Calculate the areas of the parallelograms shown below:

Let us consider a parallelogram ABDC.
Let us draw a perpendicular on base AB from C say, CE

Let length of AE = x cm.

In Δ AEC:

∠CAE = 45° and ∠CEA = 90°
⇒ ∠ACE = 45° (∵ ∠C + ∠E + ∠A = 180° )

We know that sides opposite to equal angles are also equal. So the perpendicular side opposite to the angle 45° is also equal to x cm
∴ AE = CE = x cm.

Using the Pythagoras theorem in Δ AEC ,
we get,

AC² = AE² + EC²
⇒ (2)² = (x)² + (x)²
⇒ (2)² = 2(x)²
⇒ (2) = (x)²
⇒ x = √2 cm

OR

We know that sides of any triangle of angles 45°, 45° and 90°
are in the ratio 1:1:√2.

⇒ x:x:2 = 1:1: √2
⇒ x = √2 cm

⇒ CE = √2 cm and also line segment CE is a height of parallelogram ABDC.
Also base AB = 4 cm.

Thus area of parallelogram = base × height

∴ area (ABDC) = 4 × √2 cm² = 4√2 cm²

Area = 4√2 cm²