An extremely important consideration on this problem is that the sizes are quoted in terms of the length of the diameter, but the area formula requires you to work with the radius.  So as you calculate the area of each pizza, it is very important to first divide the diameter by 2 so that you are working with the radius in the formula \(\displaystyle A=\pi r^2\).

10-inch pizza → radius of 5 → \(\displaystyle A=\pi (5^2)=25\pi\) square inches

8-inch pizza → radius of 4 → \(\displaystyle A=\pi (4^2)=16\pi\) square inches

Therefore the difference in surface area is \(\displaystyle 25\pi -16\pi =9\pi\) square inches.

To calculate an arc length, you need two pieces of information:

1) The circumference of the circle (calculated as \(\displaystyle 2\pi r\) or \(\displaystyle \pi d\), where \(\displaystyle r\) is the radius and \(\displaystyle d\) is the diameter)

2) The measure of the central angle that connects the two end points of the arc.

Then the arc length is the proportion of the circumference represented by that angle: multiply \(\displaystyle \frac{angle}{360}\) by the circumference and you have the arc length.

Here you're given the diameter (via line segment BD) as 18, so you know that the circumference is \(\displaystyle 18\pi\). And you know that angle AOB measures 40 degrees, so you can conclude that angle BOC is 140 degrees, since angles AOB and BOC must complete the straight line AC and straight lines measure 180 degrees.

Therefore your calculation is \(\displaystyle \frac{140}{360}(18\pi )\), which simplifies to \(\displaystyle \frac{7}{18}(18\pi)=7\pi\).

The distance around the outside of a circle is, of course, the circumference. A common way to test the circumference in a word problem is to use the circumference of a wheel as a straight line distance: for each revolution of the wheel, a vehicle will travel the length of the circumference.

Here you know that the diameter of the wheel is 70 centimeters, which means that the circumference, calculated as \(\displaystyle \pi d\), is \(\displaystyle 70\pi\). Note that the question asks for "approximately" the number of revolutions, and that the answer choices are spread quite far apart. This means that you can use an estimate of 3.14 or \(\displaystyle \frac{22}{7}\) for \(\displaystyle \pi\) and say that one revolution moves the wheel approximately 220 centimeters.  Since the wheel needs to cover 100000 centimeters, you should then divide 100000 centimeters by 220 to see that the answer is approximately 450.

This problem tests several of the core properties of circles:

Area = \(\displaystyle \pi r^2\)

Circumference = \(\displaystyle \pi d = 2\pi r\)

Arc length = \(\displaystyle \frac{central angle}{360}\times circumference\)

Here you're given the area, but to determine the perimeter of that sector you need to find the radius (for line segments AB and AD) and the arc length BCD. With an area of \(\displaystyle 144\pi\) that means that the radius is 12. Since you need two radii (AB and AD) to form the "legs" of the sector, that means that the straight-line legs sum to 24 (a good hint that your correct answer should include the number 24).

For the arc length, note that the central angle measures 45, and that \(\displaystyle \frac{45}{360}=\frac{1}{8}\). So the arc BD will equal one-eighth of the circumference. The circumference is \(\displaystyle 2\pi r=24\pi\), so you can find the arc by calculating  \(\displaystyle \frac{1}{8}(24\pi )=3\pi\). Adding together the two legs plus the arc, you get your answer \(\displaystyle 24+3\pi\)

To calculate an arc length, such as the length of minor arc BC here, your job is to find the proportion that that arc represents out of the total circumference. So you'll need to find 1) the circumference and 2) the measure of the central angle of that arc.

Here since you know that the area is \(\displaystyle 81\pi\), you can use the formula \(\displaystyle Area =\pi r^2\) to determine that the radius measures 9.

Then you can plug in that radius into the circumference formula, \(\displaystyle Circumference = 2\pi r\), to find that the circumference measures \(\displaystyle 18\pi .\)

From this point, you need to find the measure of angle BOC. Since angle AOC measures 180 degrees (you know that it's a straight line, because it's defined as a diameter), and angle AOB = 110, that means that BOC measures 70 degrees. So minor arc BC will be \(\displaystyle \frac{70}{360}\) of the total circumference, setting up the calculation:

\(\displaystyle \frac{70}{360}(18\pi )\)

That reduces to \(\displaystyle \frac{7}{2}\pi\) which equals \(\displaystyle 3.5\pi\).

The cross section can be found by calculating the area of the larger, diameter 24, circle (inclusive of the pipe in gray) and subtracting the area of the smaller, diameter 22, circle (everything in white inside of the pipe). To do this, use the area formula: \(\displaystyle Area = \pi r^2\).

The radius of the larger circle is 12 and of the smaller circle is 11, meaning that your areas can be calculated as:

Outer/Larger Circle: \(\displaystyle \pi (12^2)=144\pi\)

Inner/Smaller Circle: \(\displaystyle \pi (11^2)=121\pi\)

Then subtract the areas and you'll be left with just the area of the gray ring: \(\displaystyle 144\pi -121\pi =23\pi\)

While this problem asks about probability, it's essentially an area of a circle problem. The probability of hitting the bullseye is just the area of the bullseye divided by the area of the dartboard in total.

Since \(\displaystyle Area=\pi r^2\) your job becomes to find the radius of each circle. You're quoted the diameters, so to find the radius just divide the diameter by 2.

Bullseye: Diameter = 4 so Radius = 2. Area = \(\displaystyle \pi (2^2)=4\pi\)

Dartboard: Diameter = 20 so Radius = 10. Area = \(\displaystyle \pi (10^2)=100\pi\)

So the probability of hitting the bullseye is \(\displaystyle \frac{4\pi }{100\pi }\) which reduces to \(\displaystyle \frac{1}{25}\).

This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is \(\displaystyle C=2\pi r\), and the area of a circle is \(\displaystyle A=\pi r^2\).  Since you’re given the circumference of the circle as \(\displaystyle 12\pi\), you can set up an equation to solve for \(\displaystyle r\), the radius:

\(\displaystyle 12 \pi =2\pi r\)

\(\displaystyle 6=r\)

Then you can plug in 6 as the radius in the area formula:

\(\displaystyle A=\pi (6^2)=36\pi\)

This problem puts a slight twist on the area of a circle formula, \(\displaystyle A=\pi r^2\). You’re given the area of half a circle as \(\displaystyle 32\pi\), so to apply the area formula you should double \(\displaystyle 32\pi\) to account for the missing half of the circle so you can use the classic area formula. Since the area of the entire circle would be \(\displaystyle 64\pi\) 64pi, you can use the area formula to solve for the radius:

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

\(\displaystyle 64=r^2\)

\(\displaystyle 8=r\)

So the radius is 8, which means you’re ready to apply the formula for circumference of a circle. But again, you’re only dealing with half the circumference since the problem asks for the arc length of the semicircle, so while the formula for circumference i s\(\displaystyle C=2\pi r\), you can cut that in half to find the distance of half the circumference. That means you need to calculate \(\displaystyle \pi r\).  With a radius of 8, that means that your answer is \(\displaystyle 8\pi\).

This problem tests your ability to work with the two most important circle formulas (both given to you in your SAT test booklet). The circumference of a circle is \(\displaystyle C=2\pi r\), and the area of a circle is \(\displaystyle A=\pi r^2\).  Since you’re given the area of the circle as \(\displaystyle 144\pi\), you can set up an equation to solve for \(\displaystyle r\), the radius:

\(\displaystyle 144\pi =\pi r^2\)

\(\displaystyle 144=r^2\)

\(\displaystyle 12=r\)

Now you can plug in that radius of 12 into the circumference formula:

 \(\displaystyle C=2\pi r\)

\(\displaystyle C=2\pi (12)\)

\(\displaystyle C=24\pi\)