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`b)`Xét `ΔMAO` vuông tại `A:`

$$MA = \sqrt{OM^2 - OA^2} = \sqrt{(2R)^2 - R^2} = R\sqrt{3}$$ 

Xét `ΔMAO` vuông tại `A` có `AI` là đường cao:

$$OI \cdot OM = OA^2 \implies OI = \frac{OA^2}{OM} = \frac{R^2}{2R} = \frac{R}{2}$$

$$MI = OM - OI = 2R - \frac{R}{2} = \frac{3R}{2}$$

$$AI = \sqrt{OA^2 - OI^2} = \sqrt{R^2 - \left(\frac{R}{2}\right)^2} = \frac{R\sqrt{3}}{2} \implies AC = 2AI = R\sqrt{3}$$

Xét `ΔABC` vuông `C`:

$$BC = \sqrt{AB^2 - AC^2} = \sqrt{(2R)^2 - (R\sqrt{3})^2} = R$$

Xét `ΔMBC` có: `OM //// BC`

$$\frac{MK}{MB} = \frac{MI}{MO} = \frac{\frac{3R}{2}}{2R} = \frac{3}{4}$$

$${Ta} {có:}\\ S_{\Delta MIK} = \frac{1}{2} \cdot MI \cdot d(K, MI)$$

$$S_{\Delta MIK} = \frac{1}{2} \cdot MI \cdot d(K, MI)$$

Xét `ΔMIK` $\sim$ `ΔMBC`

$$\frac{S_{\Delta MIK}}{S_{\Delta MBC}} = \left(\frac{MI}{MC}\right)^2 = \left(\frac{MI}{MA}\right)^2 = \left(\frac{\frac{3R}{2}}{R\sqrt{3}}\right)^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$$

Ta có:

$$S_{\Delta ABC} = \frac{1}{2} \cdot AC \cdot BC = \frac{1}{2} \cdot R\sqrt{3} \cdot R = \frac{R^2\sqrt{3}}{2}$$

$$S_{\Delta MAC} = \frac{1}{2} \cdot MI \cdot AC = \frac{1}{2} \cdot \frac{3R}{2} \cdot R\sqrt{3} = \frac{3R^2\sqrt{3}}{4}$$

$$S_{\Delta MBC} = \frac{1}{2} \cdot S_{\Delta MAC} = \frac{3R^2\sqrt{3}}{8}$$ ( Do `OM////BC` và `I` là trung điểm `AC`)

$$S_{\Delta MIK} = \frac{3}{4} \cdot S_{\Delta MBC} = \frac{3}{4} \cdot \frac{3R^2\sqrt{3}}{8} = \frac{9\sqrt{3}}{32}R^2$$