The sides AD, BC of a trapezium ABCD are parallel and the diagonals AC and BD meet at O. If the area of triangle AOB is 3 cm square and the area of triangle BDC id=s 8 cm square. Then what is the area of the triangle AOD
Given that AD∥BC and ∆AOB=3cm2,∆BDC=8cm2 So for ∆ABC=∆BCD (they are on the same base and between the sam eparallel lines) ⇒∆AOB+∆OBC=∆BCD ⇒∆OBC=8−3=5cm2 ∆COD=∆BCD−∆OBC=8−5=3cm2 ∆ABD=∆ACD (they are on the same base and between the same parallel lines) From the property of the trapezium ∆AOB×∆COD=∆BOC×∆AOD 3×3=5×∆AOD ∆AOD=