The parallel sides of a trapezium are 20 cm and 10 cm.

Solution:

Let $\mathrm{ABCD}$ be the given trapezium in which $\mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=20 \mathrm{~cm}, \mathrm{DC}=10 \mathrm{~cm}$ and $\mathrm{AD}=\mathrm{BC}=13 \mathrm{~cm}$.

Draw $\mathrm{CL} \perp \mathrm{AB}$ and $\mathrm{CM} \| \mathrm{DA}$ meeting $\mathrm{AB}$ at $\mathrm{L}$ and $\mathrm{M}$, respectively.

Clearly, AMCD is a parallelogram.

Now,

$\mathrm{AM}=\mathrm{DC}=10 \mathrm{~cm}$

$\mathrm{MB}=(\mathrm{AB}-\mathrm{AM})$

$=(20-10) \mathrm{cm}$

$=10 \mathrm{~cm}$

Also,

$\mathrm{CM}=\mathrm{DA}=13 \mathrm{~cm}$

Therefore, $\Delta \mathrm{CMB}$ is an isosceles triangle and $\mathrm{CL} \perp \mathrm{MB}$.

$\mathrm{L}$ is the midpoint of $\mathrm{B}$.

$\Rightarrow \mathrm{ML}=\mathrm{LB}=\left(\frac{1}{2} \times \mathrm{MB}\right)$

$=\left(\frac{1}{2} \times 10\right) \mathrm{cm}$

$=5 \mathrm{~cm}$

From right $\Delta$ CLM, we have :

$\mathrm{CL}^{2}=\left(\mathrm{CM}^{2}-\mathrm{ML}^{2}\right) \mathrm{cm}^{2}$

$\Rightarrow \mathrm{CL}^{2}=\left\{(13)^{2}-(5)^{2}\right\} \mathrm{cm}^{2}$

$\Rightarrow \mathrm{CL}^{2}=(109-25) \mathrm{cm}^{2}$

$\Rightarrow \mathrm{CL}^{2}=144 \mathrm{~cm}^{2}$

$\Rightarrow \mathrm{CL}=\sqrt{144} \mathrm{~cm}$

$\Rightarrow \mathrm{CL}=12 \mathrm{~cm}$

$\therefore$ Length of $\mathrm{CL}=12 \mathrm{~cm}$

Area of the trapezium $=\left\{\frac{1}{2} \times(\mathrm{AB}+\mathrm{DC}) \times \mathrm{CL}\right\}$

$=\left\{\frac{1}{2} \times(20+10) \times 12\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 30 \times 12\right) \mathrm{cm}^{2}$

$=(15 \times 12) \mathrm{cm}^{2}$

$=180 \mathrm{~cm}^{2}$

Hence, the area of the trapezium is $180 \mathrm{~cm}^{2}$.