Find the perimeter of rectangle ABCD, AD = 4√3, BD = 8, and the angle BDA = 30 °.

Given a rectangle ABCD: AD = 4√3, BD = 8 – diagonal, ∠BDA = 30 °. Diagonal BD divides rectangle ABCD into two equal right triangles – △ BAD and △ DCB.

1. Consider △ BAD: ∠BAD = 90 °, ∠BDA = 30 °, AD = 4√3 and AB – legs, BD = 8 – hypotenuse, since it lies opposite the right angle.

Since leg AB lies opposite ∠BDA, equal to 30 °, it is equal to half of the hypotenuse BD:

AB = BD / 2 = 8/2 = 4.

2. The perimeter of a polygon is equal to the sum of the lengths of all its sides.

The perimeter of the rectangle ABCD is:

P = AB + BC + CD + AD.

The opposite sides of the rectangle are pairwise parallel and equal, that is:

AB = CD = 4;

AD = BC = 4√3.

Thus:

P = 4 + 4√3 + 4 + 4√3 = 8 + 8√3 = 8 * (1 + √3).

Answer: P = 8 * (1 + √3).