In the above figure, AB is parallel to DC, area of AOD=4 cm 2 and area BCD=7cm 2 then the area of. CDO is

The figure is a trapezium ABCD with AB parallel to DC, and the diagonals AC and BD meet at O. The key idea is that triangles on the same base and between the same pair of parallel lines have equal area.

Why D is correct: Triangles ADC and BDC share the base DC, and their apexes A and B lie on the line AB, which is parallel to DC. So ar(ADC) = ar(BDC) = 7 cm². Now ADC is split by the diagonal DB, which passes through O, into triangles AOD and DOC. So ar(ADC) = ar(AOD) + ar(DOC), giving 7 = 4 + ar(DOC), so ar(DOC) = 3 cm².

Why the others are wrong:

• A is missing its numerical value, so it cannot be the answer.
• B (5 cm²) comes from wrongly adding or mixing the given areas rather than using the equal-area property.
• C (4 cm²) just repeats the area of AOD, which is a different triangle.

This is a standard UPPSC result: diagonals of a trapezium divide it into four triangles, and the two triangles on the non-parallel sides (AOD and BOC) are linked through the equal-area rule. Mains answers on geometry-based reasoning often use exactly this shortcut.

Trap: Aspirants often try to find the ratio of AO to OC and work through similar triangles. Spot the equal-area triangles ADC and BDC first, and the answer follows in one line.