GMAT

One of the key things to remember with circles is that once you know one piece of information, you know everything about the circle itself. Additional angles and lengths inside are not always so simple, but it is possible to convert circumferences to areas, to radii and diameters without intermediate steps. This will save you time in Data Sufficiency questions.

Arc Lengths and Sector Areas

Arc Lengths (portions of perimeters) and Sectors (pie slices) seem more complicated than they really are. Both relate directly to the internal angle at the circle’s center, represented by θ in this diagram. Here are the equations:

Sector Area = A = θ/360 * πr²

Arc Length = L = θ/360 * 2πr

Literally, all we are doing is finding the fraction of the circle and applying it to either the area or circumference formula, respectively. Note there are only 2 variables. This is good to keep in mind for Data Sufficiency questions. It will be helpful to finding the solution to the following problem:

In the above diagram, a square is inscribed in a circle, which is inscribed in another square. Which is larger, the green region or the yellow region?

(1) The area of the larger square is 64.
(2) The area of the green region is 4.

(Standard Data Sufficiency answer choices apply)

Let’s start with Statement 1.

A(large square) = 64
s(large square) = 8
diameter = 8
r = 4

First, recognize that the figure is symmetrical. So while we may not explicitly be given an angle to find the sector area (not drawn) in which the yellow region resides, we do know its measure. The diagonals of a square intersect at a right angle, so we can deduce that the sector including the yellow region is 1/4 of the area of the circle. Since we know the radius…

A(large sector) = 1/4 * 16π = 4π

To find the yellow region itself, we must subtract the imaginary (not drawn) triangle from 4π. (Note that this imaginary triangle will be twice the green triangle.)

A(imaginary triangle) = 1/2 * r * r = 1/2 * 4 * 4 = 8

A(Yellow Region) = 4π – 8

The area of the green region can be found in two ways. Either we can see that it’s simply one-half of the 8 we just found, OR we can find both sides of the green triangle with the common 45-45-90 1:1:√2 formula. With a hypotenuse of 4, we derive 2√2 for each side, which yields an area of 4.

Which is greater,  4π – 8 or 4?

4π – 8. Sufficient. (We can save the calculations on the GMAT for DS questions, but it’s still good to go through it to practice for PS questions involving similar calculations.)

What about Statement 2?

If we know the area of the green triangle equals 4, and that it is an isosceles right triangle, then we can set up a simple equation to find its sides, which can be denoted x:

1/2 * x² = 4
x² = 8
x = 2√2

If x = 2√2, then the hypotenuse (r) = 2√2 * √2 = 4. From here, we follow the same logic as we did for Statement 1, and determine Statement 2 is sufficient. The answer choice is D.

One More Problem

A circular loop of wire is attached to the two straight wires of a dipole antenna at points X and Y. Point Z is the base of the antenna where the two straight wires meet. How far is point X from the center of the circular loop of wire?

(1) The perimeter of the circular loop of wire is 3 feet.
(2) The center of the circular loop of wire is 2 feet from point Z.

The key point to remember here is that all radii are equal. That is, the distance from the center of the circle (C, not drawn) to both X and Y is the same.

Statement 1: Before we mentioned any if you have any one piece of information about the circle, then you have it all. This is perfect example. If we know the circumference (perimeter) of the wire, we know the radius, regardless of the presence of Z in the diagram. Sufficient.

Statement 2: We can draw a line from Z to C, and mark that distance 2. Because both antenna are tangents to the circle, we know that angles CYZ and CXZ are both right angles, making 2 the hypotenuse. However, there is not additional information (angles or additional sides) that dictates the distance CX. You can visualize this lack of information; picture pinching point Z and dragging it “down” such that X and Y are still on the circle, but closer together. Since there is nothing restricting us from doing that, there is not enough information to accurately determine the length of either the two remaining sides of the right triangle. Insufficient. Answer Choice A.

Here are just a couple examples, but keep working hard. Givens in geometry provide a series of information, not just what’s stated. Make sure to keep that in mind for Data Sufficiency questions.

Good luck!

Read other articles in this series:

Geometry Series pt 1, Circles Inscribed in Squares
Geometry Series pt 2, Inscribed Triangles

To start off, let’s quickly review the essentials. These are formulas/concepts you must know:

1. a² + b² = c², but only when a right triangle. If you don’t know it’s a right triangle, Pythagorean theorem does not apply!
2. Common special right triangles include 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and their multiples.)
3. 45-45-90 triangles are ALWAYS in the ratio 1:1:√2
4. 30-60-90 triangle are ALWAYS in the ratio 1:√3:2
5. Angles and opposite sides are in the same relative size order, but are NOT proportional.

Let’s continue with a standard diagram in which we have an equilateral triangle inscribed in a circle, which is inscribed in a square.

The center point of all three figures (triangle, circle, square) are all the same, but this is ONLY true if the triangle is equilateral. Therefore, if given ANY piece of information about the circle, square or triangle, we can derive the rest. We draw a perpendicular line from the center to the side of the triangle.

Note that the hypotenuses of the smaller triangles are equal to the radius of the circle. We also know that the smaller triangles are each 30-60-90 because you are taking the 120-degree internal angle from the circle’s center and cutting it in two. Here are your basic conversions:

r = ½d = ½s, where s is the side of the square.
The sides of the 30-60-90 triangles become ½r : (r√3)/2 : r respectively
The side of the equilateral triangle becomes 2*(r√3)/2 = r√3

If given the area of the square, we should be able to derive essentially any other information.

Area of an Equilateral Triangle

The area of an equilateral triangle equals (s²√3)/4. Memorize this. It will save you the time of drawing a 30-60-90 triangle, solving for the base, finding the height, multiplying and dividing by 2. That was long to write, imagine how long it takes to do!

If  the area of the square = 64 and we needed to find the area of the triangle, we just use the conversions above:

d = 8
r = 4
side of triangle = 4√3

Area of triangle = [(4√3)²√3]/4 = 16*3*√3 / 4 = 4*3*√3 = 12√3

Angle Relationships

Another important rule is that the interior angle created from of two radii extending to the outside of the circle is exactly twice the measure of any angle on the circle extending to those same points.  In the image above, 2b = a. This information is never explicitly stated on tests, but will come up on DS questions over and over.

Let’s take a look at this practice question:

In the figure above, a circle is inscribed in a square. What is the area of the shaded region?

(1) The perimeter of the square equals 32.
(2) x = 35

Statement 1 seems irrelevant to the question, but we can determine r by knowing that the length of the square’s side. If s = 32/4 = 8, then d = 8 and r = 4. This is insufficient, since we do not know the interior angle.

Statement 2 provides information about x, and from this, we know that the interior angle of the shaded region is 2(35) = 70. This is insufficient, since we do not know the size of the circle.

Together, we know both the size of the circle and the degree measure of the interior sector angle.

A(shaded) = (2x/360) * πr² = 70/360 * 16π, whatever the hell that comes out to. Remember, since it’s a Data Sufficiency question, we don’t actually need to calculate the number.

There are infinite variations of these concepts. Be flexible in your reasoning, and practice makes perfect!

Good luck!

Read other articles in this series:
Geometry Series pt 1, Circles inscribed in squares

In this series, we will cover many types of geometric scenarios encountered on the GMAT. A basic knowledge of simple formulas (area, perimeter, etc.) is essential, but there are numerous shortcuts to geometry questions that will save you time. Today, we’ll explore circles inscribed in squares.

Some Things to Remember

• The center of the square is the same point as the center of the circle
• Draw lines! Depending on what the stimulus asks for, draw in lines that create simple shapes. (Squares can be turned into triangles, for example.)
• Shared angles will normally not be explicitly stated, unless necessary.
• Trust the pictures, but not too much. Inferences must be drawn from fact. Just because it looks like 90-degrees doesn’t mean it is! (Many of these common inferences will be detailed in this series.)
• Lengths cannot be negative. Be careful in DS questions that pose equations in the context of quadratic equations with two solutions. If one solution is negative and the other is positive, only the positive solution remains and the information is sufficient.

For circles:

• d=2r and all lines from the center to the exterior equal r.
• C = 2πr = πd
• A = πr²
• NEVER use 2πr² unless you are adding the areas of identical circles!
• Tangent lines create right angles with the radius that meets that tangent.
• If you know r, you know everything about the circle!
• Use π = 22/7 with caution. Remember 22/7 > π.

For squares:

• The diagonal equals s√2, since it creates 45-degree angles.
• The intersection of the diagonals creates a right angle.
• When a circle is inscribed inside a square, the side equals the diameter.

Usually, you will be provided with one bit of information that tells you a whole lot, if not everything. If given the length of the side of the square in the above image, we can actually find the length of the hypotenuse of the internal triangle (s = d = 2r, so the hypotenuse = (s√2)/2).

Shaded Areas

Find the large area and subtract the small area from it. When dealing with circles along with other figures, eliminate answer choices that ONLY have π’s in them or don’t have any at all. Typically, your answer will look like x + yπ.

The picture above depicts a circle perfectly inscribed in a square. Is the shaded region > 4?

(1) The area of the large rectangle equals 64.
(2) The perimeter of the shaded region equals 8 + 2π.

Statement 1: By knowing the area of the large square we also know the lengths of its sides. (Note that 64 is a perfect square, which should be a clue.) If the side is 8, then so is the diameter, which means the radius equals 4. In the image, we can see that the “larger figure” is the top-right square bordered by two radii and the outer border. How do we know that it’s a square? Two pieces of information: All sides are equal to 4, and the radius meets the large square at a right angle because it is a tangent. In this instance, the area of the smaller square equals 16. Since the interior angle is 90-degrees (360/4), the area of the sector of the circle can be represented by A = πr²/4. So that A = 16π/4 = 4π.

A(shaded) = A(small square) – A(sector) = 16 – 4π < 4 because 4π > 12. Sufficient.

Statement 2: The first thing that should jump out is the combination of a π-term and non- π-term. We can reasonably assume that 8 represents the two straight sides of the perimeter and the 2π the arc-length of the quarter circle, which is known because of the internal right angle. If 2π = C/4, then C = 8π. If C = 8π, then r = 4. From here, we return to the same reasoning as above:

A(shaded) = A(small square) – A(sector) = 16 – 4π < 4 because 4π > 12. Sufficient.

Each statement is sufficient, so the answer is choice D.

Two important takeaways:

1. Never assume without proof.
2. Follow the trail.

Post below with other helpful tips for your fellow GMATers.

Next Lesson: Inscribed Triangles.