Stalin sites of isosceles trapezium at 6 and 14 if the length of non parallel sides 5 centimetre find the area of trapezium

Let trapezium is ABCD.
AB and CD are parallel sides with length 6 and 14 cm.
Given that non parallel side AD=BC=5 cm
if we draw a line parallel to BC from A to E on DC,
then ar of triangle ADE=12 cm^2 (applying Heron's formula with sides 5,5 and 8)
ar(ADE)=1/2*b*h=12
=>h=6 cm
Ar of parallelogram ABCE=bh=6^2=36 cm^2

So total area of trapezium=ar(ADE)+ar(ABCE)=12+36
=48 cm^2[ans]

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)