# Find x given the area of quad ABEF=the area of quad BCDE. AC=7. DE=5. AF=x. EF=x. AF is perpendicular to FD. AF is perpendicular to AC. Info to draw diagram: draw trapezoid ACDF. Draw a segment from inbetween AC to inbetween FD. (doesn't cut either in half) I have drawn the diagram according to the instructions given by you, and I have assumed BE is the line segment that you are talking about and BE line segment is parallel to AF.

Therefore if you take quad, ABEF it is a square and its sides are,

`AB =...

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I have drawn the diagram according to the instructions given by you, and I have assumed BE is the line segment that you are talking about and BE line segment is parallel to AF.

Therefore if you take quad, ABEF it is a square and its sides are,

`AB = BE = EF = AF = x`

Therefore its area is = `x^2`

Now quad BCDE is a trapezoid with BC parallel to DE.

Its sides are,

BC = 7-x

CD = unknown

DE = 5

BE = x

Area of BCDE = `(((7-x)+5)/2) * x`

= `1/2x(12-x)`

Now according to given data, Area of BCDE = Area of ABEF

Therefore,

`1/2x(12-x) = x^2`

`12x -x^2 = 2x^2`

`3x^2 = 12x`

`3x^2-12x = 0`

`x(x-4) = 0`

The solutions are x =0 or x = 4
. But we know x!= 0

Therefore x =4.

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