To find the area of a circle you must plug the radius into \(\displaystyle r\) in the following equation.

In this case, the radius is \(\displaystyle 8\), so we plug \(\displaystyle 8\) into \(\displaystyle r\).

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

 \(\displaystyle A=64\pi\)

To find the area of a circle you must plug the radius into \(\displaystyle r\) in the following equation

In this case the radius is 9 so we plug it into \(\displaystyle r\) to get  \(\displaystyle 9^{2}=81\)

We then multiply it by \(\displaystyle \pi\) to get our answer \(\displaystyle A=81\pi\)

In order to find the circle's area, utilize the formula \(\displaystyle \dpi{100} A=\pi r^{2}\).

\(\displaystyle \dpi{100} A=\pi r^{2}\)

\(\displaystyle \dpi{100} A=\pi (6m)^{2}\)

\(\displaystyle \dpi{100} \rightarrow 36\pi m^{2}\)

In order to find the circle's area, utilize the formula \(\displaystyle \dpi{100} A=\pi r^{2}\).

\(\displaystyle \dpi{100} A=\pi r^{2}\)

\(\displaystyle A=\pi (5in)^{2}\)

\(\displaystyle \rightarrow 25\pi in^{2}\)

In order to find the circle's area, utilize the formula \(\displaystyle \dpi{100} A=\pi r^{2}\).

\(\displaystyle \dpi{100} A=\pi r^{2}\)

However, we need to convert our diameter into a radius.

\(\displaystyle D=2r\)

\(\displaystyle \pi=2r\)

Solve for \(\displaystyle r\).

\(\displaystyle r=\frac{\pi}{2}\)

Insert the radius into the area formula and solve.

\(\displaystyle A=\pi (\frac{\pi}{2})^{2}\)

\(\displaystyle A=\pi*\frac{\pi^{2}}{4}\)

\(\displaystyle A=\frac{\pi}{1}*\frac{\pi^{2}}{4}\)

\(\displaystyle \rightarrow \frac{\pi^{3}}{4}\)

The total area of the garden is the area of the outer circle - \(\displaystyle R = 40\) - minus that of the inner circle - \(\displaystyle r=20\).

\(\displaystyle A = \pi R^{2} - \pi r^{2} = \pi \cdot 40^{2} - \pi \cdot 20^{2} \approx 5,027 - 1,257 =3,770\)

To find the area of a circle you must plug the radius into the following equation

In this case the radius is \(\displaystyle 5\) so we plug it in and square it to get \(\displaystyle 5^{2}=25\)

We then multiply it by \(\displaystyle \pi\) to get our answer \(\displaystyle A=25\pi\)

Each set of two circles fits perfectly between the sides of the square. The diameters of two of the circle must equal the length of one side of the square.

\(\displaystyle 2d=s\)

Each side has a length of 8 and the diameter of a circle is twice the radius. We can use these relationships to substitute into the first equation.

\(\displaystyle d=2r\ \text{and}\ s=8\)

\(\displaystyle 2(2r)=8\)

Solve for the radius, \(\displaystyle \small r\).

\(\displaystyle r=(8)(\frac{1}{4})=2\)

The area of a circle can be calculated using the equation \(\displaystyle A=r^2\pi\). The shaded region is made of four circles, all with the same radius.

The area of one circle is given below.

\(\displaystyle A=r^2\pi=2^2\pi=4\pi\)

The area of the shaded region (four circles) would be \(\displaystyle 4(4\pi)=16\pi\).

The area of the shaded region can be found by using the areas of the individual circles. The area of the full figure is the area of circle O. The area of the center shaded area is the area of circle I. Circle M contains circle I. The area of the shaded region will be equal to \(\displaystyle A_O-A_M+A_I\).

First, find the area of each circle using \(\displaystyle \small A_{circ}=\pi r^2\).

\(\displaystyle r_O=3+3+\frac{1}{2}6=9\)

\(\displaystyle A_O=\pi (9)^2=81\pi\)

\(\displaystyle r_M=3+\frac{1}{2}6=6\)

\(\displaystyle A_M=\pi(6)^2=36\pi\)

\(\displaystyle r_I=\frac{1}{2}6=3\)

\(\displaystyle A_I=\pi(3)^2=9\pi\)

Now we can substitute into our equation.

 \(\displaystyle A_O-A_M+A_I\)

 \(\displaystyle 81\pi-36\pi+9\pi\)

\(\displaystyle 54\pi\)

The formula for area of a circle is \(\displaystyle A=\pi r^2\). Unfortunately, the problem gives us a diameter instead of a radius. The good news is that the diameter is equal to twice the length of the radius or, mathematically, \(\displaystyle d=2r\). 

Plug in the given diameter to find the radius:

\(\displaystyle d=2r\)

\(\displaystyle 10=2r\)

\(\displaystyle \frac{10}{2}=r\)

\(\displaystyle 5=r\)

Plug that into our first equation to solve:

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

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

\(\displaystyle A=25\pi\)