+2441 One way to approach this problem is to find the area of the circle and subtract the area of the minor sector and the area of the three congruent triangles.
Let's find the area of the circle:
| \(A_{\circ}=\pi*6^2\) | Use the formula for the area of a circle. |
| \(A_{\circ}=36\pi\text{ft}^2\) | |
Let's find the area of the minor sector:
| \(A_{\text{sector}}=\frac{1}{4}A_{\circ}\) | This minor sector is really just a quarter of the area of the entire circle. |
| \(A_{\text{sector}}=\frac{1}{4}*36\pi\text{ft}^2\) | |
| \(A_{\text{sector}}=9\pi\text{ft}^2\) | |
Let's find the area of the three congruent triangles that are right and isosceles:
| \(A_{\triangle}=\frac{1}{2}*6*6\) | It is not too difficult to find the area of these triangles since they are all right. |
| \(A_{\triangle}=18\text{ft}^2\) | |
| \(3A_{\triangle}=3*18\text{ft}^2\) | Thus far, I have calculated the area of one triangle, but we need to calculate the area of all three. |
| \(3A_{\triangle}=54\text{ft}^2\) |
Now that we know everything, let's find the area of the three minor segments of this particular circle:
| \(A_{\text{segments}}=A_{\circ}-A_{\text{sector}}-3A_{\triangle}\) | Do the substitution. |
| \(A_{\text{segments}}=(36\pi-9\pi-54)\text{ft}^2\) | Combine like terms. |
| \(A_{\text{segments}}=(27\pi-54)\text{ft}^2\) | Match this with the answer choices above. |
