I AB I = 10 cm
I CD I = 5 cm
I AD I = 4 cm
Niech I DM I = x
Z podobieństwa trójkatów
ABM i DCM mamy
5/10 = x/ ( x + 4)
5*( x + 4) = 10 x
5x + 20 = 10 x
5x = 20
x = 4
x = 4 cm
zatem I MA I = 4 cm + 4 cm = 8 cm
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I AB I = 10 cm
I CD I = 5 cm
I AD I = 4 cm
Niech I DM I = x
Z podobieństwa trójkatów
ABM i DCM mamy
5/10 = x/ ( x + 4)
5*( x + 4) = 10 x
5x + 20 = 10 x
5x = 20
x = 4
x = 4 cm
zatem I MA I = 4 cm + 4 cm = 8 cm
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