Find the area of a trapezium whose parallel sides are 25 cm,

Given:

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 $15 \mathrm{~cm}$.

A rough skech of the trapezium is given below:

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

This is because both have two common sides as $15 \mathrm{~cm}$ and the altitude as $\mathrm{x}$ and a right angle.

Hence, the remaining side of both the triangles will be equal.

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

Also $\mathrm{MN}=13$

Now, since $\mathrm{AB}=\mathrm{AM}+\mathrm{MN}+\mathrm{NB}$ :

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

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

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 $\mathrm{x}$, we will use the Pythagorian theorem in the right angle triangle AMD whose sides are 15,6 and $\mathrm{x}$.

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

$(15)^{2}=(6)^{2}+(\mathrm{x})^{2}$

$225=36+\mathrm{x}^{2}$

$\mathrm{x}^{2}=225-36=189$

$\therefore \mathrm{x}=\sqrt{189}=\sqrt{9 \times 21}=3 \sqrt{21} \mathrm{~cm}$

$\therefore$ Distance between the parallel sides $=3 \sqrt{21} \mathrm{~cm}$

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

$=\frac{1}{2} \times(25+13) \times(3 \sqrt{21})$

$=57 \sqrt{21} \mathrm{~cm}^{2}$