Calculate the areas of the parallelograms shown below:

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

Let length of AE = x cm and CE = y cm.

In Δ AEC:

∠CAE = 60° and ∠CEA = 90°
⇒ ∠ACE = 30° (∵ ∠C + E + ∠A = 180° )

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ x:y:2 = 1:√3:2
⇒ x = 1 cm and y = √3 cm

OR
We can consider Δ AEC as a half of equilateral Δ ACL.

Since, Δ ACL is an equilateral triangle.
We get,

AL = AC
⇒ (x + x) = 2 cm
⇒ 2x = 2 cm
⇒ x = 1 cm

Using the Pythagoras theorem in Δ AEC ,
we get,

AC² = AE² + EC²
⇒ (2)² = (1)² + (y)²
⇒ 4 - 1 = (y)²
⇒ (3) = (y)²
⇒ y = √3 cm

⇒ CE = √3 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 × √3 cm² = 4√3 cm²

Area = 4√3 cm²