$20 \cdot 15$ In a circle with a radius of 5 units, $C D$ and $A B$ are perpendicular diameters, a chord $C H$ intersects $A B$ at $K$, and is 8 units long, while the diameter $A B$ is divided into two segments, whose lengths are
(A) $1.25 ; 8.75$.
(B) $2.75 ; 7.25$.
(C) $2 ; 8$.
(D) $4 ; 6$.
(E) None of the above.
(3rd American High School Mathematics Examination, 1952)
[Solution] As shown in the figure, let $A K=x$, then $K B=10-x, O K=5-x$, and
$$
\begin{array}{l}
C K=\sqrt{C O^{2}+O K^{2}}=\sqrt{5^{2}+(5-x)^{2}}, \\
H K=8-C K=8-\sqrt{5^{2}+(5-x)^{2}} .
\end{array}
$$

By the intersecting chords theorem, we know $C K \cdot H K=A K \cdot B K$,
Substituting the above expressions and simplifying, we get
$$
16 x^{2}-160 x+175=0,
$$

Solving this, we get $x=\frac{5}{4}$ or $\frac{35}{4}$ (since $0<x<5$, we discard the latter). Therefore, the answer is $(A)$.