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.


Answers (1)
M manish painkra

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.