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 #5
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May 27, 2019
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May 27, 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
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\(\begin{array}{|rcll|} \hline \text{area}_{[BCP]} &=& \dfrac{4x}{2}=2x \\\\ \text{area}_{[ABC]} &=& \dfrac{4^2}{2} = 8 \\\\ \text{area}_{[ABE]} &=& \dfrac{4\cdot 2\sqrt{3}}{2} = 4\sqrt{3} \quad | \quad h = 2\sqrt{3} \\ \hline \text{area}_{[APE]} &=& \text{area}_{[ABE]}+\text{area}_{[BCP]} -\text{area}_{[ABC]}\\\\ &=& 4\sqrt{3}+2x -8 \\ \hline \end{array}\)

 

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

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

 

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

 

 

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May 27, 2019

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