The parallel sides of a trapezium are 25 cm and 11 cm,

Solution:

Let $\mathrm{ABCD}$ be the trapezium in which $\mathrm{AB} \| \mathrm{DC}, \mathrm{AB}=25 \mathrm{~cm}, \mathrm{CD}=11 \mathrm{~cm}, \mathrm{AD}=13 \mathrm{~cm}$ and $\mathrm{BC}=15 \mathrm{~cm}$.

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,

$\mathrm{MC}=\mathrm{AD}=13 \mathrm{~cm}$

$\mathrm{AM}=\mathrm{DC}=11 \mathrm{~cm}$

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

$=(25-11) \mathrm{cm}$

$=14 \mathrm{~cm}$

Thus, in $\Delta \mathrm{CMB}$, we have :

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

$\mathrm{MB}=14 \mathrm{~cm}$

$\mathrm{BC}=15 \mathrm{~cm}$

$\therefore \mathrm{s}=\frac{1}{2}(13+14+15) \mathrm{cm}$

$=\frac{1}{2} 42 \mathrm{~cm}$

$=21 \mathrm{~cm}$

$(\mathrm{s}-\mathrm{a})=(21-13) \mathrm{cm}$

$=8 \mathrm{~cm}$

$(\mathrm{~s}-\mathrm{b})=(21-14) \mathrm{cm}$

$=7 \mathrm{~cm}$

$(\mathrm{~s}-\mathrm{c})=(21-15) \mathrm{cm}$

$=6 \mathrm{~cm}$

$\therefore$ Area of $\Delta \mathrm{CMB}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

$=\sqrt{21 \times 8 \times 7 \times 6} \mathrm{~cm}^{2}$

$=84 \mathrm{~cm}^{2}$

$\therefore \frac{1}{2} \times \mathrm{MB} \times \mathrm{CL}=84 \mathrm{~cm}^{2}$

$\Rightarrow \frac{1}{2} \times 14 \times \mathrm{CL}=84 \mathrm{~cm}^{2}$

$\Rightarrow \mathrm{CL}=\frac{84}{7}$

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

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

$=\left\{\frac{1}{2} \times(25+11) \times 12\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 36 \times 12\right) \mathrm{cm}^{2}$

$=(18 \times 12) \mathrm{cm}^{2}$

$=216 \mathrm{~cm}^{2}$

Hence, the area of the trapezium is $216 \mathrm{~cm}^{2}$.