#### Question

The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

#### Solution

Given:

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

Its nonparallel sides are equal in length and each is equal to 10 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.

AD = BC = 10 cm

D = CN = x cm

\[\angle DMA = \angle CNB = 90^\circ\]

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

\[ \therefore AM=BN\]

Also, MN=13

Since AB = AM+MN+NB:

\[ \therefore 25=AM+13+BN\]

\[AM+BN=25-13=12 cm\]

\[Or, BN+BN=12 cm (\text{ Because AM=BN })\]

\[2 BN=12\]

\[BN=\frac{12}{2}=6 cm\]

∴ AM = BN = 6 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 x.

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

\[ {(10)}^2 {=(6)}^2 {+(x)}^2 \]

\[ {100=36+x}^2 \]

\[ x^2 =100-36=64\]

\[x=\sqrt{64}=8 cm\]

∴ Distance between the parallel sides = 8 cm

∴ Area of trapezium\[=\frac{1}{2}\times( \text{ Sum of parallel sides })\times(\text{ Distance between parallel sides })\]

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

\[ {=152 cm}^2\]