In Delta ABC,AD is a median. P and Q are the | vedclass

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(C) $1$. Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles of equal area.$\operatorname{ar}(\Delta ABD) = \frac{1}{2

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$9$

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(C) $1$. Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles of equal area.$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{2} 	imes 72 = 36 \, 	ext{cm}^2$.$2$. In $\Delta ABD$,$Q$ is the midpoint of $AD$. Therefore,$BQ$ is a median of $\Delta ABD$.$\operatorname{ar}(\Delta ABQ) = \frac{1}{2} \operatorname{ar}(\Delta ABD) = \frac{1}{2} 	imes 36 = 18 \, 	ext{cm}^2$.$3$. In $\Delta ABQ$,$P$ is the midpoint of $AB$. Therefore,$QP$ is a median of $\Delta ABQ$.$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} 	imes 18 = 9 \, 	ext{cm}^2$.

In $\Delta ABC$,$AD$ is a median. $P$ and $Q$ are the midpoints of $AB$ and $AD$ respectively. If $\operatorname{ar}(\Delta ABC) = 72 \, \text{cm}^2$,then $\operatorname{ar}(\Delta APQ) = \dots \text{cm}^2$.

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