Proste AB i A1B1 przecięto prostymi równoległymi AA1, BB1, CC1,jak na rysunku.
Link do rysunku:
http://m.onet.pl/_m/54907309cc65c7298c8e96436bc7b655,62,37.jpg
Oblicz:
a) | OC1| , jeśli | OA1 | = 1,8 , | AC1 | = 11,2 , | OC | = 5,4
b) | OB | ,jeśli | CC1| = 40 , | BB1 | = 56 , | C1B | = 120
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a)
|CA1|/|C1A|=|CO|/|C1O|
7,2/11,2=5,4/x
7,2x=60,48
x=8.4
b)
56/x=40/(120-x)
40x=6720-56x
96x=6720
x=70