The formula for the area of a sector of a circle is:

\(\displaystyle \frac{central\: angle}{360}*\pi r^2\)

The central angle given is 120 thus:

\(\displaystyle \frac{120}{360}*\pi (5cm)^2= \frac{1}{3}\pi 25cm^2 = \frac{25\pi}{3}cm^2\)

If the radius is \(\displaystyle 6m\), then the circumference of the wheel is:

\(\displaystyle C = 2r\pi = 12\pi\) 

If the wheel turns \(\displaystyle 120^{\circ}\) each minute, then it turns \(\displaystyle \frac{120^{\circ}}{360^{\circ}} = \frac{1}{3}\) of the circumference each minute.

\(\displaystyle d = 12\pi \cdot \frac{1}{3} = 4\pi\)

Thus, the wheel turns \(\displaystyle 4\pi m\) each minute.

If the area of the circle is \(\displaystyle 64\pi\:in^2\), the radius can be found using the formula for the area of a circle:
\(\displaystyle A=\pi r^2\)

For our data, this is:

\(\displaystyle 64\pi=\pi r^2\)

\(\displaystyle r^2=64\)

Therefore, \(\displaystyle r=8\:in\)

Now, the circumference of the circle is defined as:

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

For our data, this is:

\(\displaystyle C=2*8*\pi=16\pi\:in\)

Now, we know that a sector is a percentage of the total area. This percentage is easily calculated:

\(\displaystyle \frac{12\pi\:in^2}{64\pi\:in^2}=\frac{3}{16}\)

So, the length of the arc will merely be the same percentage, but now applied to the circumference:

\(\displaystyle \frac{3}{16} * 16\pi\:in = 3\pi\:in\)

Since we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees. 

Since angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees.

To begin, you should compute the complete area of the circle:

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

For your data, this is:

\(\displaystyle A=15^2\pi=225\pi\)

Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:

\(\displaystyle \frac{45}{225\pi}\)

Now, multiply this by the total \(\displaystyle 360\) degrees in a circle:

\(\displaystyle \frac{45}{225\pi}*360=22.918311805212\)

Rounded, this is \(\displaystyle 22.92^{\circ}\).

The first thing to do for this problem is to compute the total circumference of the circle. Notice that you were given the diameter. The proper equation is therefore:

\(\displaystyle C=\pi d\)

For your data, this means,

\(\displaystyle C=12\pi\)

Now, to compute the angle, note that you have a percentage of the total circumference, based upon your arc length:

\(\displaystyle \frac{13.5}{12\pi}*360=128.9155039044336\)

Rounded to the nearest hundredth, this is \(\displaystyle 128.92^{\circ}\).

The perimeter of a circle = 2 πr = πd

Therefore d = 36

For the circle to contain all 3 vertices, the hypotenuse must be the diameter of the circle. The hypotenuse, and therefore the diameter, is 5, since this must be a 3-4-5 right triangle.

The equation for the area of a circle is A = πr².

\(\displaystyle A=\pi (5/2)^2=6.25\pi\)

1. Use the area to find the radius:

 \(\displaystyle Area=\pi r^{2}\)

\(\displaystyle 49\pi=\pi r^{2}\)

\(\displaystyle 49=r^{2}\)

\(\displaystyle r=7\)

2. Use the radius to find the diameter:

\(\displaystyle d=2r=2\cdot 7=14\)

To begin, be very careful to note that the question asks about a semi-circle—not a complete circle! This means that a complete circle composed out of two of these semi-circles would have an area of \(\displaystyle 50\pi\). Now, from this, we can use our area formula, which is:

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

For our data, this is:

\(\displaystyle 50\pi=\pi r^2\)

Solving for \(\displaystyle r\), we get:

\(\displaystyle r=\sqrt{50}\)

This can be simplified to:

\(\displaystyle r = 5\sqrt{2}\)

The diameter is \(\displaystyle 2r\), which is \(\displaystyle 2 * 5\sqrt{2}\) or \(\displaystyle 10\sqrt{2}\).