Question

in the adjoining figure,ABCD is a trapezium.If AB=60,CD=77,AD=26 and BC=25,find the area of given trapezium​

Answer 1

Draw a line BC parallel to AD 

Draw a perpendicular line BE on DF

ABCD is a parallelogram.

BC=AD=25cm

CD=AB=60cm

CF=77-CD=17cm

For Δ BCF

perimeter of triangle=(25+26+17)/2

=68/2=34

By Heron's Formula of triangle=√s(s-a)(s-b)(s-c)

=√34(34-25)(34-26)(34-17)

=204cm²

Now area of ΔBCF=1/2xbase x height

=1/2 BExCF204cm²

=1/2x BEx17BE

=408/17

=24cm

Area of Trapezium =1/2(AB+DF)xBE

=1/2(60+77)x24

=1644cm²

Answer 2

Here is your answer ⤵⤵⤵

Draw a line BC parallel to AD 

Draw a perpendicular line BE on DF

ABCD is a parallelogram.

BC=AD=25cm

CD=AB=60cm

CF=77-CD=17cm

For Δ BCF

perimeter of triangle=(25+26+17)/2

=68/2=34

By Heron's Formula of triangle=√s(s-a)(s-b)(s-c)

=√34(34-25)(34-26)(34-17)

=204cm²

Now area of ΔBCF=1/2xbase x height

=1/2 BExCF204cm²

=1/2x BEx17BE

=408/17

=24cm

Area of Trapezium =1/2(AB+DF)xBE

=1/2(60+77)x24

=1644cm²

HOPE IT HELPS YOU ☺☺ !!