In Delta ABC,AD is a median and AM is an alti | vedclass

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(B) $1$. Given that $AD$ is a median of $\Delta ABC$,it divides the triangle into two triangles of equal area. Thus,$Area(\Delta ABD) = Area(\Delta AD

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(B) $1$. Given that $AD$ is a median of $\Delta ABC$,it divides the triangle into two triangles of equal area. Thus,$Area(\Delta ABD) = Area(\Delta ADC) = \frac{1}{2} 	imes Area(\Delta ABC)$.$2$. The area of $\Delta ABC = \frac{1}{2} 	imes BC 	imes AM = \frac{1}{2} 	imes 16 	imes 8 = 64\, cm^2$.$3$. Therefore,$Area(\Delta ABD) = \frac{1}{2} 	imes 64 = 32\, cm^2$.$4$. In $\Delta EBD$,$BA$ is a median because $AB = AE$ (given),which implies $A$ is the midpoint of $BE$. Thus,$DA$ is a median of $\Delta EBD$ relative to base $BE$,but more simply,$\Delta ABD$ and $\Delta EBD$ share the same altitude from $D$ to the line $BE$,and their bases $AB$ and $AE$ are equal.$5$. Since $AB = AE$,the triangles $\Delta ABD$ and $\Delta AED$ have equal areas. However,looking at $\Delta EBD$,the median $DA$ divides it into two equal areas: $Area(\Delta ABD) = Area(\Delta AED) = 32\, cm^2$.$6$. The total area of $\Delta EBD = Area(\Delta ABD) + Area(\Delta AED) = 32 + 32 = 64\, cm^2$.

In $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.

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