Let $ABCD$ be a parallelogram. The lengths of the side $AD$ and the diagonal $AC$ a $10$ cm and $20$ cm, respectively. If the angle $\angle ADC$ is equal to $30^\circ$ then the area of the parallelogram, in sq. cm, is

Given Information:

ABCD is a parallelogram

AD = 10 cm (side length)

AC = 20 cm (diagonal length)

∠ADC = 30°

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To find the area of a parallelogram, we need base × height. The challenge is finding the height.

We drop a perpendicular from point A to side DC, and call this point E.

AE becomes the height of our parallelogram because it's perpendicular to the base DC.

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Triangle DAE has:

∠ADE = 30° (given as ∠ADC)

∠AED = 90° (since AE ⊥ DC)

∠DAE = 60° (since angles in a triangle sum to 180°)

This is a 30-60-90 triangle.

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In a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2

Side opposite 30° : Side opposite 60° : Hypotenuse = 1 : √3 : 2

In triangle DAE:

AD = 10 cm (hypotenuse)

AE = side opposite 30° = $\tfrac{1}{2} \times 10 = 5$ cm

DE = side opposite 60° = $\tfrac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}$ cm

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In right triangle AEC:

AC = 20 cm (hypotenuse)

AE = 5 cm (one leg)

EC = ? (other leg)

Using $AC^2 = AE^2 + EC^2$:

$20^2 = 5^2 + EC^2$

$400 = 25 + EC^2$

$EC^2 = 375$

$EC = \sqrt{375} = \sqrt{25 \times 15} = 5\sqrt{15}$ cm

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Area of parallelogram = Base × Height

Height = AE = 5 cm

Base = DC = DE + EC = $5\sqrt{3} + 5\sqrt{15}$ cm

Area = $5 \times (5\sqrt{3} + 5\sqrt{15}) = 25(\sqrt{3} + \sqrt{15})$ cm²

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Key Takeaways:

For parallelogram area: We always look for ways to create a right triangle to find the height

30-60-90 triangles: We remember the ratio 1 : √3 : 2 - it's a huge time-saver

Perpendicular drop: This technique converts complex parallelogram problems into simpler right triangle problems

Final Answer: $25(\sqrt{3} + \sqrt{15})$ cm²