Q15 of 31 Page 162

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

It is given that:

Area (ΔAOD) = Area (ΔBOC)


Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)


Area (ΔADB) = Area (ΔACB)


We know that triangles on the same base having areas equal to each other lie between the same parallels



Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels i.e.,


AB || CD


Therefore, ABCD is a trapezium


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Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.