The parallel sides of a trapezium are 25 cm and 13 cm;

Solution:

Given:

The parallel sides of a trapezium are $25 \mathrm{~cm}$ and $13 \mathrm{~cm}$.

Its nonparallel sides are equal in length and each is equal to $10 \mathrm{~cm}$.

A rough sketch for the given trapezium is given below:

In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles.

$\mathrm{AD}=\mathrm{BC}=10 \mathrm{~cm}$

$\mathrm{D}=\mathrm{CN}=x \mathrm{~cm}$

$\angle \mathrm{DMA}=\angle \mathrm{CNB}=90^{\circ}$

Hence, the third side of both the triangles will also be equal.

$\therefore \mathrm{AM}=\mathrm{BN}$

Also, $\mathrm{MN}=13$

Since $\mathrm{AB}=\mathrm{AM}+\mathrm{MN}+\mathrm{NB}:$

$\therefore 25=\mathrm{AM}+13+\mathrm{BN}$

$\mathrm{AM}+\mathrm{BN}=25-13=12 \mathrm{~cm}$

$\mathrm{Or}, \mathrm{BN}+\mathrm{BN}=12 \mathrm{~cm}$      (Because $\mathrm{AM}=\mathrm{BN}$ )

$2 \mathrm{BN}=12$

$\mathrm{BN}=\frac{12}{2}=6 \mathrm{~cm}$

$\therefore \mathrm{AM}=\mathrm{BN}=6 \mathrm{~cm} .$

Now, to find the value of

$x$, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10,6 and $\mathrm{x}$.

$(\text { Hypotenuse })^{2}=(\text { Base })^{2}+(\text { Altitude })^{2}$

$(10)^{2}=(6)^{2}+(\mathrm{x})^{2}$

$100=36+\mathrm{x}^{2}$

$\mathrm{x}^{2}=100-36=64$

$\mathrm{x}=\sqrt{64}=8 \mathrm{~cm}$

$\therefore$ Distance between the parallel sides $=8 \mathrm{~cm}$

$\therefore$ Area of trapezium $=\frac{1}{2} \times($ Sum of parallel sides $) \times($ Distance between parallel sides $)$

$=\frac{1}{2} \times(25+13) \times(8)$

$=152 \mathrm{~cm}^{2}$