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For any triangle, inradius r=Asr = \dfrac{A}{s}, where AA is area and ss is semiperimeter. For a right triangle, a convenient formula is r=a+b−c2r = \dfrac{a + b - c}{2}, where aa and bb are legs and cc is the hypotenuse. Compute cc via Pythagoras: c=62+82=36+64=10c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10. Then r=(6+8−10)/2=4/2=2r = (6 + 8 - 10)/2 = 4/2 = 2. Option B is correct. Alternatively, compute area A=12⋅6⋅8=24A = \frac{1}{2}\cdot 6 \cdot 8 = 24 and perimeter P=6+8+10=24P = 6 + 8 + 10 = 24, so semiperimeter s=12s = 12. Then r=A/s=24/12=2r = A/s = 24/12 = 2. This problem targets GRE geometry fundamentals: right triangles, Pythagorean triples (6-8-10), and circle–triangle relationships. Common errors include mixing up inradius with circumradius (for a right triangle, circumradius =c/2=5= c/2 = 5, not asked) or incorrectly using whole perimeter instead of semiperimeter when applying r=A/sr = A/s. Knowing multiple equivalent formulas and confirming results by two methods helps avoid computational mistakes under time pressure.