Odpowiedź:
Niech
I A E I = 4 x I CE I = x
więc
I DE I = y I BE I = 4 y
Mamy
[tex]P_1 = 0,5*4 x*y *sin \alpha = 2 x*y*sin \alpha = 12[/tex]
[tex]P_2 = 0,5* x*4 y *sin \alpha = 2 x*y*sin \alpha = 12[/tex]
[tex]P_3 = 0,5*4 x*4 y *sin ( 180^o - \alpha ) = 8 x*y* sin \alpha = 4*12 = 48[/tex]
[tex]P_4 = 0,5 x*y*sin \alpha = 0,25* 2 x*y*sin \alpha = 0,25*12 = 3[/tex]
Pole trapezu
P = 12 + 12 + 48 + 3 = 75
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Odpowiedź:
Niech
I A E I = 4 x I CE I = x
więc
I DE I = y I BE I = 4 y
Mamy
[tex]P_1 = 0,5*4 x*y *sin \alpha = 2 x*y*sin \alpha = 12[/tex]
[tex]P_2 = 0,5* x*4 y *sin \alpha = 2 x*y*sin \alpha = 12[/tex]
[tex]P_3 = 0,5*4 x*4 y *sin ( 180^o - \alpha ) = 8 x*y* sin \alpha = 4*12 = 48[/tex]
[tex]P_4 = 0,5 x*y*sin \alpha = 0,25* 2 x*y*sin \alpha = 0,25*12 = 3[/tex]
Pole trapezu
P = 12 + 12 + 48 + 3 = 75
=========================
Szczegółowe wyjaśnienie: