Posted on
22
Mar 2022

## Pieces of Pi: Sectors, Arcs, and Central Angles

Welcome back to our series on GMAT circles. In the first article, we introduced the properties of radius/diameter, circumference, and area, discussing the relationships between all of these. This time we will introduce something called a central angle, which creates portions of a circle’s area and circumference called sectors and arcs, respectively.

The best way to define these things is probably with a simple visual.

A central angle is an angle created by using line segments to connect a circle’s center to two points on its edge. A sector is the part of a circle’s area bounded by this central angle, and an arc is the part of a circle’s circumference between the two points used to draw the angle. A 90-degree central angle creates both a 90-degree sector and a 90-degree arc. An important note is that the lines used to form the central angle are radii of the circle.

As a further illustration, think about a pizza (something I do regularly). The pizza is a circle, the pieces are sectors separated by central angles, the crust is the circle’s circumference, and each piece’s section of crust is an arc. From this, you can see that any central angle creates both a sector and an arc that correspond to one another. When you pull a piece from a pizza or cut out a piece from a pie, you use a central angle to create a sector with an arc on its rounded edge.

To represent these things mathematically, we consider a circle to be like a 360-degree central angle. In this setup, the fractional relationship of a central angle to 360 corresponds to two things:

1. The fractional relationship of the resulting sector to the circle’s total area
2. The fractional relationship of the resulting arc to the circle’s total circumference

Since 90 is ¼ of 360, the area of a 90-degree sector is ¼ of its circle’s total area, and the length of a 90-degree arc is ¼ of its circle’s circumference.

To show all of this algebraically, let’s use the variable x for the degree measure of a central angle:

x / 360 = arc length / circumference
x / 360 = sector area / circle area

Most pizzas are divided into 8 slices. This means that each slice has a central angle of 360/8 = 45° and that each slice is ⅛ of the area of the entire pizza.

### Examples:

1. What is the central angle for three slices of pizza?
The central angle formed by 3 slices of pizza is 3 * 360 / 8 = 135 degrees.

2. What’s the area of a slice if the diameter is 20cm and there are six slices?
The area of a slice of pizza is 1/6 of its pizza’s total area. So, the area of a pizza can be found by using this formula A= π*r2 = 3.14*102 = 314cm
The area of a slice of pizza is  314/ 6 = 52.33cm

Keep in mind that you may have to consider this relationship in either direction. You may be given some info about the whole circle and then tasked with concluding something about a sector or an arc. Or you may be given some info about a sector or an arc and then tasked with concluding something about the whole circle. You may even be given info about both the whole circle (its area or circumference) and a sector or arc and then tasked with calculating the central angle. Each of these represents a perspective shift, and when doing a problem form, you can rewrite the problem from each of these perspectives to make sure you can fully navigate problems of this sort.

## Pieces of Pi: Official GMAT Problems

Now for some official GMAT problems. Let’s start with two straightforward sector problems, one problem solving and one data sufficiency.

### Problem-Solving Problem

The annual budget of a certain college is to be shown on a circle graph. If the size of each sector of the graph is to be proportional to the amount of the budget it represents, how many degrees of the circle should be used to represent an item that is 15 percent of the budget?

A. 15°
B. 36°
C. 54°
D. 90°
E. 150°

From the question, we can tell that the “circle graph” mentioned here is what we usually call a “pie chart,” a handy way to show the breakdown of a whole (like a budget) into its various parts. If we want to represent 15% of the budget, we need a sector with a central angle using 15% of the (360) available degrees in the circle. 0.15 * 360 = 54, so the correct answer is C. Piece of cake. Or piece of pie.

### Now for a DS pie chart problem

TOTAL EXPENSES FOR THE FIVE DIVERSIONS OF COMPANY H

The figure represents a circle graph of Company’s H total expenses broken down by the expenses for each of its five diversions. If O is the center of the circle and if Company H’s total expenses are \$5,400,000, what are the expenses for Division R?

1. x = 94
2. The total expenses for Division S and T are twice as much as the expenses for Division R.

Once again, this pie chart (which the GMAT apparently prefers to call a “circle graph”) is being used to represent a budget breakdown. Here we are told that the value represented by the whole circle is \$5,400,000. We can think of this value as the area of the circle. We are asked for the expenses for division R, or in circle terms, the area of sector R.

#### Statement 1: x = 94

This is the measure of the central angle bounding the sector whose area we need to know (sector R). Since we already know the area of the whole circle, the measure of this central angle is the final piece of the puzzle. (Area of sector R = 94/360 * \$5,400,000) Statement 1 is sufficient.

#### Statement 2: The total expenses for Divisions S and T are twice as much as the expenses for Division R.

This statement relates the total of two unknown sectors to another unknown sector. Given this statement alone, we don’t know the relationship of any of these sectors to the whole circle, so we can’t solve for any of their areas. Statement 2 is insufficient.

Statement 1 is sufficient.
Statement 2 is insufficient.

## Pieces of Pi: More Difficult Problems

Let’s ratchet up the difficulty a bit with another sector problem that involves more smoke and mirrors.

The figure consists of three identical circles that are tangent to each other. If the area of the shaded region is 64√3 – 32*π, what’s the radius of each circle?

A. 4
B. 8
C. 16
D. 24
E. 32

(tangent means just touching and not overlapping)

The problem mentions only three circles and a shaded region, but the graphic includes something more: an equilateral triangle drawn by connecting the centers of the three circles. You can solve this problem without knowing anything about the formula for the area of an equilateral triangle (although you should know this formula).  It should occur to you that the area of the shaded region could be expressed as the area of the triangle minus the combined area of those three sectors, which matches the given expression 64√3 – 32*π. So the (irrelevant) area of the triangle is 64√3, and the area of the three sectors is 32*π.

You might start by trying to get the area of a single sector by dividing 32*π by 3. But 32 won’t divide nicely by 3, which should signal you to try something else. If you can’t go from the combined area of the three sectors to the area of one sector, maybe you can go from the combined area of the three sectors to the area of one circle. You might use your spatial reasoning and conclude that the three sectors together look like they make up half a circle. Or you might recall that each interior angle of an equilateral triangle measures 60 degrees. Therefore each of these sectors is ⅙ (60/360) of a whole circle, and the three of them together do indeed make up half a circle (3 * ⅙ = ½).

Now, if the sectors’ combined area is 32*π, and this is half a circle, then the area of the whole circle is 2 * 32*π = 64. Having found the area of a circle, we can now solve for the radius.

A = 64 = π*r2
64 = r2
r = 8

And the correct answer choice is B.

### Arc Length Problem

Let’s try one more problem, this time focusing on arc length.

The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4*π/3 what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E.  6

A note about points that “lie on a circle.” This always means that the points are on the edge or perimeter of the circle.

It may be helpful to visualize or even draw out what has been described here.

This is a good opportunity to introduce some terminology. We see that line segment RU connects two points on the circle. Such a line segment is called a chord. If the line continues on to either side of the circle so that the circle is “skewered,” the line is called a secant (the GMAT does not expect you to know this term). When a chord or secant passes through the circle’s center, it creates a diameter. A line outside a circle that just touches the circle at one point is called a tangent.

If you aren’t sure how to calculate the length of a chord like RU, start with what you know. We are given the length of arc RTU (4*π/3) and the radius of the circle. A good step is to calculate the circumference of the circle so that we can see how it relates to arc RTU.

C = 2*π*r
C = 2*π*4
C = 8*π

The circumference is 8*π, and arc RTU is 4*π/3. 8*π/6 = 4*π/3. Therefore arc RTU represents ⅙ of the circumference of the circle, and its corresponding central angle is 60 degrees (360/6). Drawing out this information helps us to see its relevance.

The central angle and chord RU form an equilateral triangle. Since the radius of the circle is 4, chord RU also has length 4, and the correct answer is D.

This concludes our second article on the GMAT’s treatment of circles. Next time we will look at another kind of angle inside a circle: an inscribed angle, and at the related topic of inscribed polygons.

Contributor: Elijah Mize (Apex GMAT Instructor)

Posted on
15
Mar 2022

## Circles On The GMAT 101 – Area, Circumference, and Pi

Circles on the GMAT function like any other GMAT quant topic: the list of “knowledge bits” you need is short, but the questions creatively combine and/or disguise these few “knowledge bits” to create complex problems.

In this first article, we will discuss the most basic properties of circles and the formulas that relate to these properties. The properties in question are area, circumference, and radius/diameter.

## Circle Properties: Radius & Diameter

I treat radius (r) and diameter (D) together because they essentially express the same thing, and because the relationship between them is so simple.

Diameter tells you how “wide” a circle is at its widest point. If you draw a straight line all the way across a circle, hitting the center on the way, that line is a diameter, and the length of that line is the diameter of the circle. The radius of the circle is half of this line – or any straight line drawn from the center of the circle to a point on its edge.

This is the most basic property of a circle. The two next most basic properties involve bringing in a specific number, one of the most famous numbers in mathematics: pi.

## Circle Properties: Pi

Pi is the name of a Greek letter that looks (in its lowercase form) like this: π.

The value represented by this character is irrational (the numbers after the decimal never stop), but for most purposes and for the GMAT, 3.14 is enough. If you happen to be working with fractions or if you prefer fractions to decimals, 22/7 is a rather precise way to express pi. Rounded to the thousandths place, pi is 3.142, and 22/7 is 3.143. That’s close enough for jazz and certainly close enough for GMAT quant.

Interestingly – and for reasons relating to math beyond what is required for the GMAT – this same value can be combined with the radius/diameter to calculate both the circumference (C)  and area (A) of the circle. Circumference is the distance around the circle (its perimeter), and area is the space inside the circle.

C = π*D     or     C = 2*π*r
A = π*r²

The two options for the circumference equation are, in fact, equivalent, since D = 2r. You may be wondering why mathematicians split the 2 and the r around the π, instead of just saying π2r. One reason is simply the conventions that have developed for algebraic notation; it “looks wrong” to have the 2 in the middle after the π. But another reason is a (non-GMAT math) connection between the circumference and area equations. Both equations have a π, an r, and a 2. In the area equation, the 2 functions as an exponent on the r. In the circumference equations, it functions as a coefficient multiplying the expression. Area is pi times the square of the radius. Circumference is pi times the radius doubled.

Examples:

1. If a radius is 3, what’s the area of the circle?

A = π*r²= π*3² ≈ 28.27433

2. If the area of a circle is 25π, what is the diameter of the circle?

A = π*
r² = 25*π÷π
r = √25 = 5

3. If a radius is 4, what’s the circumference of the circle?

C = 2*r*π
C = 2*4*π
C = 8*π = 25.13274

## Official GMAT Problem

In all likelihood, none of this looks new. But like we said at the beginning, GMAT quant can get creative with common mathematical knowledge. Take a look at this official GMAT problem, and try to answer it before reading on:

In the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle?

A. 4
B. 3
C. 2
D. √3
E. √2

### Understanding the Problem

The only properties at play here are area and circumference, perhaps with radius/diameter as a “bridge” between the two, but the answer to the problem may not be immediately obvious. Part of this is due to a common GMAT technique – the removal of numbers to make the problem more abstract.

This problem disguises the relationship between the two circles by referring not to the area of each circle, but to the area of the shaded region. A helpful preliminary deduction is that if, as the problem says, the area of the shaded region is 3 times the area of the smaller circular region, then the area of the larger circle is 4 times the area of the smaller circle (comprising the 3 “parts” in the shaded region and the 1 “part” in the smaller circle).

So we have the factor relating the circles’ areas: 4. But we were asked about circumference, not area. How do we get from area to circumference? Well, both the area and circumference equations have the radius in them, so one option is to pick two values for area, with one being 4 times the other, solve for the radius of each circle, and then plug each of these radius values into the circumference equation.

### Solving the Problem

Let’s say the area of the larger circle is 16, and the area of the smaller circle is 4. Since π is common to all the circle equations, in this case it is irrelevant, and we should just remove it instead of keeping it as “dead weight” to move around algebraically.

Large Circle                             Small Circle
A = π*r²                                     A = π*r²
16 = π*r²                                   4 = π*r²
r = 4                                         r = 2

So when the areas of two circles are related by a factor of 4, their radii (plural for radius) are related by a factor of 2: that is, the square root of 4. This makes sense, since radius is a linear value and area is a square value. This is just like how when you double the length of the side of a square, you quadruple its area (since 2² = 4). When you triple the length of the side of a square, you make its area nine times what it was before (since 3² = 9). The factor of increase for area is the square of the factor of increase for the linear measure.

C = 2*π*r
C = 2*π*4
C = 2*π*2

Since we are only concerned about the factor relating the circumferences of the circles, we can ignore whatever is common to both (2*π), leaving us with 4 and 2. Since 4 is twice as much as 2, the circumference of the larger circle is twice the circumference of the smaller circle, and the correct answer choice is C.

### The Final Step

That last step (and any real calculation) is avoided with two insights:

1. There is a square relationship between area and radius. Since the area of the larger circle is 4 times that of the smaller circle, the radius of the larger circle is the square root of 4 (2) times the radius of the smaller circle.
2. There is a linear relationship between circumference and radius. Since the radius of the larger circle is twice that of the smaller circle, the circumference of the larger circle is also twice that of the smaller circle.

When you move beyond rote memorization and understand the circle equations, these insights happen naturally and lead you to the correct answer choice quickly, with virtually no calculation.

This concludes the first article in our series on the GMAT’s treatment of circles. Next, we will dive into what happens when you use something called a central angle to mark off only a part of the area (a sector) and circumference (an arc) of a circle.

Contributor: Elijah Mize (Apex GMAT Instructor)