Let $A = (F_{n+4}, 0)$, $B = (0, F_{n+2})$ and $C = (F_{n+3}, F_n)$. Notice that
$$AB = \sqrt{F_{n+4}^2 + F_{n+2}^2} = V_{n+2}, \quad BC = \sqrt{F_{n+3}^2 + (F_{n+2} - F_n)^2} = \sqrt{F_{n+3}^2 + F_{n+1}^2} = V_{n+1}$$
and $CA = \sqrt{(F_{n+4} - F_{n+3})^2 + F_n^2} = \sqrt{F_{n+2}^2 + F_n^2} = V_n$. Its area equals $\frac{|D|}{2}$, where
$$\begin{aligned} D &= \begin{vmatrix} F_{n+4} & 0 & 1 \\ 0 & F_{n+2} & 1 \\ F_{n+3} & F_n & 1 \end{vmatrix} = F_{n+4}F_{n+2} - F_{n+3}F_{n+2} - F_nF_{n+4} \\ &= F_{n+4}(F_{n+2} - F_n) - F_{n+3}F_{n+2} = F_{n+4}F_{n+1} - F_{n+3}F_{n+2} \\ &= \begin{vmatrix} F_{n+1} & F_{n+2} \\ F_{n+3} & F_{n+4} \end{vmatrix} \end{aligned}$$
Let $A_{n+1} = \begin{pmatrix} F_{n+1} & F_{n+2} \\ F_{n+3} & F_{n+4} \end{pmatrix}$. An easy induction proves that $A_n = \begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix}$.
$(\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix})^n$. Indeed, $A_0 = \begin{pmatrix} F_0 & F_1 \\ F_2 & F_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^0$ and
$$\begin{aligned} A_{n+1} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} &= \begin{pmatrix} F_{n+1} & F_{n+2} \\ F_{n+3} & F_{n+4} \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1} + F_{n+2} \\ F_{n+4} & F_{n+3} + F_{n+4} \end{pmatrix} \\ &= \begin{pmatrix} F_{n+2} & F_{n+3} \\ F_{n+4} & F_{n+5} \end{pmatrix} = A_{n+2} \end{aligned}$$
Hence
$$D = |A_{n+1}| = \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} \cdot \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}^{n+1} = (0 \cdot 2 - 1 \cdot 1) \cdot (0 \cdot 1 - 1 \cdot 1)^{n+1} = (-1)^{n+2}$$
and area $ABC = \frac{|D|}{2} = \frac{1}{2}$.