A) \[ar(ABCD)=ar(ADEF)\]
B) \[ar(ABCD)=ar(\Delta ADP)+ar(DEFP)\]
C) \[ar(\Delta ADP)=\frac{1}{2}[ar(ABCD)]\]
D) \[ar(ABCP)=ar(DEFP)\]
Correct Answer: C
Solution :
\[\Delta ADP\]is on the same base as parallelogram ABCD but not between the same parallels AD and BE.You need to login to perform this action.
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