Wiedząc, że |AB|=5, |BC|=4 oraz |BD| / |DE|= 5/1. Oblicz obwód równoległoboku BDEF.
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|BC| = 4
|BD| / |DE| = 5 / 1
|BD| = 5x = |EF|
|DE| = x = |BF|
|CF| = 4 - x
ΔABC jest podobny do ΔEFC na mocy zasady kąt, kąt, kąt.
|EF| / |AB| = |CF| / |BC|
5x / 5 = (4 - x) / 4 Mnożę na krzyż:
5x * 4 = 5(4 - x)
20x = 20 - 5x
25x = 20
x = 4/5
5x = 4
Obw = 2 * (x + 5x) = 2 * 6x = 12x = 12 * 4/5 = 48/5 = 9 3/5 = 9,6