In the given figure, ABCD and EFGD are two parallelogram and G is the mid-point of CD. Then ar(ΔDPC)=12ar(∥EFGD). Write True or False and justify your answer:
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Solution
As ΔDPC and ∥gmABCD are on the same base DC and between the same parallels AB and DC, So
ar(ΔDPC)=12ar(∥gm,ABCD) ....(1)
As, ar(EFGD)ar(ABCD)=DG×hDC×h=DG2DG=12 (G is the mid point of DC)