Area of the given rhombus = 498.83 ft²
Solution:
Given figure is a rhombus ABCD.
BE = 12 ft and ∠BAE = 30°
Property of rhombus:
Diagonals bisect each other at right angles.
In ΔAEB, ∠BAE = 30°, ∠AEB = 90° and BE = 12 ft
![$\sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}}](/tpl/images/0513/9769/2e459.png)
![$\sin30^\circ=\frac{\text{BE}}{\text{AB}}](/tpl/images/0513/9769/f3bf4.png)
![$\frac{1}{2} =\frac{\text{12}}{\text{AB}}](/tpl/images/0513/9769/8bce9.png)
Do cross multiplication, we get
AB = 24 ft
Using Pythagoras theorem,
![\text {Adjacent}^{2}+\text{Opposite}^{2}=\text{ Hypotenuse }^{2}](/tpl/images/0513/9769/ba76e.png)
![AE^2+BE^2=AB^2](/tpl/images/0513/9769/d7ba4.png)
![AE^2+12^2=24^2](/tpl/images/0513/9769/873cd.png)
![AE^2+144=576](/tpl/images/0513/9769/41688.png)
![AE^2=432](/tpl/images/0513/9769/c699a.png)
Taking square root on both sides, we get
ft
AC = ![2\times 12\sqrt{3}=24\sqrt {3}](/tpl/images/0513/9769/2631a.png)
Area of the rhombus = ![\frac{1}{2}\times d_1 \times d_2](/tpl/images/0513/9769/0b3e2.png)
                  ![$=\frac{1}{2}\times 24 \times 24\sqrt3](/tpl/images/0513/9769/67ab4.png)
                  ![=288\sqrt{3}](/tpl/images/0513/9769/d8af4.png)
                  = 498.83 ft²
Area of the given rhombus = 498.83 ft²