In Delta ABC,AD is a median,M and N are the m | vedclass

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(B) $1$. In $\Delta ABD$,$AM$ is a median because $M$ is the midpoint of $BD$. Therefore,$\operatorname{ar}(ADM) = \operatorname{ar}(ABM) = \frac{1}{2

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(B) $1$. In $\Delta ABD$,$AM$ is a median because $M$ is the midpoint of $BD$. Therefore,$\operatorname{ar}(ADM) = \operatorname{ar}(ABM) = \frac{1}{2} \operatorname{ar}(ABD)$.$2$. In $\Delta ADM$,$AN$ is a median because $N$ is the midpoint of $MD$. Therefore,$\operatorname{ar}(AND) = \operatorname{ar}(ANM) = \frac{1}{2} \operatorname{ar}(ADM)$.$3$. Given $\operatorname{ar}(AND) = 20\, cm^2$,we have $\operatorname{ar}(ADM) = 2 	imes 20 = 40\, cm^2$.$4$. Since $\operatorname{ar}(ADM) = \frac{1}{2} \operatorname{ar}(ABD)$,then $\operatorname{ar}(ABD) = 2 	imes 40 = 80\, cm^2$.$5$. In $\Delta ABC$,$AD$ is a median,so $\operatorname{ar}(ABC) = 2 	imes \operatorname{ar}(ABD) = 2 	imes 80 = 160\, cm^2$.

In $\Delta ABC$,$AD$ is a median,$M$ and $N$ are the midpoints of $BD$ and $MD$ respectively. If $\operatorname{ar}(AND) = 20\, cm^2$,then $\operatorname{ar}(ABC) = \dots cm^2$.

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