Question

Find the height of a trapezium in which parallel sides are 25 cm 77 cm and non-parallel sides and 26 cm and 60 cm. Given the area of the trapezium as 1644 cm2.

Answer 1

$\bf\purple{Given:-}$

ABCD is a parallelogram.

BC=AD=25cm

CD=AB=60cm

CF=77-CD=17cm

$\bf\purple{Draw:-}$

Draw a line BC parallel to AD

Draw a perpendicular line BE on DF

$\bf\purple {For \: ∆BCF:-}$

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 Δ BCF=1/2xbase x height

=1/2 BExCF204cm²

=1/2x BE x 17BE

=408/17

=24cm

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

=1/2(60+77) x 24

=1644cm²

$\tt\small\boxed{★Good \: luck★}$