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(c) $80 \mathrm{~cm}^{2}$

From the given trapezium, we find:

$D C=A L=7 \mathrm{~cm} \quad[$ since $D A \perp A B$ and $C L \perp A B]$

From the right $\Delta$ CBL, we have :

$C L^{2}=C B^{2}-L B^{2}$

$\Rightarrow C L^{2}=(10)^{2}-(6)^{2}$

$\Rightarrow C L^{2}=100-36$

$\Rightarrow C L^{2}=64$

$\Rightarrow C L=\sqrt{64}$

$\Rightarrow C L=8 \mathrm{~cm}$

Area of the trapezium $=\left\{\frac{1}{2} \times(7+13) \times 8\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 20 \times 8\right) \mathrm{cm}^{2}$

$=80 \mathrm{~cm}^{2}$