The figure below shows a parallelogram ABCD with BD=4 m and DC=5 m. Calculate the area of parallelogram ABCD.

Solution:

ΔADB is congruent to ΔDBC as the diagonal of a parallelogram divides it into two congruent triangles.

Now in right-angled triangle DBC:

∴ (BC)² + (BD)² = (DC)²          (Pythagoras theorem)

∴ (BC)² + 16 = 25

∴ (BC)² = 25 - 16 = 9

∴ BC = 3 m

Now, Area of ΔDBC = $\frac{1}{2}$ ​× DB × BC

= $\frac{1}{2}$ ​× 4 × 3

= 6 m²

Since congruent triangles have equal areas 

So, the area of ΔADB is also 6 m²

∴ Area of parallelogram ABCD = Area of ΔDBC + Area of ΔADB

= 6 + 6

= 12 m²