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

Q : 16    In Fig.\small 9.29,  \small ar(DRC)=ar(DPC)  and \small ar(BDP)=ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

            
 

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Given,
ar(\DeltaDPC) = ar(\DeltaDRC) ..........(i)
and ar(\DeltaBDP) = ar(ARC)............(ii)

from equation (i),
Since \DeltaDRC and \DeltaDPC lie on the same base DC and between same parallels.
\therefore  CD || RP (opposites sides are parallel)

Hence quadrilateral DCPR is a trapezium

Now, by subtracting eq(ii) - eq(i) we get

ar(\DeltaBDP) - ar(\DeltaDPC) = ar(\DeltaARC) - ar(\DeltaDRC)
ar(\DeltaBDC) = ar(\DeltaADC) (Since theya are on the same base DC)
\RightarrowAB || DC

Hence ABCD is a trapezium.

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