1. **State the problem:** We have an equilateral triangle inscribed in a circle with radius $8$ inches. We need to find the area inside the circle but outside the triangle, rounded to the nearest tenth. 2. **Formulas and important rules:** - Area of a circle: $$A_{circle} = \pi r^2$$ - For an equilateral triangle inscribed in a circle, the radius $r$ relates to the side length $s$ by: $$r = \frac{s}{\sqrt{3}}$$ - Area of an equilateral triangle: $$A_{triangle} = \frac{\sqrt{3}}{4} s^2$$ 3. **Find the side length $s$ of the triangle:** Given $$r = 8$$, $$8 = \frac{s}{\sqrt{3}} \implies s = 8 \sqrt{3}$$ 4. **Calculate the area of the triangle:** $$A_{triangle} = \frac{\sqrt{3}}{4} (8 \sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 64 \times 3 = \frac{\sqrt{3}}{4} \times 192$$ Simplify: $$A_{triangle} = 48 \sqrt{3}$$ 5. **Calculate the area of the circle:** $$A_{circle} = \pi \times 8^2 = 64 \pi$$ 6. **Find the area outside the triangle but inside the circle:** $$A = A_{circle} - A_{triangle} = 64 \pi - 48 \sqrt{3}$$ 7. **Numerical approximation:** $$64 \pi \approx 64 \times 3.1416 = 201.1$$ $$48 \sqrt{3} \approx 48 \times 1.732 = 83.1$$ 8. **Final answer:** $$201.1 - 83.1 = 118.0$$ The area of the region inside the circle but outside the triangle is approximately **118.0 square inches**.