ABCD is a trapezium where AD parallel BC and O is the mid point of side CD. Than prove that ar(AOB)=1/2ar(ABCD)
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Given ABCD is a trapezium in which AD∥BC and O is the midpoint of CD.
To prove area∆AOB is half the area of trapezium ABCD
Construction OL parallel to DA and CB.Join BD and let BD intersect OL at M
DA∥OL∥CBNow in ∆DBC,OM∥CB and O is the midpoint of DC.Hence M is the midpoint of DB and also MO =12BC (by converse of midpoint theorem)In ∆DAB,M is the midpoint of DB and ML∥DAHence L is the midpoint of AB and also ML=12DA (by converse of midpoint theorem)∴MO+ML =12BC+12DA=12(BC+DA)And form the figure MO+ML = OLHence OL=12(BC+DA)Now area∆AOBarea trapezium ABCD=12×AB×OL12×(DA+CB)×AB=OL(DA+CB)=12(BC+DA)(DA+CB)=12Hence area∆AOB =12area trapezium ABCD
To prove area∆AOB is half the area of trapezium ABCD
Construction OL parallel to DA and CB.Join BD and let BD intersect OL at M
DA∥OL∥CBNow in ∆DBC,OM∥CB and O is the midpoint of DC.Hence M is the midpoint of DB and also MO =12BC (by converse of midpoint theorem)In ∆DAB,M is the midpoint of DB and ML∥DAHence L is the midpoint of AB and also ML=12DA (by converse of midpoint theorem)∴MO+ML =12BC+12DA=12(BC+DA)And form the figure MO+ML = OLHence OL=12(BC+DA)Now area∆AOBarea trapezium ABCD=12×AB×OL12×(DA+CB)×AB=OL(DA+CB)=12(BC+DA)(DA+CB)=12Hence area∆AOB =12area trapezium ABCD
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