Question

In the trapezium ABCD, AD 11 BC, AB - 25 cm,
BC - 76 cm, CD - 39 cm, AD - 20 cm, find its
area
plsss answer correctly​

Answer 1

Step-by-step explanation:

Draw DL perpendicular to BC and DM parallel to AB, meeting BC at L and M respectively.

Clearly,ABMD is a parallelogram.

Therefore MD=AB=25 cm and BM=AD=20cm.

MC=(BC-BM)=(76-20)cm=56cm.

Thus,in∆DMC,we have:

DM=25 cm,MC=56cm and CD=39 cm.

Therefore semi-perimeter(s)=[(25+56+39)]÷2cm=60 cm.

therefore area of∆ DMC=✓60(60-25)(60-56)(60-39)

=420 sq.cm

Therefore,1÷2×MC×DL=1÷2×56×DL=420 sq.cm

Therefore DL=15 cm.

now,

area of trap.ABCD={1/2(AD+BC)×DF}={1/2(20+76)×15}

=720 sq.cm