The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. find the area of the trapezium.

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

AB = 25 cm, CD = 11 cm and AD= 13 cm, BC=15cm

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

Clearly, AMCD is a parallelogram.

Now,

MC = AD = 13cm

AM = DC =11cm

MB = (AB—Am)

= (25—11) = 14 cm

Thus, in CMB, we have:

CM = 13 cm

MB = 14 cm

BC = 15 cm

Here let ML = X, hence LB = 14 – X and let CL = Y cm

Now in CML, using Pythagoras theorem

CL² = (CM2 – ML²)

Y² = (132 – X²) eq – 1

Again in CLB, using Pythagoras theorem

CL² = (CB2 – LB²)

Y² = (152 – (14—X) ²) eq – 2

Sub eq 1 in 2, we get

(132 – X²) = (152 – (14—X) ²)

169 – X² = 225 – (196 + X² – 28 X)

169 – X² = 225 – 196 – X² + 28 X

28X = 169 + 196 – 225 + X² – X²

28X = 140

X = 5 cm

Now substitute X value in eq –1

That is Y² = (132 – X²)

Y² = (132 – 5²)

Y² = (169 – 25)

Y² = 144

Y = 12 cm

Therefore CL = 12 cm that is height of the trapezium = 12 cm

Therefore

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

Therefore Area of trapezium = × (25 + 11) × 12 = 216 cm².