Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

The area of an annulus is

where is the radius of the larger circle, and is the radius of the smaller circle.

The image below shows the rectangle inscribed in the circle. Dividing the rectangle into two triangles allows us to find the diameter of the circle, which is equal to the length of the line we drew. Using a2+b2= c2 we get 62 + 82 = c2. c2 = 100, so c = 10. The area of a circle is . Radius is half of the diameter of the circle (which we know is 10), so r = 5.

To convert degrees to radians, we need to remember the following formula.

.

Now lets substitute for degrees.

The area of the circle is (30/2)2*3.14 (π) = 706.5, since the shaded region is simply the area difference between the circle and the star, it’s 706.5-345 = 361.5

If you answered that the whirlpool’s area is 18.84 feet and therefore fits, you are incorrect because 18.84 is the circumference of the whirlpool, not the area.

If you answered that the area of the whirlpool is 56.52 feet, you multiplied the area of the whirlpool by 1.5 and assumed that that was the correct area, not the legal limit.

If you answered that the area of the backyard was smaller than the area of the whirlpool, you did not calculate area correctly.

And if you thought the area of the backyard was 30 feet, you found the perimeter of the backyard, not the area.

The correct answer is that the area of the whirlpool is 28.26 feet and, when multiplied by 1.5 = 42.39, which is smaller than the area of the backyard, which is 56 square feet.

If the area of the square is , then that means that each side is since the area of a square is given by .

In the figure, the side of the square is equal to the diameter of the semicircle. Therefore, the semicircle's radius is .

The area of the semicircle is given by:

The area of the entire figure is the sum of the two component areas.

Our total area is .

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 circumference of a circle = πd where d = diameter. Therefore, this circle’s diameter must equal 16. Knowing that diameter = 2 times the radius, we can determine that the radius of this circle = 8. Halving the radius would give us a new radius of 4. To find the area of this new circle, use the formula A=πr² where r = radius. Plug in 4 for r. Area will equal 16π.