Question

plz answer fast and right​

Answer 1

${}{ \huge{ \underline{ \purple{Solution - }}}}$

It is given that,

Area(∆DRC)=Area(∆DPC)

As ∆DRC and ∆DPC lies 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, ∆DBC and ∆ADC 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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Answer 2

$\Huge{\underline{\boxed{\bf{\green{Answer\::}}}}}$

• $\large{\underline{\underline{\textbf{Given\::}}}}$

● Area (∆ DRC) = Area (∆ DPC)

● Area (∆ BDP) = Area (∆ ARC)

• $\large{\underline{\underline{\textbf{To\:prove\::}}}}$

● Both quadrilaterals ABCD and DCPR are trapeziums.

• $\large{\underline{\underline{\textbf{Solution\::}}}}$

Here it is given that Area (∆ DRC) = Area (∆ DPC).

So, as ∆ DRC and ∆ DPC lies on the same base which is DC and have equal areas, therefore they must lie between the same parallel lines.

DC || RP

DCPR is a trapezium. It is also given that :

Area of (∆ BDP) = Area (∆ ARC)

Area of (∆ BDP) - (∆ DPC) = Area (∆ ARC) - (∆ DRC)

Area (∆ BDC) = (∆ ADC)

Area (∆ BDP) = (∆ ARC)

Area (∆ BDP) - Area (∆ DPC) = Area (∆ ARC) - Area (∆ DRC)

Area (∆ BDC) = Area (∆ ADC)

So, ∆ BDC and ∆ ADC are on same base which is CD and also have equal areas, they must be lie between the same parallel lines.

$\huge{\boxed{\bf{\red{AB\:||\:CD}}}}$

Therefore,

ABCD is a trapezium.