3. A street with two lanes, each 12 ft wide, goes through a semicircular tunnel with radius 15 ft . How high is the tunnel at the edge of each lane? 4. A window is to be constructed as shown, with its upper boundary the arc of a circle having radius 4 ft and center at the midpoint of base AD. If the vertical side is to be \( 3 / 4 \) as long as the base, find the dimensions (vertical side and base) of this window.

To determine the dimensions of the window, we'll analyze the given information step by step. ### Problem 4 Recap: - **Upper Boundary:** Arc of a circle with radius \( 4 \) ft. - **Center of Circle:** Midpoint of base \( AD \). - **Vertical Side:** \( \frac{3}{4} \) as long as the base. ### Step-by-Step Solution: 1. **Define Variables:** - Let \( b \) = length of the base \( AD \). - Let \( v \) = length of the vertical sides. Given that \( v = \frac{3}{4}b \). 2. **Coordinate System Setup:** - Place the midpoint of \( AD \) at the origin \( O(0, 0) \). - Thus, points \( A \) and \( D \) are at \( \left( -\frac{b}{2}, 0 \right) \) and \( \left( \frac{b}{2}, 0 \right) \), respectively. - The circle has the equation \( x^2 + y^2 = 16 \) (since radius \( r = 4 \) ft). 3. **Finding the Height \( v \):** - The vertical sides extend from \( A \) and \( D \) upwards to the circle. - At \( A \) (\( x = -\frac{b}{2} \)), the intersection with the circle occurs at: \[ y = \sqrt{16 - \left( \frac{b}{2} \right)^2} \] - Therefore, \( v = \sqrt{16 - \frac{b^2}{4}} \). 4. **Establishing the Relationship:** - Given \( v = \frac{3}{4}b \), set up the equation: \[ \frac{3}{4}b = \sqrt{16 - \frac{b^2}{4}} \] 5. **Solving for \( b \):** - Square both sides to eliminate the square root: \[ \left( \frac{3}{4}b \right)^2 = 16 - \frac{b^2}{4} \] \[ \frac{9}{16}b^2 = 16 - \frac{b^2}{4} \] - Multiply all terms by \( 16 \) to eliminate denominators: \[ 9b^2 = 256 - 4b^2 \] - Combine like terms: \[ 13b^2 = 256 \] \[ b^2 = \frac{256}{13} \] \[ b = \frac{16\sqrt{13}}{13} \text{ ft} \] 6. **Calculating \( v \):** - Using \( v = \frac{3}{4}b \): \[ v = \frac{3}{4} \times \frac{16\sqrt{13}}{13} = \frac{12\sqrt{13}}{13} \text{ ft} \] ### Final Dimensions: - **Base (\( AD \)):** \( \frac{16\sqrt{13}}{13} \) feet (approximately **4.43 ft**). - **Vertical Sides:** \( \frac{12\sqrt{13}}{13} \) feet (approximately **3.33 ft**). ### Summary: - **Base:** \( \boxed{\dfrac{16\,\sqrt{13}}{13}\ \text{feet}} \) - **Vertical Sides:** \( \boxed{\dfrac{12\,\sqrt{13}}{13}\ \text{feet}} \) Alternatively, in decimal form: - **Base:** Approximately **4.43 ft** - **Vertical Sides:** Approximately **3.33 ft**