Rhombus

The formula for the area of a rhombus is $a^2 \sin \theta$ where $a$ is the side length and $\theta$ is the included angle. A rhombus has equal sides so the length of one side is $\dfrac{8}{4}=2$ . So the area is $2^2 \sin 60=4 \sin 60$ . By using a calculator the desired answer is $(4 \sin 60)^2 = \boxed{12}$

Perimeter is 8 so each side is 2.

The given rhombus can be viewed as the join of two equilateral triangles. Hence, area of rhombus: $b= 2 \times (\frac{\sqrt3(side)^2}{4})$

$b=2\sqrt3$

$\displaystyle \Rightarrow b^2 = \boxed{12}$

Alternatively, we can view two adjacent sides of the rhombus as vectors $\vec{p}$ and $\vec{q}$ of length 2 each, making an angle $60^{\circ}$ or $120^{\circ}$ with each other.

So, area of rhombus:

$b = \vec{p} \times \vec{q}$

$\displaystyle \Rightarrow b = |\vec{p} ||\vec{q}| \sin(60^{\circ} \text{ or } 120^{\circ})$

$\displaystyle \Rightarrow b = (2)(2)(\frac{\sqrt3}{2}) = 2\sqrt3$

$\displaystyle \Rightarrow b^2 = \boxed{12}$

Perimeter of the rhombus is 8 so each side is 2. We are given than one angle is 60 degrees. Adjacent angles in a rhombus are supplementary, i.e sum to 180 degrees and opposite angles are equal. Therefore the angles in the rhombus are 60,120,60,120 degrees.

In a rhombus, diagonals bisect at 90 degrees and each diagonal also
bisects the interior angle. So consider one triangle formed by the two diagonals and a side.

The triangle has angles 30,60 and 90 degrees. $\sin(30)=\frac{1}{2}$ therefore one diagonal is $2 \times 1 = 2$ . Also, $\sin(60)=\frac{\sqrt{3}}{2}$ therefore the other diagonal is $2 \times \sqrt{3} = 2 \sqrt{3}$ .

Area of a rhombus is half of the product of the diagonals. So area of the rhombus = $\frac{1}{2} \times 2 \times 2 \sqrt{3} = 2\sqrt{3}$ .

We need to find the square of the area = $(2 \sqrt{3})^2=\boxed{12}$ .