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.

EF = 58/7

Step-by-step explanation:

If you extend both BC and AD to meet at P you get 3 similar triangles PAB, PEF, and PCD. Let's call BF = 3x and FC = 4x. Therefore the following proportions can be expressed:

BP/7 = (BP+7x)/10

10BP = 7*(BP+7x)

10BP = 7BP+49x

3BP = 49x

BP = (49/3)x

BP/7 = (BP+3x)/EF

EF = (BP+3x)*7/BP

EF = ((49/3)x + 3x)*7/((49/3)x)

EF = ((58/3)x*7*3)/(49x)

EF = 58/7

answer:   rlly

step-by-step explanation:

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