We know that the formula for the area of a circle is πr². Therefore, we must set 36π equal to this formula to solve for the radius of the circle.

36π = πr²

36 = r²

6 = r

Find the total area of the circle, then use the area formula to find the radius.

Area of section A = section B = section C

Area of circle X = A + B + C = 12π+ 12π + 12π = 36π

Area of circle =  where r is the radius of the circle

36π = πr²

36 = r²

√36 = r

6 = r 

To Find your answer, we would use the formula:  C=2πr. We are given that C = 29.5. Thus we can plug in to get  [29.5]=2πr and then multiply 2π to get 29.5=(6.28)r.  Lastly, we divide both sides by 6.28 to get 4.70=r.   (The information given of 22 ounces is useless) 

For the circle to be tangent to the y-axis, it must have its outer edge on the axis. The center is 8 units from the edge.

We know that the formula for the area of a circle is πr². Therefore, we must set 49π equal to this formula to solve for the radius of the circle.

49π = πr²

49 = r²

7 = r

To answer this question we need to find the radius of the circle given the circumference of \(\displaystyle 32\pi\).

The equation for a circle's circumference is:

\(\displaystyle circumference=\pi \cdot diameter\)

We can plug our circumference into this equation to find the diameter.

\(\displaystyle circumference=\pi \cdot diameter\)

\(\displaystyle 32\pi=\pi\cdot diameter\)

We can now divide both sides by \(\displaystyle \Pi\)

\(\displaystyle \frac{32\pi }{\pi }=\frac{\pi \cdot diameter}{\pi }\)

\(\displaystyle 32=diameter\)

So our diameter is \(\displaystyle 32\:ft\). To find the radius from the diameter, we use the following equation:

\(\displaystyle radius=\frac{diameter}{2}\)

So, for this data:

\(\displaystyle radius=\frac{32}{2}=16\)

Therefore, the radius of our circle is \(\displaystyle 16\:ft\).

The perimeter of the polygon is 46. Think of this polygon as a rectangle with two of its corners "flipped" inwards. This "flipping" changes the area of the rectangle, but not its perimeter; therefore, the top and bottom sides of the original rectangle would be 12 units long \(\displaystyle \dpi{100} \small (10+2=12)\). The left and right sides would be 11 units long \(\displaystyle \dpi{100} \small (8+3=11)\). Adding all four sides, we find that the perimeter of the recangle (and therefore, of this polygon) is 46.

Fill in the missing sides by thinking about the entire figure as a big rectangle. In the figure below, the large rectangle is outlined in blue and the missing numbers are supplied in red.

To find the perimeter of an \(\displaystyle n\)-sided polygon with a side length of \(\displaystyle a\), simply multiply the side length by the number of sides:
\(\displaystyle 4\textup{cm}*13=52\textup{cm}\)

Congruent sides \(\displaystyle \overline{AB}\), \(\displaystyle \overline{BC}\), and \(\displaystyle \overline{CD}\), and the diagonal \(\displaystyle \overline{AD}\) form an isosceles trapezoid. 

\(\displaystyle \angle ABC\) and \(\displaystyle \angle BCD\). being angles of a nine-sided regular polygon, have measure

\(\displaystyle m \angle ABC = m\angle BCDC = \frac{180 ^{\circ } (9-2)}{9} =140^{\circ }\)

The other two angles are supplementary to these:

\(\displaystyle m \angle DAB = m \angle ADC = 180 ^{\circ }- 140^{\circ } = 40 ^{\circ }\)

The length of one side of the nonagon is one-ninth of 500, so

\(\displaystyle AB = BC =CD= \frac{500}{9} \approx 55.56\)

The trapezoid formed is below (figure NOT drawn to scale):

Altitudes \(\displaystyle \overline{BM}\) and \(\displaystyle \overline{CN}\) to the base have been drawn, so

\(\displaystyle AD = AM + MN + ND\)

\(\displaystyle AD = AM + BC+ AM\)

\(\displaystyle AD = 2 \cdot AM + 55.56\)

\(\displaystyle \frac{AM}{AB} = \cos 40^{ \circ}\)

\(\displaystyle AM=AB \cos 40^{ \circ} \approx 55.56 \cdot 0.7660 \approx 42.56\)

\(\displaystyle AD \approx 2 \cdot 42.56 + 55.56 \approx 140.68\)

This makes 140 the best choice.