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(C) In $\Delta ABC$, $P$ and $Q$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively.Therefore, by the Midpoint Theorem, $PQ \parall

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) In $\Delta ABC$, $P$ and $Q$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively.Therefore, by the Midpoint Theorem, $PQ \parallel BC$ and $PQ = \frac{1}{2} BC$.This implies that $\Delta APQ \sim \Delta ABC$ by the $AA$ similarity criterion.The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Thus, $\frac{	ext{Area}(\Delta APQ)}{	ext{Area}(\Delta ABC)} = \left( \frac{AP}{AB} ight)^2$.Since $P$ is the midpoint of $AB$, $\frac{AP}{AB} = \frac{1}{2}$.Therefore, $\frac{12 \sqrt{3}}{	ext{Area}(\Delta ABC)} = \left( \frac{1}{2} ight)^2 = \frac{1}{4}$.Hence, $	ext{Area}(\Delta ABC) = 12 \sqrt{3} 	imes 4 = 48 \sqrt{3}$.

In $\Delta ABC$, $P$ and $Q$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. If the area of $\Delta APQ = 12 \sqrt{3}$, then the area of $\Delta ABC = \ldots$ (in $\sqrt{3}$)

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