Leiame diagonaalide pikkused:

$$\vec{AC} = \vec{AB} + \vec{BC} + \vec{CD} = (6; -2) + (1; -3) + (-5; 1) = (2; -4)$$ $$\vec{BD} = \vec{BC} + \vec{CD} = (1; -3) + (-5; 1) = (-4; -2)$$ $$\vert{\vec{AC}}\vert = \sqrt{2^2 + (-4)^2} = 2\sqrt{5}$$ $$\vert{\vec{BD}}\vert = \sqrt{(-4)^2 + (-2)^2} = 2\sqrt{5}$$

Leiame kujundi ABC pindala:

$$S=\sqrt{p(p-a)(p-b)(p-c)}$$ $$p=\frac{a+b+c}{2}$$ $$p=\frac{\sqrt{40} + \sqrt{10} + 2\sqrt{5}}{2}$$ $$S \approx 6.6$$

Seega kujundi ABCD diagonaalide pikkused on võrdsed ja kujundi pool pindala on 6.6.