Math, asked by moharchat8, 11 months ago

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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Answered by CHILLIEST

Step-by-step explanation:

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

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Answered by mehek2440

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