In Delta ABC,AD is a median. If operatorname{ | vedclass

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(B) median of a triangle divides it into two triangles of equal areas.Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triang

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$106$

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(B) median of a triangle divides it into two triangles of equal areas.Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles,$\Delta ADB$ and $\Delta ADC$,such that $\operatorname{ar}(ADB) = \operatorname{ar}(ADC)$.Given that $\operatorname{ar}(ADB) = 53 \, cm^2$,therefore $\operatorname{ar}(ADC) = 53 \, cm^2$.The area of $\Delta ABC$ is the sum of the areas of $\Delta ADB$ and $\Delta ADC$.$\operatorname{ar}(ABC) = \operatorname{ar}(ADB) + \operatorname{ar}(ADC) = 53 \, cm^2 + 53 \, cm^2 = 106 \, cm^2$.

In $\Delta ABC$,$AD$ is a median. If $\operatorname{ar}(ADB) = 53 \, cm^2$,then find $\operatorname{ar}(ABC)$ in $cm^2$.

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