### Theory:

A special quadrilateral is nothing but a rhombus. To find the area of a rhombus, we can use the same triangulation method as used in finding the area of a general quadrilateral.

A rhombus after triangulation is given below. From the figure given above, we can come to the following inferences.

$$ABCD$$ is a rhombus with diagonals $$AC$$ and $$BD$$.

Let $$AC$$ be $$d_1$$ and $$BD$$ be $$d_2$$. The diagonals $$d_1$$ and $$d_2$$ intersect at $$O$$.

Also, as per the properties of a rhombus, the diagonals bisect each other.

$$\text{Area of the rhombus}$$ $$ABCD =$$ $$(\text{Area of}$$ $$\triangle ABD))$$ $$+$$ $$(\text{Area of}$$ $$\triangle BCD)$$

$$=$$ $$\frac{1}{2} \times BD \times OA$$ $$+$$ $$\frac{1}{2} \times BD \times OC$$

$$=$$ $$\frac{1}{2} \times BD \times (OA + OC)$$

$$=$$ $$\frac{1}{2} \times BD \times AC$$

[Since $$AC = OA + OC$$]

$$=$$ $$\frac{1}{2} \times d_1 \times d_2$$

[Since we have assumed that $$AC = d_1$$, and $$BD = d_2$$]

Therefore, $$\text{area of a rhombus} = \frac{1}{2} \times d_1 \times d_2$$ square units or the area of the rhombus is half the product of its diagonals.