4. Among the following four figures, which one has the largest area?
(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
(C)
4.【Analysis and Solution】First, calculate the circumradius $R$ of $\triangle ABC$, we have $R=\frac{AC}{2 \sin B}=\frac{\sqrt{2}}{2 \sin 45^{\circ}}=1$.
Now compare $(A)$ with $(C)$, of course, the area of $(C)$ is larger. Then, by the formula for the area of a convex quadrilateral, which is half the product of the diagonals and the sine of the angle between them, we can compare the area of (B) with that of (C)
$$
\frac{1}{2} \cdot \sqrt{3} \cdot \sqrt{2} \sin 75^{\circ}<\frac{1}{2} \cdot 2.5 \cdot 2.5 \sin 90^{\circ} .
$$

So (C) is greater than $(\mathrm{B})$, then compare $(\mathrm{C})$ with (D).
The area of (C) is $\pi$, and the area of (D) is $\frac{6.25}{2}=3.125$, so the area of $(\mathrm{C})$ is the largest.