Let \(\displaystyle s\) be the sidelength of the square is the square and \(\displaystyle r\) be the radius of the circle. Then since the areas of the circle and the square are equal, we can set up this equation:

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

We find the ratio of \(\displaystyle s\) to \(\displaystyle r\) - that is, \(\displaystyle \frac{s}{r}\) - as follows:

\(\displaystyle s^{2 }\div r^{2}= \pi r^{2} \div r^{2}\)

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

\(\displaystyle \sqrt{\frac{s^{2 } }{r^{2}}} =\sqrt{ \pi}\)

\(\displaystyle \frac{s}{r} =\sqrt{\pi } = \frac{\sqrt{\pi }}{1}\)

The correct ratio is \(\displaystyle \sqrt{\pi } \textrm{ to }1\).

Let \(\displaystyle r\) be the radius of circle C. Then Circle B has radius equal to twice this, or \(\displaystyle 2r\), and Circle A has radius equal to three times that, or \(\displaystyle 3 (2r) = 6r\). The circumference of a circle is equal to \(\displaystyle 2 \pi\) times the radius, so the circumferences of the three circles are:

A: \(\displaystyle 2 \pi \cdot 6r = 12 \pi r\)

B: \(\displaystyle 2 \pi \cdot 2r = 4 \pi r\)

C: \(\displaystyle 2 \pi r\)

These circumferences add up to \(\displaystyle 99\pi\), so set up and solve the equation:

\(\displaystyle 12 \pi r + 4\pi r + 2 \pi r =99 \pi\)

\(\displaystyle 18 \pi r =99 \pi\)

\(\displaystyle 18 \pi r \div 18 \pi =99 \pi \div 18 \pi\)

\(\displaystyle r =5 \frac{1}{2}\)

To begin this problem, we need to recognize that the distance between points L and K is our diameter. Segment LK passes from one point on circle Q through the center, to another point on circle Q. Sounds like a diameter to me! Use distance formula to find the length of LK.

\(\displaystyle d=\sqrt{(x-x')^2+(y-y')^2}\)

Plug in our points and simplify:

\(\displaystyle d=\sqrt{(9--4)^2+(3-5)^2}=\sqrt{13^2+2^2}=\sqrt{173}\approx13.15\)

Now, don't be fooled into choosing 13.15. That is our diameter, so our radius will be half of 13.15, or 6.575. This rounds to 6.58

A close reading of the question reveals that we are given the circumference and asked to find the radius.

Circumference formula:

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

So, 

\(\displaystyle 34 \pi=2 \pi r\)

\(\displaystyle r=34\pi\div2 \pi=17\)

So 17 feet

Using the formula for the circumference of a circle, we can solve for its radius. Plugging in the given value for the circumference, we have:

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

\(\displaystyle 42\pi=2\pi r\)

\(\displaystyle r=\frac{42\pi }{2\pi }=21\)

Using the formula for the area of a circle, we can solve for its radius. Plugging in the given value for the area of the circle, we have:

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

\(\displaystyle 25\pi=\pi r^2\)

\(\displaystyle r^2=25\)

\(\displaystyle r=5\)

To solve, use the formula for the area of a circle, \(\displaystyle A={\pi}r^2\), and solve for \(\displaystyle r\).

\(\displaystyle 64\pi=\pi{r^2}\Rightarrow 64=r^2\Rightarrow r=8\)

The circumference of a circle, in terms of radius, can be found by: \(\displaystyle C=2r \pi\)

We are told the circumference with respect to \(\displaystyle \pi\) so we can easily solve for the radius:

\(\displaystyle 42 \pi = 2 r \pi\)

Notice how the \(\displaystyle \pi\) cancels out

\(\displaystyle 42 = 2r\)

\(\displaystyle r=21\)

Solving this problem is rather straightforward, we just need to remember the circumference of circle is found by \(\displaystyle C=2r\pi\) and in this case, we're given the circumference in terms of \(\displaystyle \pi\). 

\(\displaystyle 27\pi = 2 r \pi\)

Notice how the \(\displaystyle \pi\) cancel out

\(\displaystyle 27=2r\)

\(\displaystyle r=13.5\)

A circle can be divided into three congruent arcs that measure

\(\displaystyle \frac{1}{3} \cdot 360 ^{\circ } = 120 ^{\circ }\).

If the three (congruent) chords are constructed, the figure will be an equilateral triangle. The figure is below, along with the altitudes of the triangle:

Since \(\displaystyle AB = x + 2\), it follows by way of the 30-60-90 Triangle Theorem that 

\(\displaystyle BM = \frac{1}{2} AB = \frac{1}{2} (x+2) = \frac{1}{2} x+1\)

and

\(\displaystyle AM = \frac{ \sqrt{3}}{2} \cdot BM = \frac{ \sqrt{3}}{2} \cdot \left ( \frac{1}{2} x+1 \right ) = \frac{\sqrt{3}}{4} x+ \frac{ \sqrt{3}}{2}\)

The three altitudes of an equilateral triangle split each other into segments that have ratio 2:1. Therefore,

\(\displaystyle OA = \frac{2}{3} \cdot AM\)

\(\displaystyle =\frac{2}{3} \cdot \left ( \frac{\sqrt{3}}{4} x+ \frac{ \sqrt{3}}{2} \right )\)

\(\displaystyle = \frac{\sqrt{3}}{6} x+ \frac{ \sqrt{3}}{3}\)