Question

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

Answer 1

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Answer 2

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$\huge\mathfrak{Solution:-}$

⏩ ar(∆DRC) = ar(∆DPC), [Given]

But they are on the same base DC.

Therefore,

∆DRC and ∆DPC must lie between the same parallels.

So,

DC || RP

i.e., one pair of opposite sides of quadrilateral DCPR is parallel.

Therefore,

DCPR is a trapezium.

Also,

ar (∆BDP) = ar(∆ARC) → [Given] ....→(i)

and

ar (∆DPC) = ar (∆DRC) → [Given]...→(ii)

Now, on subtracting (ii) from (i), we get,

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

=> ar (∆BDC) = ar (∆ADC)

But they are on the same base DC.

Therefore,

∆BDC and ∆ADC must lie between the same parallels.

So,

AB || DC

i.e., One pair of opposite sides of quadrilateral ABCD is parallel.

Therefore,

ABCD is a trapezium.

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