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.

Question

Wyatt

Mathematics

15 October, 19:56

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.

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Answers (1)

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Douglas Boyer · Answer 1

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