11. A car's rear windshield wiper rotates \( 125^{\circ} \). The total length of the wiper mechanism is 25 inche and the length of the wiper blade is 14 inches. Find the area wiped by the wiper blade. (Hint: You need to subtract two areas, to account for the part of the sector not wiped by the blade, near the base of the mechanism.)

We model the area wiped by the wiper as the difference between two sectors. The full wiper mechanism has a length (radius) of 25 inches, but the wiper blade is only 14 inches long. Thus, the region actually wiped is the area of the sector of a circle with radius 25 inches minus the area of the sector (near the base) that is not reached by the blade. This inner sector has a radius of \[ 25 - 14 = 11 \text{ inches}. \] Both sectors have the same central angle of \( 125^\circ \). 1. The area of a sector of a circle with radius \( r \) and central angle \( \theta \) (in degrees) is given by \[ \text{Area} = \frac{\theta}{360^\circ} \pi r^2. \] 2. For the outer sector (radius 25 inches) the area is \[ A_{\text{outer}} = \frac{125}{360} \pi (25)^2 = \frac{125}{360} \pi (625). \] 3. For the inner sector (radius 11 inches) the area is \[ A_{\text{inner}} = \frac{125}{360} \pi (11)^2 = \frac{125}{360} \pi (121). \] 4. The area wiped by the blade is the difference between these two sectors: \[ A = A_{\text{outer}} - A_{\text{inner}} = \frac{125}{360} \pi \left(625 - 121\right). \] Simplify the expression inside the parentheses: \[ 625 - 121 = 504. \] Thus, \[ A = \frac{125}{360} \pi \cdot 504. \] 5. To simplify, note that \[ \frac{504}{360} = \frac{504 \div 72}{360 \div 72} = \frac{7}{5}. \] Therefore, \[ A = \frac{125}{1} \pi \cdot \frac{7}{5} = \frac{125 \times 7}{5} \pi. \] Since \[ \frac{125}{5} = 25, \] we obtain: \[ A = 25 \times 7 \pi = 175 \pi. \] Thus, the area wiped by the wiper blade is \[ \boxed{175\pi \text{ square inches}}. \]