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

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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.

THANK YOU GUYS FOR ANY HELP AT ALL!   May 26, 2019

#1
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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   May 26, 2019
#2
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Thank you soooooooo much! I get it now!!!!!!!!!

alskdj  May 26, 2019
#3
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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
. $$\begin{array}{|rcll|} \hline \text{area}_{[BCP]} &=& \dfrac{4x}{2}=2x \\\\ \text{area}_{[ABC]} &=& \dfrac{4^2}{2} = 8 \\\\ \text{area}_{[ABE]} &=& \dfrac{4\cdot 2\sqrt{3}}{2} = 4\sqrt{3} \quad | \quad h = 2\sqrt{3} \\ \hline \text{area}_{[APE]} &=& \text{area}_{[ABE]}+\text{area}_{[BCP]} -\text{area}_{[ABC]}\\\\ &=& 4\sqrt{3}+2x -8 \\ \hline \end{array}$$

$$\mathbf{x=\ ?}$$

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

$$\begin{array}{|rcll|} \hline \text{area}_{[APE]} &=& 4\sqrt{3}+2x -8 \quad | \quad x=2\sqrt{3} - 2 \\ &=& 4\sqrt{3}+2(2\sqrt{3} - 2) -8 \\ &=& 8\sqrt{3} -12 \\ \mathbf{\text{area}_{[APE]}}&=& \mathbf{4(2\sqrt{3} -3)} \\ \hline \end{array}$$ May 27, 2019