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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.

Respuesta :

Answer:

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

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