In right $\triangle \mathrm{BMC}$,
By pythagoras theorem,
$\mathrm{BC}^2=\mathrm{BM}^2+C M^2$
$\Rightarrow \mathrm{CM}^2=\mathrm{BC}^2-\mathrm{BM}^2$
$\Rightarrow \mathrm{CM}^2=5^2-4^2$
$\Rightarrow \mathrm{CM}^2=25-16$
$\Rightarrow C M^2=9$
$\Rightarrow \mathrm{CM}=3 \mathrm{~cm}$
In right $\triangle \mathrm{ALD}$
By pythagoras theorem,
$A D^2=A L^2+D L^2$
$\Rightarrow D L^2=A D^2-A L^2$
$\Rightarrow D L^2=5^2-4^2$
$\Rightarrow D L^2=25-16$
$\Rightarrow D L^2=9$
$\Rightarrow D L=3 \mathrm{~cm}$
Since $AMML$ forms a rectangle, the opposite sides are equal.
So, $\mathrm{AB}=\mathrm{LM}=7 \mathrm{~cm}$
Area of a trapezium $=\frac{1}{2} \times$ (Sum of the parallel sides) $\times$ Distance between the parallel sides
$=\frac{1}{2} \times(\mathrm{AB} \times \mathrm{CD}) \times \mathrm{AL}$
$=\frac{1}{2} \times(3+7+3) \times 4$
$=40 \mathrm{~cm}^2$
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