Dillesh4096
In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area of ABCD? (in unit\(^2\))
A. \(3\sqrt{3}/8\)
B. \(3\sqrt{3}/4\)
C. \(3\sqrt{3}/2\)
D. \(3\sqrt{3}\)
E. \(6\sqrt{3}\)
Solution:One can draw the two diagonals of the rhombus (AC and BD) and divide the rhombus into 4 congruent right triangles. The diagonals of the rhombus are perpendicular, and they bisect each other. Furthermore, they bisect the angles of the rhombus. Since the rhombus has angles that measure 120 and 60 degrees, we see that when they are bisected, they become 60 and 30 degrees, respectively. Therefore, each of the 4 triangles is a 30-60-90 triangle. Since AC = 3 and angle ABC = 120 degrees, AC must be the longer diagonal, Therefore, when AC is bisected, it must be the longer leg of the 30-60-90 triangle, i.e., opposite the 60 degree angle. In other words, the longer leg has length 3/2 and the shorter leg must be (3/2)/√3 = √3/2. Since the area of a right triangle is half the product of its legs, the area of one 30-60-90 triangle is 1/2 x 3/2 x √3/2 = 3√3/8. Since the rhombus has 4 such triangles, the area of the rhombus is 4 x 3√3/8 = 3√3/2.
Answer: C