# M is the midpoint of the side AB of a parallelogram ABCD.

Question:

is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm2, find ar(∆ABC).

Solution:

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Join AC.
AC divides parallelogram ABCD into two congruent triangles of equal areas.

$\operatorname{ar}(\triangle \mathrm{ABC})=\operatorname{ar}(\triangle \mathrm{ACD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$

M is the midpoint of AB. So, CM is the median.

$\mathrm{CM}$ divides $\triangle \mathrm{ABC}$ in two triangles with equal area.

$\operatorname{ar}(\triangle \mathrm{AMC})=\operatorname{ar}(\triangle \mathrm{BMC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$

$\operatorname{ar}(\mathrm{AMCD})=\operatorname{ar}(\triangle \mathrm{ACD})+\operatorname{ar}(\triangle \mathrm{AMC})=\operatorname{ar}(\triangle \mathrm{ABC})+\operatorname{ar}(\triangle \mathrm{AMC})=\operatorname{ar}(\triangle \mathrm{ABC})+\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$

$\Rightarrow 24=\frac{3}{2} \operatorname{ar}(\triangle \mathrm{ABC})$

$\Rightarrow \operatorname{ar}(\triangle \mathrm{ABC})=16 \mathrm{~cm}^{2}$