# 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. 