In a rhombus ABCD, AC = 3 units and angle ABC = 120°. What is the area

Bunuel please help with this. I'm getting 4.5 root3
Got it buy 30-60-90 triangle. Where am i wrong?

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Can someone please help on how to get from 3/2 (60°) to square root 3/2 (30°)? For the 30-60-90 Triangle

Can someone please help on how to get from 3/2 (60°) to square root 3/2 (30°)? For the 30-60-90 Triangle[/quote]

30-60-90 Right angle Triangle Rule: Sides will be in Ratio. x:x√3:2.

Now as angle P is 90 degree and in triangle ABP angle B is 60 degree and angle A is 30 degrees, We can use the Ratios and Find out that side PB is √3/2.

area=d_1*d_2/2
ac=diagonal_1=3
bd=diagonal_2=?

abc=120, then opposite adc=120
dcb=(360-240)/2=60, opposite dab=60

perpendicular from d_1 to abc forms a right triangle
30-60-90 x:x√3:2x
x√3 is half d_1, x√3=3/2, x=3/2√3=√3/2
x is half d_2, so twice x is d_2=2√3/2=√3

area=3*√3/2

Ans (C)

The main trick here is to know that 3/2√3=√3/2
The rest is a pretty basic 30-60-90 triangle.

3/2√3:3/2:3/√3

2*(3/2√3*3/2) = 9/2√3 = 3√3/2

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

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