The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.

Let ABCD be the given trapezium in which AB|| DC,

AB = 20 cm, DC = 10 cm and AD=BC=13cm

Draw CL AB and CM || DA meeting AB at L and M, respectively.

Clearly, AMCD is a parallelogram.

Now,

AM = DC =10cm

MB = (AB-Am)

= (20-10) = 10 cm

Also,

CM = DA = 13cm

Therefore, CMB is an isosceles triangle and CL MB.

And L is midpoint of B.

⇒ML = LB = = = 5 cm

From right CLM, we have:

CL² = (CM2 – ML²)

CL² = (132 – 5²)

CL² = (169 – 25)

CL² = 144

CL = 12

Therefore length of CL is 12 cm that is height of trapezium is 12 cm

There fore

We know that area of trapezium is × (sum of parallel sides) × height

Therefore Area of trapezium = × (20 + 10) × 12 = 180 cm².