sector area = ∏ * r² * degrees/360 = ∏ * 4² * 86/360 = 172/45 * ∏

We need to begin by finding the area of the following sector:

If ∠BAC = 120°, then the area of sector BAC is equal to 120 / 360 = 1/3 of the entire circle.  Since AC is 12 and is a radius, we know the total area is pi *12² = 144*pi.  The sector is then 144*pi / 3 = 48*pi.

Now, we need to find the area of the triangle ABC.  Since AB and AC are equal (both being radii of our circle), we have an icoseles triangle.  If we drop a height from ∠BAC, we have the following triangle ABC:

We can use the 30-60-90 rule to find the height and base.  One half of the base (x) is found by the following ratio:

√(3) / 2 = x / 12

Solving for x, we get: x = 6 * √(3).  Therefore, the base is 12 * √(3).

To find the height (y), we  use the following ratio:

1 / 2 = y / 12

Solving for y, we get y = 6.

Therefore, the area of the triangle = 0.5 * 6 * 12 * √(3) = 3 * 12 * √(3) = 36 * √(3)

The area of the shaded region is then 48*pi - 36 * √(3).

In this case, the minute hand is pointing at 3 while the hour hand is pointing at some value past 4. Now, since 15 minutes represents ¹/₄ of an hour, the hour hand will be ¹/₄ the distance between 4 and 5. The total number of degrees between minutes is ³⁶⁰/₁₂ = 30°; therefore, the hour hand is ³⁰/₄ or 7.5° past the 4 o'clock point. Now, between 3 and 4, there are 30°.  Add to that the 7.5°.  The total distance is therefore 37.5°.  

To find the degrees between hands on a clock, you must first remember that a clockface is a circle, and therefore has an internal angle sum of 360 degrees. Since a clockface has a sum of 360 degrees and there are 12 numbers on a clockface, the degree difference between each number is 360 divided by 12, or 30 degrees.  

You can use this 30 degrees to figure out the distance between the minute and hour hands at the time given.  At 5:40pm, the minute hand will be on the 8, as each number indicates five minutes on the clock.  

It is the hour hand where this calculation gets tricky. While it would be easy to assume that the hour hand would be on the 5 at this time, that's actually incorrect.  The hour hand slowly creeps towards the next number throughout the hour, moving in appropriate increments to reflect the fraction of the hour that has past.  Since 40 minutes is 2/3 of an hour, then the hour hand is 2/3 of the way towards the number 6 on 5:40pm. 

Thus, when calculating the distance, you must find the difference between 5 2/3 and 8 on the clock.  

\(\displaystyle 8 -5\frac{2}{3}=2\frac{1}{3}\) 

Multiply this number by 30 (the degrees between each number on the clock, and thus the way we must convert this number to degrees), and you get 70 degrees.

\(\displaystyle 2\frac{1}{3}\cdot 30=\frac{7\cdot 30}{3}=\frac{210}{3}=70^\circ\)

Our clock looks roughly like this:

Now, between every number on the clock, there are \(\displaystyle \frac{360}{12}\) or \(\displaystyle 30\) degrees. Therefore, from \(\displaystyle 12\) to \(\displaystyle 4\), there are \(\displaystyle 4*30=120\) degrees. To find the arc length, you use the equation:

\(\displaystyle \frac{degrees}{360}*Circumference\)

Now, we know:

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

We know that \(\displaystyle r=5\). Therefore, we can write our equation:

\(\displaystyle \frac{120}{360}*2*\pi*5=\frac{1}{3}*10\pi=\frac{10\pi}{3}\)

Our clock looks roughly like this:

Now, between every number on the clock, there are \(\displaystyle \frac{360}{12}\) or \(\displaystyle 30\) degrees. We have a little trickier math to do, however. Let's subdivide the clock into \(\displaystyle 24\) subsections instead. Each of these will have \(\displaystyle 15\) degrees. Now, between \(\displaystyle 6\) and \(\displaystyle 10\), there are \(\displaystyle 8\) such subsections. Since the hour hand is directly in the middle of \(\displaystyle 10\) and \(\displaystyle 11\), there is one more such subsection. Therefore, we have \(\displaystyle 9\) total subsections or \(\displaystyle 9*15=135\) degrees. To find the arc length, you use the equation:

\(\displaystyle \frac{degrees}{360}*Circumference\)

Now, we know:

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

We know that \(\displaystyle r=6\).  Therefore, we can write our equation:

\(\displaystyle \frac{135}{360}*2*\pi*6=\frac{3}{8}*12\pi=\frac{9\pi}{2}\)

Our clock looks roughly like this:

Now, between every number on the clock, there are \(\displaystyle \frac{360}{12}\) or \(\displaystyle 30\) degrees. Therefore, from \(\displaystyle 12\) to \(\displaystyle 10\) there are \(\displaystyle 2\) of these sectors. Therefore, there are \(\displaystyle 2*30=60\) degrees. To find the arc length, you use the equation:

\(\displaystyle \frac{degrees}{360}*Circumference\)

Now, we know:

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

We know that \(\displaystyle r=3\). Therefore, we can write our equation:

\(\displaystyle \frac{60}{360}*2*\pi*3=\frac{1}{6}*6\pi=\pi\)

We know the three angles in a triangle add up to 180 degrees, and all three angles are 60 degrees in an equilateral triangle. 

A hexagon has six sides, and we can use the formula degrees = (# of sides – 2) * 180. Then degrees = (6 – 2) * 180 = 720 degrees. Each angle is 720/6 = 120 degrees.

Quantity B is greater.

Begin with Quantity A. We know the measure of one angle in an equilateral triangle is 60. Therefore, double the angle is 120 degrees.

For the hexagon, use the formula for the sum of the interior angles:

\(\displaystyle Sum = (n-2)\cdot180\)

where n= number of sides in a regular polygon

\(\displaystyle =(6-2)\cdot 180\)

\(\displaystyle =4\cdot180\)

\(\displaystyle =720\)

If the sum of the interior angles of a regular hexagon is \(\displaystyle 720\) degrees, then one angle is \(\displaystyle 120\) degrees.

The two quantities are equal.

The formula for the area of a regular pentagon is given by the equation:

\(\displaystyle A=\frac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2\)

where a is represented by the length of one side. Let's begin by finding the side length of the regular pentagon. If the perimeter is 40, then we can divide by 5 (the number of sides) to find a side length of 8.

If we plug in 8 into the equation:

\(\displaystyle A=\frac{1}{4}\sqrt{5(5+2\sqrt{5})}(8)^2\)

\(\displaystyle A=\frac{1}{4}\sqrt{5(5+2\sqrt{5})}64\)

\(\displaystyle A=\frac{64}{4}\sqrt{5(5+2\sqrt{5})}\)

\(\displaystyle A=16\sqrt{5(5+2\sqrt{5})}\)