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In trapezium ABCD (AB parallel to DC), diagonals meet at O. Since ar(ABD) = ar(ABC), subtracting the common triangle ABO gives:

Class 9Areas — Figures on the Same BaseGeometry
  • Aar(AOD) = ar(BOC)
  • Bar(AOD) = 2 ar(BOC)
  • Car(AOD) = (1/2) ar(BOC)
  • Dar(AOD) = ar(ABO)

Step-by-Step Solution

ar(ABD) = ar(ABO) + ar(AOD) and ar(ABC) = ar(ABO) + ar(BOC). Since ar(ABD) = ar(ABC), subtracting ar(ABO) gives ar(AOD) = ar(BOC).
Answer: A) ar(AOD) = ar(BOC)

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