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
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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
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