Equation of circle
$$
x^2+y^2-2 x-4 y-20=0
$$

Centre of circle $(1,2)$, radius $=5$






Equation of tangents at $(1,7)$ is
$$
\begin{aligned}
& \text { (1) } x+7 y-(x+1)-2(y+7)-20=0 \\
& \Rightarrow x+7 y-x-1-2 y+7-20=0 \Rightarrow 5 y=35 \\
& \Rightarrow \quad y=7
\end{aligned}
$$

Equation of tangent at $(4,-2)$
$$
\begin{aligned}
& 4 x-2 y-(x+4)-2(y-2)-20=0 \\
& \Rightarrow 4 x-2 y-x-4-2 y+4-20=0 \\
& \Rightarrow 3 x-2 y=20
\end{aligned}
$$

Tangents $y=7$ and $3 x-2 y=20$ intersect at $C$.
So, coordinate of $C(16,7)$
Length of $B C=\sqrt{(16-1)^2+(7-7)^2}$
$$
B C=15
$$

Length of $C D=\sqrt{(16-4)^2+(7+2)^2}=\sqrt{144+81}$
$$
C D=15
$$

So, area of $A B C D=$ area of $\triangle A B C+$ area of $\triangle A C D$
$$
=\frac{1}{2} \times 15 \times 5+\frac{1}{2} \times 15 \times 5=75
$$