A circular path 4 feet wide has an inner diameter of 250 feet. How much farther is it around the outer edge of the path than around the inner edge? Round to nearest hundredth. Use 3.14 for π.

50.24 ft

Step-by-step explanation:

In this question, we are asked to calculate the difference in distance between two seemingly eccentric circles having a distance of 4ft apart.

In this question, we must identify that we have two different circles. A smaller inner circle and a bigger outer circle. We must also identify that the difference between the body of the circumference of the first circle and that of the second circle is 4ft.

In the introduction to the solution, I said eccentric circles. What I mean by this is that they both have same centers. If they were not eccentric, the 4ft would not be a constant distance around.

Let's call the radius of the inner circle r and that of the outer circle R. From the question, the radius of the inner circle r is 250 feet. This means that the radius of the outer circle will be 258 feet (4 feet on both sides)

Mathematically, the circumference of a circle can be calculated as 2 * π * Radius

In this particular case, since we are trying to find the difference between the inner edge and the outer edge, this distance is equivalent to;

(R-r) * 2 * π

= (258-250) * 2 * π

16 * π = 50.24 ft (50.24ft to the nearest hundredth)

25.12

Step-by-step explanation:

We are given that;

Width of the circular path = 4 ft

Inner diameter of circular path = 250 ft

Thus, the inner radius of circular path = 250/2 = 125 ft

Thus, the outer radius will be;

125 + 4 = 129 ft

We know that formula for circumference = 2πr

Thus,

Circumference of inner circular path = 2π (125) = 250π

Circumference of outer circular path = 2π (129) = 258π

Thus, difference = 258π - 250π = 8π

Now, we are told that π = 3.14

Thus, difference = 8 x 3.14 = 25.12