Explanation: Visualizing the rectangle inside the circle (corners touching the circumference of the circle and the center of the rectangle is the center of the circle) you will see that the rectangle can be divided into 8 congruent right triangles, with the hypotenuse as the radius of the circle. Calculating the radius you divide each side of the rectangle by two for the sides of each right triangle (giving 6 and 8). The hypotenuse (by pythagorean theorem or just knowing right triangle sets) the hypotenuse is give as 10. Area of a circle is given by πr². 10² is 100, so 100π is the area.

The circle is inscribed in a square when it is drawn within the square so as to touch in as many places as possible. This means that the side of the square is the same as the diameter of the circle.

Let \(\displaystyle \pi =3.14\) 

\(\displaystyle A_{square}= s^{2} = (6)^{2} = 36 in^{2}\)

\(\displaystyle A_{circle}= \pi r^{2}=3.14\cdot3^{2}=3.14\cdot9=28.26 in^{2}\)

So the approximate difference is in area \(\displaystyle 8\ \textup{in}^{2}\)

The length of 20 represents the diameters of both circles. Each circle has a diameter of 10 and since radius is half of the diameter, each circle has a radius of 5. The area of a circle is A = πr² . The area of one circle is 25π. The area of both circles is 50π. The area of the rectangle is (10)(20) = 200. 200 - 50π gives you the area of the paper after the two circles have been cut out. π is about 3.14, so 200 – 50(3.14) = 43.

The area of an annulus is

\(\displaystyle \pi (R^{2}-r^{2})\)

where \(\displaystyle R\) is the radius of the larger circle, and \(\displaystyle r\) is the radius of the smaller circle.

\(\displaystyle \pi (25^{2}-10^{2})\)

\(\displaystyle \pi (625-100)\)

\(\displaystyle 525\pi\)

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 \(\displaystyle A=\pi r^2\) . Radius is half of the diameter of the circle (which we know is 10), so r = 5.

\(\displaystyle A=\pi *5^2=25\pi\)

The following diagram will help to explain the solution:

We are searching for the surface area of the shaded region.  We can multiply this by the depth (1.5 feet) to find the total volume of this area.

The radius of the outer circle is 44 feet.  Therefore its area is 44²π = 1936π.  The area of the inner circle is 40²π = 1600π.  Therefore the area of the shaded area is 1936π – 1600π = 336π.  The volume is 1.5 times this, or 504π.

The area of a circle can be solved using the equation \(\displaystyle A=\pi r^{2}\) 

The area of a circle with radius 4 is \(\displaystyle \pi 4^{2}=16\pi\) while the area of a circle with radius 2 is \(\displaystyle \pi 2^{2}=4\pi\). \(\displaystyle 16\pi \div 4\pi =4\)

The formula for the circumference of a circle is \(\displaystyle c = 2*\pi*r\). Because we are given the circumference we substitute \(\displaystyle 24\pi\) for \(\displaystyle c\) and solve for \(\displaystyle r\), yielding \(\displaystyle r = 12\).

Next, we need to plug in our value for \(\displaystyle r\) into the formula for the area of a circle. 

\(\displaystyle A = \pi * r^{2}\) and get 

\(\displaystyle A = 144\pi\)

To find the area of a circle, we use the formula 

\(\displaystyle A=\pi r^2\) 

where r is the radius.

However, the problem does not give us the circles radius. In order to solve for the area we must find the radius using the circumference. Circumference of a circle follows the equation 

\(\displaystyle C=2\pi r\),

so since we know the circumference we can manipulate the equation and plug in our value to solve for radius.

\(\displaystyle r=\frac{C}{2\pi}=\frac{16\pi}{2\pi}=8\).

Now that we know the radius we simply plug it into the area formula to solve for our final answer. 

\(\displaystyle A=\pi (8)^2=64\pi\)