Wzór na pole czworokąta:
[tex]P = \frac{1}{2}\cdot d_1\cdot d_2 \cdot sin\alpha\\d_1 = 6 \ cm\\d_2 = 15 \ cm[/tex]
[tex]a) \ P = \frac{1}{2}\cdot 6 \ cm\cdot 15 \ cm\cdot sin45^{o} = 45 \ cm^{2}\cdot\frac{\sqrt{2}}{2}=\boxed{22,5\sqrt{2} \ cm^{2}}[/tex]
[tex]b) \ P = \frac{1}{2}\cdot d_1\cdot d_2\cdot sin135^{o} = \frac{1}{2}\cdot 6 \ cm\cdot15 \ cm \cdotsin135^{o}=45 \ cm^{2}\cdot sin(180^{o}-45^{o})=\\\\=45 \ cm^{2}\cdot sin45^{o} = 45 \ cm^{2}\cdot\frac{\sqrt{2}}{2} = \boxed{22,5\sqrt{2} \ cm^{2}}[/tex]
[tex]c) \ P = \frac{1}{2}\cdot 6 \ cm \cdot 15 \ cm \cdot sin60^{o} = 45 \ cm^{2}\cdot\frac{\sqrt{3}}{2} = \boxed{22,5\sqrt{3} \ cm^{2}}[/tex]
[tex]d) \ P = \frac{1}{2}\cdot 6 \ cm \cdot 15 \ cm \cdot sin150^{o} = 45 \ cm^{2}\cdot sin(180^{o}-30^{o}) = 45 \ cm^{2}\cdot sin30^{o} =\\\\=45 \ cm^{2}\cdot\frac{1}{2} = \boxed{22,5 \ cm^{2}}[/tex]
