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# plz help

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G is the centroid of equilateral Triangle ABC. D, E, and F are midpoints of the sides as shown. P, Q, and R are the midpoints of Line AG, Line BG, and Line CG, respectively. If AB=sqrt(3) what is the perimeter of hexagon DREPFQ? Dec 9, 2018

#1
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G is the centroid of equilateral Triangle ABC. D, E, and F are midpoints of the sides as shown.
P, Q, and R are the midpoints of Line AG, Line BG, and Line CG, respectively.
If AB=sqrt(3) what is the perimeter of hexagon DREPFQ? $$\text{Let AB=\sqrt{3}} \\ \text{Let CE=\frac12\sqrt{3}} \\ \text{Let Point C=(0,0)} \\ \text{Let Point A=(\sqrt{3},0)} \\ \text{Let Point B=(\frac12\sqrt{3},\frac32)} \\$$

$$\begin{array}{|rcll|} \hline BE^2 + \left(\frac12 \sqrt{3} \right)^2 &=& (\sqrt{3})^2 \\ BE^2 &=& 3 - \frac34 \\ BE^2 &=& \frac94 \\ \mathbf{BE} & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}$$

$$\text{Let Point B=(\frac12\sqrt{3},\frac32)}$$

$$\begin{array}{|rcll|} \hline G &=& \left( \frac{0+\sqrt{3}+\frac12\sqrt{3}}{3}, \frac{0+\sqrt{0}+\frac32}{3} \right) \\ G &=& \left( \frac12\sqrt{3}, \frac12\right) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline s &=& \frac12 \\ \text{perimeter}_{\text{of hexagon }DREPFQ} &=& 6s \\ &=& 6\cdot \frac12 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array}$$ Dec 10, 2018

#1
+10

G is the centroid of equilateral Triangle ABC. D, E, and F are midpoints of the sides as shown.
P, Q, and R are the midpoints of Line AG, Line BG, and Line CG, respectively.
If AB=sqrt(3) what is the perimeter of hexagon DREPFQ? $$\text{Let AB=\sqrt{3}} \\ \text{Let CE=\frac12\sqrt{3}} \\ \text{Let Point C=(0,0)} \\ \text{Let Point A=(\sqrt{3},0)} \\ \text{Let Point B=(\frac12\sqrt{3},\frac32)} \\$$

$$\begin{array}{|rcll|} \hline BE^2 + \left(\frac12 \sqrt{3} \right)^2 &=& (\sqrt{3})^2 \\ BE^2 &=& 3 - \frac34 \\ BE^2 &=& \frac94 \\ \mathbf{BE} & \mathbf{=} & \mathbf{\frac32} \\ \hline \end{array}$$

$$\text{Let Point B=(\frac12\sqrt{3},\frac32)}$$

$$\begin{array}{|rcll|} \hline G &=& \left( \frac{0+\sqrt{3}+\frac12\sqrt{3}}{3}, \frac{0+\sqrt{0}+\frac32}{3} \right) \\ G &=& \left( \frac12\sqrt{3}, \frac12\right) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline s &=& \frac12 \\ \text{perimeter}_{\text{of hexagon }DREPFQ} &=& 6s \\ &=& 6\cdot \frac12 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array}$$ heureka Dec 10, 2018