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 circumference of a circle is equal to the diameter of the circle times \(\displaystyle \pi\). The diameter is equal to twice the radius so:

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

The radius of the first can be solved as follows:

\(\displaystyle 80\pi=2\pi r_{1}\)

\(\displaystyle r{_{1}}=40\)

Likewise for the second circle:

\(\displaystyle 40\pi=2\pi r_{2}\)

\(\displaystyle r_{2}=20\)

The are of a circle is given by the following formula:

\(\displaystyle A=\pi r^2\).

The ratio of the larger area to the smaller area can be found as follows:

\(\displaystyle \frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}=\frac{\pi (40)^2}{\pi(20)^2}=\frac{1600\pi}{400\pi}\)

Cancelling out \(\displaystyle \pi\) and dividing gives the correct answer of 

\(\displaystyle 4\)

This question asks you to apply the concept of area in finding both the area of a circle and square. Since the cirlce is inscribed in the square, we know that its diameter (two times the radius) is the same length as one side of the square. Since we are given the radius, \(\displaystyle 4x\), we can find the area of both the circle and square.

Square:

^{\(\displaystyle 2(4x) \cdot 2(4x) = 4 \cdot 16x^{}2 = 64x^{}2\)}  

This gives us the area for the entire square.

The bottom half of the square has area \(\displaystyle \frac{1}{2} \cdot 64x^{}2 = 32x^{}2\).

Now that we have this value, we must find the area that the circle occupies. The area of a circle is given by \(\displaystyle \pi \cdot r^{}2\).

So the area of this circle will be \(\displaystyle \pi \cdot 16x^{}2\).

The bottom half of the circle has half that area:

\(\displaystyle \rightarrow \pi \cdot 8x^{}2\)

Now that we have both our values, we can subtract the bottom half of the circle from the bottom half of the square to give us the shaded region:

\(\displaystyle 32 x^{}2 - \pi \cdot 8x^{}2\)