In Delta ABC,points P and Q are the points of | vedclass

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(D) Let the base $BC$ of $\Delta ABC$ be divided into three equal parts by points $P$ and $Q$. Thus,$BP = PQ = QC = \frac{1}{3} BC$.Since all three tr

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$1:3$

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(D) Let the base $BC$ of $\Delta ABC$ be divided into three equal parts by points $P$ and $Q$. Thus,$BP = PQ = QC = \frac{1}{3} BC$.Since all three triangles $\Delta ABP$,$\Delta APQ$,and $\Delta AQC$ share the same vertex $A$ and have equal bases $(BP = PQ = QC)$,their heights with respect to these bases are identical.The area of a triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Since the heights are the same and the bases are equal,$\operatorname{ar}(\Delta ABP) = \operatorname{ar}(\Delta APQ) = \operatorname{ar}(\Delta AQC)$.Therefore,$\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta ABP) + \operatorname{ar}(\Delta APQ) + \operatorname{ar}(\Delta AQC) = 3 	imes \operatorname{ar}(\Delta APQ)$.This implies $\frac{\operatorname{ar}(\Delta APQ)}{\operatorname{ar}(\Delta ABC)} = \frac{1}{3}$.Thus,the ratio is $1:3$.

In $\Delta ABC$,points $P$ and $Q$ are the points of trisection of $BC$. Then,$\operatorname{ar}(\Delta APQ) : \operatorname{ar}(\Delta ABC) = \dots$

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