UPSC CDS 2 2019 Math Paper

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=‌

9

5

=1.8cm2