2008 AMC 12B 第 9 题

The radius through the arc midpoint $C$ perpendicularly bisects chord $AB,$ meeting it at $D$ with $AD = 3$

Find the center-to-chord distance $OD,$ then $DC = 5 - OD,$ and apply the Pythagorean theorem to $\triangle ADC$

Let $O$ be the center and $D$ the point where $\overline{OC}$ meets $\overline{AB}.$ Since $C$ is the midpoint of arc $AB,$ $\overline{OC}$ is the perpendicular bisector of the chord, so $AD = 3.$

In right triangle $ADO,$ $OD = \sqrt{5^2 - 3^2} = 4,$ so $DC = OC - OD = 5 - 4 = 1.$

Then in right triangle $ADC,$ $AC = \sqrt{AD^2 + DC^2}$ $= \sqrt{3^2 + 1^2}$ $= \sqrt{10}.$

Thus, the correct answer is A.