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

  • A
    $36$
  • B
    $106$
  • C
    $336$
  • D
    $128$

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In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$

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