In Delta ABC,D is the midpoint of overline{BC | vedclass

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(C) Let the side length of the equilateral triangle $\Delta ABC$ be $s_1 = BC = a$.Since $D$ is the midpoint of $\overline{BC}$,the side length of the

Vedclass · Accepted Answer

$4:1$

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(C) Let the side length of the equilateral triangle $\Delta ABC$ be $s_1 = BC = a$.Since $D$ is the midpoint of $\overline{BC}$,the side length of the equilateral triangle $\Delta BDE$ is $s_2 = BD = \frac{a}{2}$.The area of an equilateral triangle with side length $s$ is given by the formula $Area = \frac{\sqrt{3}}{4} s^2$.Therefore,the area of $\Delta ABC$ is $Area(ABC) = \frac{\sqrt{3}}{4} a^2$.The area of $\Delta BDE$ is $Area(BDE) = \frac{\sqrt{3}}{4} (\frac{a}{2})^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{1}{4} \cdot Area(ABC)$.The ratio of the areas of $\Delta ABC$ to $\Delta BDE$ is $\frac{Area(ABC)}{Area(BDE)} = \frac{Area(ABC)}{\frac{1}{4} Area(ABC)} = \frac{4}{1}$.Thus,the ratio is $4:1$.

In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$. $\Delta ABC$ and $\Delta BDE$ are equilateral triangles. Then,the ratio of the areas of $\Delta ABC$ and $\Delta BDE$ is.......

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