2. The lengths of the four sides of a cyclic quadrilateral in sequence are 5, 10, 11, and 14. Then the area of this quadrilateral is ( ).
(A) $78 \frac{1}{2}$
(B) $97 \frac{1}{2}$
(C) 90
(D) 102
2.C.

As shown in Figure 3, let $AB=5$, $BC=10$, $CD=11$, $DA=14$. Since $5^{2}+14^{2}=221=10^{2}+11^{2}$, that is, $AB^{2}+AD^{2}=BC^{2}+CD^{2}$, hence $BD$ is the diameter, thus, $\angle A$ and $\angle C$ are both right angles. Therefore, the area $S$ of quadrilateral $ABCD$ is $S=\frac{1}{2} \times 5 \times 14+\frac{1}{2} \times 10 \times 11=90$.