13.60 Among the four figures below, the one with the largest area is
(A) $\triangle A B C: \angle A=60^{\circ}, \angle B=45^{\circ}, A C=\sqrt{2}$.
(B) Trapezoid: the lengths of the two diagonals are $\sqrt{2}$ and $\sqrt{3}$, and the angle between them is $75^{\circ}$.
(C) Circle: radius is 1.
(D) Square: the length of the diagonal is 2.5.
(China High School Mathematics Competition, 1981)
[Solution] Directly calculating the area of $\triangle A B C$ in $(A)$ is quite cumbersome. Instead, we can first calculate the circumradius $R$ of $\triangle A B C$, which is
$$
R=\frac{A C}{2 \sin B}=\frac{\sqrt{2}}{2 \sin 45^{\circ}}=1 .
$$

Now, comparing $(A)$ with $(C)$, it is clear that the area of $(C)$ is larger. Using the formula for the area of a quadrilateral,
$$
S_{\text {quadrilateral }}=\frac{1}{2} l_{1} l_{2} \sin \varphi,
$$
(where $l_{1}, l_{2}$ are the diagonals, and $\varphi$ is the angle between the diagonals), we can directly compare $(B)$ and $(D)$.
$$
\begin{array}{ll}
\because & \frac{1}{2} \cdot \sqrt{3} \cdot \sqrt{2} \cdot \sin 75^{\circ}<\frac{1}{2} \cdot 2.5 \cdot 2.5 \cdot \sin 90^{\circ} . \\
\therefore & (D) \text { is greater than }(B) .
\end{array}
$$

Finally, comparing $(C)$ with $(D)$, it is easy to see that $(C)$ is the largest. Therefore, the answer is $(C)$.