Answer=152 cm^2

solution,

ABCD is a trapezium in which AB = 25 cm , CD= 13cm ( AB | | CD ) , and BC=AD = 10 cm. , draw perpendiculars CP and DQ from C and D on AB.

• PB = QA as BC = AD. =10 cm. Let PB =QA = x cm.

• AB = QA+QP+PB. , putting AB=25 cm., PB= QA=x cm. , and QP = CD=13 cm.

• 25= x +13+x. => 2x= 25–13 =>. x=12/2= 6 cm.

In right angled triangle CPB , CP^2+PB^2 = BC^2.

1. or. CP^2 + 6^2 = 10^2.

1. or. CP^2 = 100 -36 =64.

1. or. CP = 8 cm.

Area of trapezium ABCD = 1/2.(AB+CD)×CP = 1/2.(25+13)×8. cm^2.

= 1/2.(38×8) = 152 cm^2.