Question

ABCD is trapezium with AD||BC. Point O is the mid-point of side CD. Show that ar(traingle AOB)= 1/2 ar(ABCD)

Answer 1

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 =1/2BC ⎜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=1/2DA ⎛⎝⎜by converse of midpoint theorem⎞⎠⎟∴MO+ML =1/2BC+1/2DA=1/2⎛⎝⎜BC+DA⎞⎠⎟And form the figure MO+ML = OLHence OL=1/2⎛⎝⎜BC+DA⎞⎠⎟Now area∆AOBarea trapezium ABCD=1/2×AB×OL1/2×(DA+CB)×AB=OL(DA+CB)=12(BC+DA)(DA+CB)=1/2Hence area∆AOB =1/2area trapezium ABCD

hope this helped you