$21 \cdot 4$ A square with a side length of 8 needs to be completely covered by rectangles whose diagonal length is $\sqrt{65}$. The minimum number of such rectangles required is
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] As shown in the figure, divide the square into two $4 \times 7$ and one $1 \times 8$ rectangles, whose diagonal lengths are
$$
\sqrt{4^{2}+7^{2}}=\sqrt{65}, \sqrt{1^{2}+8^{2}}=\sqrt{65} \text {. }
$$

At this point, they exactly cover the entire square.
The diagonal length of a rectangle with a diagonal of $\sqrt{65}$ cannot cover two points with a distance greater than $\sqrt{65}$. Thus, if two rectangles with a diagonal length of $\sqrt{65}$ are used to cover an $8 \times 8$ square, then $A C$ or $B, D$ cannot be covered simultaneously by one rectangle, meaning that each can cover at most two of the four points $A, B, C, D$.

Assume one covers points $A, D$, and the other covers points $B, C$. However, in this case, the midpoint $E$ of $D C$ cannot be covered by them, because $\sqrt{8^{2}+4^{2}}=\sqrt{80}>\sqrt{65}$. In summary, at least three such rectangles are needed.
Therefore, the answer is $(B)$.