In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point

Maybe not the most simple way....but

Let A(--2, 4)    B = (2, 4)    C  = (2, 0)    D = (-2, 0)

The height of the equilateral triangle will  be    side/ 2 * sqrt (3)  = 4/2 * sqrt (3) = 2 sqrt (3)

So point E will be located at  (0, 4 - 2sqrt (3) )

And using points B and E we can write the equation for the line containg the segment BE

The slope is    [ 4 -(4- 2sqrt (3)) ] / [ 2 - 0]  =  2sqrt(3) / 2  =  sqrt (3)

So....the equation of this line  is

y =  sqrt (3) ( x - 2) + 4

y = sqrt (3)x - 2sqrt(3) + 4

y = sqrt (3)x + (4 - 2sqrt (3) )       (1)

And we can find the equation of the line that joins AC

The slope is   [ 4 - 0 ] / [ -2 - 2 ]  =   4/-4 = - 1

So....the equation of this line is

y = -(x - 0) + 2

y = -x + 2       (2)

Set (1)  = (2) to find the x coordinate of P

sqrt (3) x + (4 - 2sqrt (3) )  = - x + 2        rearrange

x ( sqrt (3) + 1)  =  2 - 4 + 2sqrt (3)

x ( sqrt (3) + 1)  =  2sqrt (3) - 2

x  =  [ 2sqrt ( 3) - 2 ]

       _____________        rationalize the denominator

            sqrt (3) + 1

        [ 2sqrt (3) - 2 ] [ sqrt (3) - 1]             6 - 2sqrt (3) - 2sqrt (3) + 2

x =  ________________________  =   _______________________  =

        [sqrt (3)+ 1] [ sqrt (3) - 1 ]                    3 -  1

8 - 4sqrt (3)                4 - 2sqrt (3)

_________       =

    2

So the y coordinate of P  =

y = -( 4- 2sqrt (3))+ 2  =   2sqrt (3) - 2

So  the  area  of triangle APE  =   Area of triangle ABE - Area of triangle APB

Area of triangle  AEB  = (1/2)base * (height)  =  (1/2) (4) (  2sqrt (3))  = 2 (2sqrt (3))  = 

4sqrt (3)  units^2

And the area of triangle  APB  = (1/2) base (height) = (1/2)(4) [4 - (2sqrt (3) - 2 ) ]  =

2 ( 6 - 2sqrt (3) )  =  12 - 4sqrt (3)   units^2

So...the area of triangle APE  =  [ 4sqrt (3)]  - [ 12 - 4sqrt (3) ]  = [8sqrt (3) - 12] units^2 =

4 [ 2sqrt (3) - 3 ]  units^2  

In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral.
Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.
Find the area of triangle APE in simplest radical form.

\(\begin{array}{|rcll|} \hline \text{area}_{[BCP]} &=& \dfrac{4x}{2}=2x \\\\ \text{area}_{[ABC]} &=& \dfrac{4^2}{2} = 8 \\\\ \text{area}_{[ABE]} &=& \dfrac{4\cdot 2\sqrt{3}}{2} = 4\sqrt{3} \quad | \quad h = 2\sqrt{3} \\ \hline \text{area}_{[APE]} &=& \text{area}_{[ABE]}+\text{area}_{[BCP]} -\text{area}_{[ABC]}\\\\ &=& 4\sqrt{3}+2x -8 \\ \hline \end{array}\)

\(\mathbf{x=\ ?}\)

\(\begin{array}{|lrcll|} \hline (1): & \tan(45^\circ) &=& \dfrac{x}{CQ} \quad | \quad \tan(45^\circ) = 1 \\ & 1 &=& \dfrac{x}{CQ} \\ & \mathbf{CQ} &=& \mathbf{x} \\ \hline (2): & \tan(30^\circ) &=& \dfrac{x}{4-CQ} \quad | \quad CQ=x \\\\ & \tan(30^\circ) &=& \dfrac{x}{4-x} \quad | \quad \tan(30^\circ) = \dfrac{\sqrt{3}}{3} \\\\ & \dfrac{\sqrt{3}}{3} &=& \dfrac{x}{4-x} \\\\ & (4-x)\sqrt{3} &=& 3x \\ & 4\sqrt{3}-x\sqrt{3} &=& 3x \\ & x\sqrt{3} + 3x&=& 4\sqrt{3} \\ & x(3+\sqrt{3})&=& 4\sqrt{3} \\\\ & x &=& \dfrac{4\sqrt{3}} {3+\sqrt{3}} \\\\ & x &=& \dfrac{4\sqrt{3}} {3+\sqrt{3}}\times \dfrac{(3-\sqrt{3})}{(3-\sqrt{3})} \\\\ & x &=& \dfrac{4\sqrt{3}(3-\sqrt{3})} {9-3} \\\\ & x &=& \dfrac{4\sqrt{3}(3-\sqrt{3})} {6} \\\\ & x &=& \dfrac{2\sqrt{3}(3-\sqrt{3})} {3} \\\\ & x &=& \dfrac{2\sqrt{3}\cdot 3 } {3} - \dfrac{2} {3} \cdot 3 \\\\ & \mathbf{x} &=& \mathbf{2\sqrt{3} - 2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{area}_{[APE]} &=& 4\sqrt{3}+2x -8 \quad | \quad x=2\sqrt{3} - 2 \\ &=& 4\sqrt{3}+2(2\sqrt{3} - 2) -8 \\ &=& 8\sqrt{3} -12 \\ \mathbf{\text{area}_{[APE]}}&=& \mathbf{4(2\sqrt{3} -3)} \\ \hline \end{array}\)