Respuesta :
GE = 3.35, AG = 6.7, AE = 10.05
DG = 3.145, GC = 6.29, DC = 9.435
BG = 2.982, GF = 1.491, BF = 4.473
Explanation:Given: distance of the centriod to the vertex is twice the distance from centroid to the midpoint on the opposite side:
centroid: (14/3, 4/3)
[tex]\begin{gathered} AG\text{ = 2GE} \\ BG\text{ = 2GF} \\ GC\text{ = 2DG} \end{gathered}[/tex][tex]\begin{gathered} \text{Centriod = G = (14/3, 4/3) } \\ E\text{ = midpoint of BC} \\ B\text{ (}2,\text{ 0) and C (8, -4)} \\ \text{Midpoint = }\frac{1}{2}(x_1+x_2),\text{ }\frac{1}{2}(y_1+y_2) \\ \text{midpoint = }\frac{1}{2}\text{(2 + 8), }\frac{1}{2}(0-4) \\ \text{midpoint = 5, -2} \\ \\ GE\text{ = distance from G to E} \\ dis\tan ce\text{ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ =\text{ }\sqrt[]{(5-\frac{14}{3})^2+(-2-\frac{4}{3})^2}\text{ = }\sqrt[]{0.1111+11.1111} \\ GE=\text{ }3.35 \\ AG\text{ = 2GE = 2(3.35)} \\ \text{AG = }6.7 \\ AE\text{ = GE + AG} \\ AE\text{ = 3.35 + 6.7 } \\ AE\text{ = 10.05} \end{gathered}[/tex][tex]\begin{gathered} D\text{ = midpoint of AB} \\ A(4,\text{ 8), B(2, 0)} \\ \text{midpoint = }\frac{1}{2}(4\text{ + 2), }\frac{1}{2}(8+0) \\ \text{midpoint = 3, 4} \\ DG\text{ = distance from D to G} \\ dis\tan ce\text{ = }\sqrt[]{(3-\frac{14}{3})^2+(4-\frac{4}{3})^2} \\ \text{distance = }\sqrt[]{2.7778+7.1111}\text{ = }3.145 \\ \text{DG = 3}.145 \\ \\ GC\text{ = 2DG = 2(3.145)} \\ GC\text{ = }6.29 \\ DC\text{ = DG + GC} \\ DC\text{ = 3.145 + 6}.29 \\ DC\text{ = }9.435 \end{gathered}[/tex][tex]\begin{gathered} F\text{ is the midpoint of AC:} \\ A(4,\text{ 8) , C(8, -4)} \\ \text{midpoint = }\frac{1}{2}(4+8),\text{ }\frac{1}{2}(8+(-4)) \\ \text{midpoint = 6, 2} \\ \text{Distance GF from G}(\frac{14}{3},\text{ }\frac{\text{4}}{3}\text{) to F(6, 2)} \\ \text{Distance = }\sqrt[]{(2-\frac{4}{3})^2+(6-\frac{14}{3})^2} \\ \text{Distance = }\sqrt[]{\text{0.4445+1.7777}}\text{ = 1.491} \\ G\text{F = 1.491} \\ \\ BG\text{ = 2GF = 2(1.491)} \\ BG\text{ = 2.982} \\ BF\text{ = BG + GF} \\ BF\text{ = 2.982 + 1.491} \\ BF\text{ = 4.473} \end{gathered}[/tex]