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?

NerdyKid Dec 9, 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

#1**+10 **

Best Answer

**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