Question

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

Answer 1

Step-by-step explanation:

Given,

ar(DRC) = ar(DPC) .....(i)

& ar(BDP) = ar(ARC).....(ii)

To Prove,

Quadrilaterals ABCD and DCPR are trapeziums.

Proof:

On subtracting eq (I) from eq(ii),

 ar(△BDP) – ar(△DPC) = ar(△ARC)- ar(△DRC)

 ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC).

Since , these two triangles have the same base DC.

Therefore, they must lying between the same parallel lines.

Thus, AB ∥ CD

So, ABCD is a trapezium.

also,

ar(DRC) = ar(DPC).

Since , these two triangles have the same base DC.

Therefore, they must lying between the same parallel lines.

Thus, DC ∥ PR

Hence, DCPR is a trapezium.

Hope it helps!

Answer 2

A(∆DRC) =A(∆DPC)...... (1)

∆DRC and ∆DPC are on the same base DC and A(∆DRC) =A(∆DPC)

$\therefore$ $\large\tt\purple{DC||RP}$

($\because$ Triangles on the same base (or equal base) and having equal areas lie between the same parallels)

$\large\tt\underline\red{Hence,DCPR\:is\:a\:trapezium}$

and A(∆ARC) = A(∆BDP)...... (2)

subtracting (1) & (2)

A(∆ARC) - A(∆DRC) = A(∆BDP) - A(∆DPC)

A(∆ADC) = A(∆BDC)

∆ADC and ∆BDC are on the same base DC and A(∆ADC) = A(∆BDC)

$\therefore$$\large\tt\purple{AB||DC}$

($\because$ Triangles on the same base (or equal base) and having equal areas lie between the same parallels)

$\large\tt\underline\red{Hence,ABCD\:is\:a\:trapezium}$