In Delta ABC,medians overline{AD} and overlin | vedclass

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(D) Since $G$ is the centroid of $\Delta ABC$,it divides the median $\overline{AD}$ in the ratio $2:1$. Therefore,$\frac{GD}{AD} = \frac{1}{3}$.In $\D

Vedclass · Accepted Answer

$10.8$

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(D) Since $G$ is the centroid of $\Delta ABC$,it divides the median $\overline{AD}$ in the ratio $2:1$. Therefore,$\frac{GD}{AD} = \frac{1}{3}$.In $\Delta ADE$,since $\overline{GK} \parallel \overline{DE}$,by the Basic Proportionality Theorem,we have $\frac{GD}{AD} = \frac{EK}{AE} = \frac{1}{3}$.This implies $AE = 3 	imes EK = 3 	imes 1.8 = 5.4$.Since $\overline{BE}$ is a median to $\overline{AC}$,$E$ is the midpoint of $\overline{AC}$.Therefore,$AC = 2 	imes AE = 2 	imes 5.4 = 10.8$.

In $\Delta ABC$,medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$. $A$ line through $G$ and parallel to $\overline{DE}$ intersects $\overline{AC}$ at $K$. If $EK = 1.8$,then $AC = \ldots$

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