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.
4. (C)

First, calculate the circumradius $R$ of $\triangle A B C$, we have
$$
R=\frac{A C}{2 \sin B}=\frac{\sqrt{2}}{2 \sin 45^{\circ}}=1 .
$$

Now compare (A) with (C), of course, (C) has a larger area. 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 areas of (B) and (D):
$$
\frac{1}{2} \cdot \sqrt{3} \cdot \sqrt{2} \sin 75^{\circ}<\frac{1}{2} \cdot 2.5 \cdot 2.5 \cdot \sin 90^{\circ} .
$$

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