Area of Figure with Three Circles

The area of one overlapping section is comprised of one 60º circle sector (A + B) and another smaller section (B). Since $60º$ is one-sixth of the circle, the area of (A + B) is $\frac{1}{6} \cdot \pi (5)^2 = \frac{25}{6} \pi$ .

Now area B is equal to the 60º circle sector (A + B) minus the equilateral triangle (A). We have just found area (A + B), but we don't know the area of the equilateral triangle. One way to do this is to first find its height. This is just $\sqrt{5^2 - (5/2)^2}= \frac{5 \sqrt 3}{2}$ by Pythagoras, so the area is $\frac{1}{2} \cdot 5 \cdot \frac{5 \sqrt 3}{2} = \frac{25 \sqrt 3}{4}$ . So area B is (A + B) - A, or just $\frac{25}{6} \pi - \frac{25 \sqrt 3}{4}$ . Other methods such as using the formula $\text{Area} = \frac{1}{2} ab \sin C$ , or using trigonometry to find the height of the triangle give the same answer.

The area of one overlapping section is now (A + B) + B, which equals $\frac{25}{6} \pi + \frac{25}{6} \pi - \frac{25 \sqrt 3}{4} = 2 \cdot \frac{25}{6} \pi - \frac{25 \sqrt 3}{4}$ . Now back to the original problem. If we add the areas of the three circles together, then we will have double-counted four of the overlapping sections. Consequently, we need to subtract the area of the four overlapping sections from the 3 circles to get the area of the entire figure. This is just $3 \cdot \pi (5)^2 - 4 \left(2 \cdot \frac{25}{6} \pi - \frac{25 \sqrt 3}{4} \right) = 174.20 \cdots$ , which when rounded to the nearest integer gives $\boxed{174}$ .

The three circles' area is 75π with 4 overlapping sections. Each of the overlapping sections are equal to (25π/3) - [2.5* (sqrt(18.75))] 75π - {(25π/3) - [2.5* (sqrt(18.75))]} * 4 = 174