# ABCD rectangle, AB = 3, point K belongs to AB, point M belongs to CD, angle KAC = 30 degrees

**ABCD rectangle, AB = 3, point K belongs to AB, point M belongs to CD, angle KAC = 30 degrees, AKCM rhombus, Find AK.**

1. From △ ABC: ∠ABC = 90 ° (since ABCD is a rectangle, and all its corners are straight), ∠BAC (aka ∠KAC) = 30 °, AB = 3 and BC – legs, AC – hypotenuse.

Since BC lies opposite an angle equal to 30 °, then according to the properties of a right-angled triangle:

BC = AC / 2.

By the Pythagorean theorem:

AC² = AB² + BC²;

AC² = 3² + (AC / 2) ²;

AC² = 9 + AC² / 4;

AC² – AC² / 4 = 9;

(4 * AC² – AC²) / 4 = 9;

(3 * AC²) / 4 = 9;

AC² = (4 * 9) / 3 (proportional);

AC² = 36/3;

AC² = 12;

AC = √12;

AC = 2√3.

2. АКСМ – rhombus: AC = 2√3 and KM – diagonals. The diagonals of the rhombus intersect at right angles and the intersection point is halved. AC and KM meet at point O, then:

AO = AC / 2 = 2√3 / 2 = √3.

From △ AKO: ∠AOK = 90 °, ∠KAO (aka ∠KAC) = 30 °, AO = √3 and KO – legs, AK – hypotenuse.

Since KO lies opposite an angle equal to 30 °, then according to the properties of a right-angled triangle:

KO = AK / 2.

By the Pythagorean theorem:

AK² = AO² + KO²;

AK² = (√3) ² + (AK / 2) ²;

AK² = 3 + AK² / 4;

AK² – AK² / 4 = 3;

(4 * AK² – AK²) / 4 = 3;

(3 * AK²) / 4 = 3;

AK² = (4 * 3) / 3;

AK² = 4;

AK = √4;

AK = 2.

Answer: AK = 2.