Q. 164.3( 230 Votes )

# In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Answer :

It is given that:

Area (ΔDRC) = Area (ΔDPC)

As ΔDRC and ΔDPC lie on the same base DC and have equal areas, therefore, they must lie between the same parallel lines

DC || RP

Therefore, DCPR is a trapezium. It is

Also given that:

Area (ΔBDP) = Area (ΔARC)

Area (BDP) − Area (ΔDPC) = Area (ΔARC) − Area (ΔDRC)

Area (ΔBDC) = Area (ΔADC)

Since ΔBDC and ΔADC are on the same base CD and have equal areas, they must lie between the same parallel lines

AB || CD

Therefore,

**ABCD is a trapezium.**

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PREVIOUSDiagonals 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.NEXTParallelogram 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.

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