Question

In the adjoining figure, AP \\RD and ABCD is a parallelogram, prove that area(PQRDB)= area(PQRCA).​

Answer 1

Answer:

Draw PR

As ABCD is a parallelogram, AB = CD ------------------> (I)

Given, AP || RD ⇒ AP || RC

Now, AP + RC = AB + BP + RC = BP + CD + RC = BP + RD

Let AP + RC = BP + RD = a

The trapeziums APRC and DRPB have the same height

Let the common height be h

Area of trapezium = (1/2) x Height x (Sum of parallel sides)

Now,

         ar(PQRDB)

              = ar(ΔPQR) + ar(DRPB)

              = ar(ΔPQR) + $\frac{1}{2}$ x h x (BP + RD)

              = ar(ΔPQR) + $\frac{1}{2}$ x h x a

And,

         ar(PQRCA)

              = ar(ΔPQR) + ar(APRC)

              = ar(ΔPQR) + $\frac{1}{2}$ x h x (AP + RC)

              = ar(ΔPQR) + $\frac{1}{2}$ x h x a

∵ ar(PQRDB) = ar(ΔPQR) + $\frac{1}{2}$ x h x a and ar(PQRCA) = ar(ΔPQR) + $\frac{1}{2}$ x h x a,

                   ar(PQRDB) = ar(PQRCB)

      Hence, proved

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