Question

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersects CD at M. If ar (DMB) = 7 cm², find the area of parallelogram ABCD.

Answer 1

Consider △ ADM and △ PCM

From the figure we know that ∠ ADM and ∠ PCM are alternate angles

∠ ADM = ∠ PCM

We know that AD = BC = CP

It can be written as

AD = CP

∠ AMD and ∠ PMC are vertically opposite angles

∠ AMD = ∠ PMC

By ASA congruence criterion

△ ADM ≅ △ PCM

So we get

Area of △ ADM = Area of △ PCM

DM = CM (c. p. c. t)

We know that BM is the median of △ BDC

So we get

Area of △ DMB = Area of △ CMB

We get

Area of △ BDC = 2 (Area of △ DMB)

By substituting the value

Area of △ BDC = 2 × 7

By multiplication

Area of △ BDC = 14 cm²

We know that

Area of parallelogram ABCD = 2 (Area of △ BDC)

So we get

Area of parallelogram ABCD = 2 × 14

By multiplication

Area of parallelogram ABCD = 28 cm²

Therefore, area of parallelogram ABCD is 28 cm².