In trapezoid $ABCD$, $\overline{AB}$ is parallel to $\overline{CD}$, $AB = 7$ units, and $CD = 10$ units. Segment $EF$ is drawn parallel to $\overline{AB}$ with $E$ lying on $\overline{AD}$ and $F$ lying on $\overline{BC}$. If $BF:FC = 3:4$, what is $EF$? Express your answer as a common fraction.
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
www.jiskha.com/questions/1798144/point-g-is-the-midpoint-of-the-median-xm-of-xyz-point-h-is-the-midpoint-of-xy-and-point
Stop Cheating. Really?