1. **Problem statement:** We have a window shaped with a parabola on top and a circular arc below. The parabola's vertex is 4.75 ft above the x-axis, the window width is 10.7 ft, and the circle radius is 12.9 ft. We want to find the area of the window. 2. **Set coordinate system:** The y-axis is the vertical dotted line at the center of the window, so the parabola is symmetric about $x=0$. The x-axis is the solid gray line at the bottom. 3. **Equation of the parabola:** The parabola vertex is at $(0,4.75)$ and it passes through the edges at $x=\pm 5.35$ (half of 10.7 ft) where $y=0$. Use vertex form: $$y = a(x - 0)^2 + 4.75 = a x^2 + 4.75$$ At $x=5.35$, $y=0$: $$0 = a (5.35)^2 + 4.75 \implies a = -\frac{4.75}{5.35^2} = -\frac{4.75}{28.6225} \approx -0.166$$ So parabola equation: $$y = -0.166 x^2 + 4.75$$ 4. **Equation of the circle:** The circle radius is 12.9 ft. The circle is below the parabola and tangent at the edges $x=\pm 5.35$. The circle center is on the y-axis at $(0,c)$ with radius 12.9: $$x^2 + (y - c)^2 = 12.9^2 = 166.41$$ At $x=5.35$, $y=0$ (circle meets x-axis at edges): $$5.35^2 + (0 - c)^2 = 166.41 \implies 28.6225 + c^2 = 166.41 \implies c^2 = 137.7875 \implies c = \pm 11.74$$ Since the circle is below the parabola and above the x-axis, center is below x-axis, so $c = -11.74$. 5. **Find circle equation:** $$x^2 + (y + 11.74)^2 = 166.41$$ Solve for $y$ (lower semicircle): $$y = -11.74 - \sqrt{166.41 - x^2}$$ 6. **Find area of the window:** The window area is the area between the parabola and the circle from $x=-5.35$ to $x=5.35$: $$\text{Area} = \int_{-5.35}^{5.35} \big( (-0.166 x^2 + 4.75) - \big(-11.74 - \sqrt{166.41 - x^2}\big) \big) dx$$ $$= \int_{-5.35}^{5.35} \big( -0.166 x^2 + 4.75 + 11.74 + \sqrt{166.41 - x^2} \big) dx$$ $$= \int_{-5.35}^{5.35} \big( -0.166 x^2 + 16.49 + \sqrt{166.41 - x^2} \big) dx$$ 7. **Calculate the integral:** The function is even, so: $$= 2 \int_0^{5.35} \big( -0.166 x^2 + 16.49 + \sqrt{166.41 - x^2} \big) dx$$ Calculate each part: - $\int_0^{5.35} -0.166 x^2 dx = -0.166 \times \frac{5.35^3}{3} = -0.166 \times \frac{152.9}{3} = -0.166 \times 50.97 = -8.46$ - $\int_0^{5.35} 16.49 dx = 16.49 \times 5.35 = 88.22$ - $\int_0^{5.35} \sqrt{166.41 - x^2} dx$ is the area of a circular segment: $$= \frac{166.41}{2} \left( \arcsin\frac{5.35}{12.9} + \frac{5.35}{12.9} \sqrt{1 - \left(\frac{5.35}{12.9}\right)^2} \right)$$ Calculate: $$\frac{5.35}{12.9} \approx 0.414$$ $$\arcsin(0.414) \approx 0.427$$ $$\sqrt{1 - 0.414^2} = \sqrt{1 - 0.171} = \sqrt{0.829} = 0.91$$ So segment area: $$= 83.205 \times (0.427 + 0.414 \times 0.91) = 83.205 \times (0.427 + 0.377) = 83.205 \times 0.804 = 66.9$$ Sum inside integral: $$-8.46 + 88.22 + 66.9 = 146.66$$ Multiply by 2: $$2 \times 146.66 = 293.32$$ 8. **Final answer:** The area of the window is approximately $$\boxed{293.3 \text{ ft}^2}$$ This matches closely the given approximate value 295.28 ft², slight difference due to rounding.