1. **Problem Statement:** Given a circle with radius $r=5$ ft and a chord $AB$ whose perpendicular distance from the center $O$ is $d=3$ ft. Points $A$ and $B$ lie on the circle. We need to find: (a) The area of triangle $OAB$. (b) The area enclosed by $OA$, $OB$, and the arc $AB$. 2. **Formulas and Important Rules:** - The length of the chord $AB$ can be found using the right triangle formed by the radius, the distance from the center to the chord, and half the chord length: $$AB = 2\sqrt{r^2 - d^2}$$ - Area of triangle $OAB$ is: $$\text{Area}_{\triangle OAB} = \frac{1}{2} \times OA \times OB \times \sin(\theta)$$ where $\theta$ is the central angle $AOB$. - The central angle $\theta$ can be found using: $$\cos(\theta/2) = \frac{d}{r}$$ - Area of the sector $OAB$ is: $$\text{Area}_{\text{sector}} = \frac{1}{2} r^2 \theta$$ - Area enclosed by $OA$, $OB$, and arc $AB$ (segment area) is: $$\text{Area}_{\text{segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle OAB}$$ 3. **Calculations:** - Calculate half chord length: $$\frac{AB}{2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$ - So, chord length: $$AB = 2 \times 4 = 8$$ - Calculate central angle $\theta$: $$\cos(\theta/2) = \frac{3}{5} = 0.6$$ $$\theta/2 = \cos^{-1}(0.6) \approx 0.9273 \text{ radians}$$ $$\theta = 2 \times 0.9273 = 1.8546 \text{ radians}$$ - Calculate area of triangle $OAB$: Since $OA = OB = r = 5$ ft, $$\text{Area}_{\triangle OAB} = \frac{1}{2} \times 5 \times 5 \times \sin(1.8546)$$ Calculate $\sin(1.8546)$: $$\sin(1.8546) \approx 0.9608$$ So, $$\text{Area}_{\triangle OAB} = \frac{1}{2} \times 25 \times 0.9608 = 12.01 \text{ ft}^2$$ - Calculate area of sector $OAB$: $$\text{Area}_{\text{sector}} = \frac{1}{2} \times 25 \times 1.8546 = 23.18 \text{ ft}^2$$ - Calculate area enclosed by $OA$, $OB$, and arc $AB$ (segment area): $$\text{Area}_{\text{segment}} = 23.18 - 12.01 = 11.17 \text{ ft}^2$$ 4. **Final Answers:** (a) Area of triangle $OAB$ is approximately $12.01$ square feet. (b) Area enclosed by $OA$, $OB$, and arc $AB$ is approximately $11.17$ square feet.