8. The diagram below shows a quadrilateral $\mathrm{ABCD}$ where $\mathrm{AB}=10, \mathrm{BC}=6, \mathrm{CD}=8$ and $\mathrm{DA}=2$. The diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at the point $\mathrm{O}$ and that $\angle \mathrm{COB}=$ $45^{\circ}$. Find the area of the quadrilateral $\mathrm{ABCD}$.
(A) 28
(B) 29
(C) 30
(D) 31
(E) 32
8 .
(D)

Area of quadrilateral $=\frac{1}{2}(x u+y u+y v+x v) \sin 45^{\circ}$.
Using Cosine Rule, we have
$$
\begin{aligned}
x^{2}+u^{2}+2 x u \cos 45^{\circ} & =10^{2} \\
u^{2}+y^{2}-2 u y \cos 45^{\circ} & =6^{2} \\
y^{2}+v^{2}+2 y v \cos 45^{0} & =8^{2} \\
x^{2}+v^{2}-2 x v \cos 45^{\circ} & =2^{2}
\end{aligned}
$$
(1) $-(2)+(3)-(4)$ :
$$
2(x u+y u+y v+v x) \cos 45^{0}=10^{2}-6^{2}+8^{2}-2^{2}
$$

Hence area of quadrilateral $=\frac{100-36+64-4}{4}=31$.