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Cut and paste does not work here, as you can see, but I think I followed your coding.
extend both CB and DA to meet at P
let FB = 3x, let CF = 4x, let BP= a, let FE = k
(we want k)
you now have 3 similar triangles
PBA, PFE, and PCD
a : a+3x : a+7x = 7 : k : 10
a/7 = (a+7x)/10
10a = 7a + 49x
3a = 49x
a = 49x/3
a/7 = (a+3x)/k
ak = 7a + 21x
k = (7a +21x)/a
= (7(49x/3 + 21x)/(49x/3)
= ( 343x/3 + 21x)/(49x/3)
= (406x/3) / (49x/3)
= 406/49
= 58/7
so EF = 58/7 or appr 8.29
check my arithmetic