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