The original radius would be 4, making the new radius 8 and by the area of a circle (A=π(r)²) the new area would be 64π.

The area of a circle is πr². 

The diameter of the first circle = 6” so radius of the first circle = 3” so the area = π * 3² = 9π in²

The radius of the second circle = 8” so the area = π * 8² = 64π in²

The area of the interstitial space = area of the first circle – area of the second circle.

Area = 64π in²  - 9π in² = 55π in²

The area of a circle is A=πr². Since the radius was tripled 144π =π(3r)². Divide by π and then take the square root of both sides of the equal sign to get 12=3r, and then r=4. The diameter (d) is equal to twice the radius so d= 2(4) = 8.

We know that the equation for the area of a circle is π r². To solve this problem, we pick radii for Circles A and B, making sure that Circle A’s radius is three times Circle B’s radius, as the problem specifies. Then we will divide the resulting areas of the two circles. For example, if we say that Circle A has radius 6 and Circle B has radius 2, then the ratio of the area of Circle A to B is: (π 6²)/(π 2²) = 36π/4π. From here, the π's cancel out, leaving 36/4 = 9.

         Area of a circle = A = πr²

         R = 1/2d = ½(10) = 5

         A = 5²π = 25π

The diameter of the circle is the length of a side of the square. Therefore, first solve for the length of the square's sides. The area of the square is:

A = s² or 1089 = s². Taking the square root of both sides, we get: s = 33.

Now, based on this, we know that 2r = 33 or r = 16.5. The area of the circle is πr² or π16.5² = 272.25π.

The diameter of the circle is the length of a side of the square. Therefore, first solve for the length of the square's sides. The area of the square is:

A = s² or 32 = s². Taking the square root of both sides, we get: s = √32 = √(2⁵) = 4√2.

Now, based on this, we know that 2r = 4√2 or r = 2√2. The area of the circle is πr² or π(2√2)² = 4 * 2π = 8π.

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.)  Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 88 inches for our problem. Its total area would be 88 * 88 or 7744 in².

 Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr² or π * 44² = 1936π in². Therefore, the area remaining would be 7744 – 1936π. The cost of the waste would be 0.25 * (7744 – 1936π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 4 from our subtraction. This would give us: 0.25 * 4 * (1936 – 484π). Since 0.25 is equal to 1/4, 0.25 * 4 = 1. Therefore, our final answer is: 1936 – 484π dollars.

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 50 inches for our problem. Its total area would be 50 * 50 or 2500 in².

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr² or π * 25² = 625π in². Therefore, the area remaining would be 2500 - 625π. The cost of the waste would be 0.2 * (2500 – 625π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 5 from our subtraction. This would give us: 0.2 * 5 * (500 – 125π). Since 0.2 is equal to 1/5, 0.2 * 625 = 125. Therefore, our final answer is: 500 – 125π dollars.