0)^2} = sqrt{1+9} = sqrt{10} \ \|BC|= sqrt{(-5-1)^2+(-1+3)^2} = sqrt{36+4} = sqrt{40} \ \ \P= sqrt{10} cdot sqrt{40} = sqrt{400} =20[/latex]