Let $\displaystyle r$ be the radius of Circle A. Then its area is $\displaystyle \pi r^{2}$.

The area of Circle B is $\displaystyle 4\pi r^{2} =2^{2}\pi r^{2} = \pi (2r)^{2}$, so the radius of Circle B is twice that of Circle A; by a similar argument, the radius of Circle C is twice that of Circle B, or $\displaystyle 2 (2r ) = 4r$.

(a) Twice the radius of circle B is $\displaystyle 2 (2r) = 4r$.

(b) The sum of the radii of Circles A and B is $\displaystyle r + 4r = 5r$.

This makes (b) greater.

Every hour, the tip of the minute hand travels the circumference of a circle. Between noon and 1:45 PM, one and three-fourths hours pass, so the tip travels $\displaystyle 1 \frac{3}{4}$ or $\displaystyle \frac{7}{4}$ times this circumference. The length of the minute hand is the radius of this circle $\displaystyle r$, and the circumference of the circle is $\displaystyle C = 2 \pi r$, so the distance the tip travels is $\displaystyle \frac{7}{4}$ this, or

$\displaystyle \frac{7}{4} C = \frac{7}{4} \cdot 2 \pi r = \frac{7 \pi r }{2}$

Set this equal to $\displaystyle \frac{14\pi}{3}$ feet:

$\displaystyle \frac{7 \pi r }{2} = \frac{14\pi}{3}$

$\displaystyle \frac{7 \pi r }{2} \times \frac{2}{7 \pi}= \frac{14\pi}{3}\times \frac{2}{7 \pi}$

$\displaystyle r= \frac{2}{3}\times \frac{2}{1} = \frac{4}{3} = 1 \frac{1}{3}$ feet.

This is equivalent to 1 foot 4 inches.

Betwen noon and 2:30 PM, the minute hand has made two and one-half revolutions; that is, the tip of minute hand has traveled the circumference of its circle two and one-half times. Therefore, 

$\displaystyle C \cdot 2 \frac{1}{2} = 40 \pi$

$\displaystyle C = 40 \pi \div 2 \frac{1}{2} = \frac{40 \pi}{1} \div \frac{5}{2} = \frac{40 \pi}{1} \cdot \frac{2}{5}= \frac{80 \pi}{5} = 16\pi$ feet.

The radius of this circle $\displaystyle r$ is the length of the minute hand. We can use the circumference formula to find this:

$\displaystyle 2 \pi r = C$

$\displaystyle 2 \pi r = 16 \pi$

$\displaystyle r = 16 \pi \div 2 \pi = 8$

The minute hand is eight feet long, which is less than three yards (nine feet), so (B0 is greater.

Recall for this question that the formulae for the area and circumference of a circle are, respectively:

$\displaystyle A=\pi * r^2$

$\displaystyle C=2*\pi*r$

For our two quantities, we have:

Quantity A

$\displaystyle 64\pi = \pi * r^2$

Therefore, $\displaystyle r^2 = 64$

Taking the square root of both sides, we get: $\displaystyle r = 8$

Quantity B

$\displaystyle 50\pi = 2 * \pi * r$

Therefore, $\displaystyle r = 25$

Therefore, quantity B is greater.

Recall for this question that the formulae for the area and circumference of a circle are, respectively:

$\displaystyle A=\pi * r^2$

$\displaystyle C=2*\pi*r$

For our two quantities, we have:

Quantity A

$\displaystyle 144\pi = \pi * r^2$

Therefore, $\displaystyle r^2 = 144$

Taking the square root of both sides, we get: $\displaystyle r = 12$

Quantity B

$\displaystyle 24\pi = 2 * \pi * r$

Therefore, $\displaystyle r = 12$

Therefore, the two quantities are equal.

Given that the radius is equal to half the diameter, the value of the radius would be equal to $\displaystyle 5x^{2}+4x-2$ divided by 2. This gives us:

$\displaystyle \frac{5x^{2}+4x-2}{2}$

$\displaystyle 2.5x^{2}+2x-1$

The relation between the area of a circle $\displaystyle A$ and its radius $\displaystyle r$ is given by the formula

$\displaystyle A = \pi r^{2}$

Since 

$\displaystyle A= 81x^{2}$:

$\displaystyle 81x^{2} = \pi r^{2}$

We solve for $\displaystyle r$:

$\displaystyle \pi r^{2} = 81x^{2}$

$\displaystyle \frac{\pi r^{2} }{\pi}= \frac{81x^{2}}{\pi}$

$\displaystyle r^{2} = \frac{81x^{2}}{\pi}$

$\displaystyle \sqrt{r^{2} }= \sqrt{\frac{81x^{2}}{\pi}}$

Since $\displaystyle r$ is positive, as is $\displaystyle x$:

$\displaystyle r =\frac{ \sqrt{81} \cdot \sqrt{x^{2}}}{\sqrt{\pi}} = \frac{9x}{\sqrt{\pi}}$

The area of a circle with radius $\displaystyle r$ is $\displaystyle A = \pi r^{2}$. Therefore, the areas of the circles with radii 4 and 8, respectively, are

$\displaystyle A_{1} = \pi \cdot 4^{2} = 16 \pi$

and 

$\displaystyle A_{2} = \pi \cdot 8^{2} = 64\pi$

The areas form an arithmetic sequence, the common difference of which is 

$\displaystyle 64 \pi - 16 \pi = 48 \pi$.

The circles will have areas:

$\displaystyle 16 \pi, 64 \pi, 112 \pi, 160 \pi , 208 \pi$

Since the area of the largest circle is $\displaystyle 208 \pi$, we can find the radius as follows:

The radius can be calculated now:

$\displaystyle \pi r^{2} = 208 \pi$

$\displaystyle r^{2} = 208$

$\displaystyle r = \sqrt{208} = \sqrt{16} \cdot \sqrt{13} = 4 \sqrt{13}$

If we call the radius of Circle A $\displaystyle r$, then the radius of Circle B is $\displaystyle \frac{2}{3} r$.

The areas of the circles are:

(a) $\displaystyle \pi r^{2}$

(b) $\displaystyle \pi \left( \frac{2}{3} r \right ) ^{2} = \frac{4}{9} \pi r^{2}$

Twice the area of Circle B is

 $\displaystyle 2 \cdot \pi \left( \frac{2}{3} r \right ) ^{2} =2 \cdot \frac{4}{9} \pi r^{2} = \frac{8}{9} \pi r^{2} < \pi r^{2}$,

making (a) the greater number.

If the radius of Circle 1 is $\displaystyle r$, then the square will have sidelength equal to the diameter of the circle, or $\displaystyle 2r$. Circle 2 will have as its diameter the length of a diagonal of the square, which by the $\displaystyle 45 ^{\circ }-45 ^{\circ }-90^{\circ }$ Theorem is $\displaystyle \sqrt{2}$ times that, or $\displaystyle 2r \sqrt{2}$. The radius of Circle 2 will therefore be half that, or $\displaystyle r \sqrt{2}$.

The area of Circle 1 will be $\displaystyle A = \pi r^{2}$. The area of Circle 2 will be $\displaystyle A = \pi \left ( r\sqrt{2} \right ) ^{2} = \pi \left ( r^{2} \cdot 2 \right ) = 2 \pi r^{2}$, twice that of Circle 1.