Question

answer the 2nd Question

Answer 1

hlo

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Answer 2

given,           ABCD a parallelogram in which          E,F,G,H, are mid points of AB,BC,CD,AD. Then to prove,           ar(EFGH) = ar(ABCD) Construction, Draw HF parallel to AB and CD Proof:          AB is parallel and equal to HF . Therefore, ABFH is a parallelogram Since, triangle EFH and parallelogram ABFH lies on the same base HF and between same parallels AB and HF  Therefore, ar(EFH) = 1/2 ar (ABFH)                            ( 1) Now, DC is parallel and equal to HF. Therefore,DCFH is a parallelogram Since, triangle GFH and parallelogram DCFH lies on the same base HF and between same parallels DC and HF. Therefore, ar(GFH) = 1/2 ar (DCFH)                            (2) From 1 and 2 we get,                  ar(EFH) + ar (GFH) = 1/2 ar (ABFH) + 1/2 ar (DCFH)                 ar (EFGH) = 1/2 ( ar(ABFH) + ar(DCFH))                 ar (EFGH) = 1/2 ( ar(ABCD))                 ar (EFGH) = 1/2 ar (ABCD)                     hence,,  proved..