Posted on
22
Jul 2021

## GMAT Quant Syllabus 2021-2022

Author: Apex GMAT
Contributor: Altea Sollulari
Date: 22 July, 2021

We know what you’re thinking: math is a scary subject and not everyone can excel at it. And now with the GMAT the stakes are much higher, especially because there is a whole section dedicated to math that you need to prepare for in order to guarantee a good score. There is good news though, the GMAT is not actually testing your math skills, but rather your creative problem solving skills through math questions. Furthermore, the GMAT only requires that you have sound knowledge of high school level mathematics. So, you just need to practice your fundamentals and learn how to use them to solve specific GMAT problems and find solution paths that work to your advantage.

The Quantitative Reasoning section on the GMAT contains a total of 31 questions, and you are given 62 minutes to complete all of them. This gives you just 2 minutes to solve each question, so in most cases, the regular way of solving math equations that you were taught in high school will not cut it. So finding the optimal problem solving process for each question type is going to be pivotal to your success in this section. This can seem a daunting start, so our expert Apex GMAT instructors recommend that you start your quant section prep with a review of the types of GMAT questions asked in the test and math fundamentals if you have not been using high school math in your day to day life.

## What types of questions will you find in the GMAT quant?

There are 2 main types of questions you should look out for when preparing to take the GMAT exam:

### Data Sufficiency Questions

For this type of GMAT question, you don’t generally need to do calculations. However, you will have to determine whether the information that is provided to you is sufficient to answer the question. These questions aim to evaluate your critical thinking skills.

They generally contain a question, 2 statements, and 5 answer choices that are the same in all GMAT data sufficiency questions.

Here’s an example of a number theory data sufficiency problem video, where Mike explains the best way to go about solving such a question.

### Problem Solving Questions

This question type is pretty self-explanatory: you’ll have to solve the question and come up with a solution. However, you’ll be given 5 answer choices to choose from. Generally, the majority of questions in the quant section of the GMAT will be problem-solving questions as they clearly show your abilities to use mathematical concepts to solve problems.

Make sure to check out this video where Mike shows you how to solve a Probability question.

## The main concepts you should focus on

The one thing that you need to keep in mind when starting your GMAT prep is the level of math you need to know before going in for the Quant section. All you’ll need to master is high-school level math. That being said, once you have revised and mastered these math fundamentals, your final step is learning how to apply this knowledge to actual GMAT problems and you should be good to go. This is the more challenging side of things but doing this helps you tackle all the other problem areas you may be facing such as time management, confidence levels, and test anxiety

Here are the 4 main groups of questions on the quant section of the GMAT and the concepts that you should focus on for each:

### Algebra

• Algebraic expressions
• Equations
• Functions
• Polynomials
• Permutations and combinations
• Inequalities
• Exponents

### Geometry

• Lines
• Angles
• Triangles
• Circles
• Polygons
• Surface area
• Volume
• Coordinate geometry

### Word problems

• Profit
• Sets
• Rate
• Interest
• Percentage
• Ratio
• Mixtures

Check out this Profit and Loss question.

### Arithmetic

• Number theory
• Percentages
• Basic statistics
• Power and root
• Integer properties
• Decimals
• Fractions
• Probability
• Real numbers

Make sure to try your hand at this GMAT probability problem.

## 5 tips to improve your GMAT quant skills?

1. Master the fundamentals! This is your first step towards acing this section of the GMAT. As this section only contains math that you have already studied thoroughly in high-school, you’ll only need to revise what you have already learned and you’ll be ready to start practicing some real GMAT problems.
2. Practice time management! This is a crucial step as every single question is timed and you won’t get more than 2 minutes to spend on each question. That is why you should start timing yourself early on in your GMAT prep, so you get used to the time pressure.
3. Know the question types! This is something that you will learn once you get enough practice with some actual GMAT questions. That way, you’ll be able to easily recognize different question types and you’ll be able to use your preferred solution path without losing time.
4. Memorize the answer choices for the data sufficiency questions! These answers are always the same and their order never changes. Memorizing them will help you save precious time that you can spend elsewhere. To help you better memorize them, we are sharing an easier and less wordy way to think of them:
5. Make use of your scrap paper! There is a reason why you’re provided with scrap paper, so make sure to take advantage of it. You will definitely need it to take notes and make calculations, especially for the problem-solving questions that you will come across in this GMAT question.
• Only statement 1
• Only statement 2
• Both statements together
• Either statement
• Neither statement
Posted on
21
Jul 2021

## GMAT 3D Geometry Problem

In this problem we’re going to take a look at 3D objects and in particular a special problem type on the GMAT that measures the longest distance within a three-dimensional object. Typically, they give you rectangular solids, but they can also give you cylinders and other such objects. The key thing to remember about problems like this one is that effectively we’re stacking Pythagorean theorems to solve it – we’re finding triangles and then triangles within triangles that define the longest distance.

This type of problem is testing your spatial skills and a graphic or visual aid is often helpful though strictly not necessary. Let’s take a look at how to solve this problem and because it’s testing these skills the approach is generally mathematical that is there is some processing because it’s secondary to what they’re actually testing.

#### GMAT 3D Geometry Problem Introduction

So, we have this rectangular solid and it doesn’t matter which way we turn it – the longest distance is going to be between any two opposite corners and you can take that to the bank as a rule: On a rectangular solid the opposite corners will always be the longest distance. Here we don’t have any way to process this central distance so, what we need to do is make a triangle out of it.

Notice that the distance that we’re looking for along with the height of 5 and the hypotenuse of the 10 by 10 base will give us a right triangle. We can apply Pythagoras here if we have the hypotenuse of the base. We’re working backwards from what we need to what we can make rather than building up. Once you’re comfortable with this you can do it in either direction.

#### Solving the Problem

In this case we’ve got a 10 by 10 base. It’s a 45-45-90 because any square cut in half is a 45-45-90 which means we can immediately engage the identity of times root two. So, 10, 10, 10 root 2. 10 root 2 and 5 makes the two sides. We apply Pythagoras again. Here it’s a little more complicated mathematically and because you’re going in and out of taking square roots and adding and multiplying, you want to be very careful not to make a processing error here.

Careless errors abound particularly when we’re distracted from the math and yet we need to do some processing. So, this is a point where you just want to say “Okay, I’ve got all the pieces, let me make sure I do this right.” 10 root 2 squared is 200 (10 times 10 is 100, root 2 times root 2 is 2, 2 times 100 is 200). 5 squared is 25. Add them together 225. And then take the square root and that’s going to give us our answer. The square root of 225 is one of those numbers we should know. It’s 15, answer choice A.

Okay guys for another 3D and Geometry problem check out GMAT 680 Level Geometry Problem – No Math Needed! We will see you next time.

Posted on
14
Jul 2021

## GMAT Geometry Problem

Hey guys, top level geometry problems are characterized typically by stringing a whole bunch of different rules together and understanding how one thing relates to the next thing, to the next thing. Until you get from the piece of information you started with to the conclusion. We’re going to start out by taking a look at this problem using the z equals 50° and seeing how that information goes down the line.

But afterwards we’re going to see a super simple logical pathway utilizing a graphic scenario that makes the z equals 50° irrelevant. To begin with we’re being asked for the sum of x and y and this will come into play on the logical side. We need the sum not the individual amounts but let’s begin with the y. We have a quadrilateral and it has parallel sides which means the two angles z and y must equal 180°. That’s one of our geometric rules. If z is 50° that means y is 130° and we’re halfway there.

Next we need to figure out how x relates and there are several pathways to this. One way we can do it is drop. By visualizing or dropping a third parallel line down, intersecting x, so on the one hand we’ll have 90 degrees. We’ll have that right angle and on the other we’ll have that piece. Notice that the parallel line we dropped and the parallel line next to z are both being intersected by the diagonal line going through which means that that part of x equals z. So we have 50° plus 90° is 140°. 130° from the y, 140° from the x, gives us 270°.

Another way we can do this is by taking a look at the right triangle that’s already built in z is 50° so y is 1 30°. now the top angle in the triangle must then be 180° minus the 130° that is 50°. it must match the z again we have the parallel lines with the diagonal coming through then the other angle the one opposite x is the 180° degrees that are in the triangle minus the 90° from the right triangle brings us to 90° minus the 50° from the angle we just figured out means that it’s 40° which means angle x is 180° flat line supplementary angles minus the 40° gives us 140° plus the 130° we have from y again we get to 270°.

#### Graphic Solution Path

Now here’s where it gets really fun and really interesting. We can run a graphic scenario here by noticing that as long as we keep all the lines oriented in the same way we can actually shift the angle x up. We can take the line that extends from this big triangle and just shift it right up the line until it matches with the y. What’s going to happen there, is we’re going to see that we have 270° degrees in that combination of x and y and that it leaves a right triangle of 90°, that we can take away from 360° again to reach the 270°.

Here the 50° is irrelevant and watch these two graphic scenarios to understand why no matter how steep or how flat this picture becomes we can always move that x right up and get to the 270°. That is the x and y change in conjunction with one another as z changes. You can’t change one without the other while maintaining all these parallel lines and right angles. Seeing this is challenging to say the least, it requires a very deep understanding of the rules and this is one of those circumstances that really points to weaknesses in understanding most of what we learn in math class in middle school, in high school. Even when we’re prepping only scratches the surface of some of the more subtle things that we’re either allowed to do or the subtle characteristics of rules and how they work with one another and so a true understanding yields this very rapid graphic solution path.

#### Logical Solution Path

The logical solution path where immediately we say x and y has to be 270° no matter what z is and as you progress into the 80th, 90th percentile into the 700 level on the quant side this is what you want to look for during your self prep. You want to notice when there’s a clever solution path that you’ll overlook because of the rules. Understand why it works and then backtrack to understand how that new mechanism that you discovered fits into the framework of the rules that we all know and love. Maybe? I don’t know if we love them! But they’re there, we know them, we’re familiar with them, we want to become intimate. So get intimate with your geometry guys put on some al green light some candles and I’ll see you next time.

If you enjoyed this problem, try other geometry problems here: GMAT Geometry.

Posted on
16
Feb 2021

## Triangles With Other Shapes

As discussed before, now that we’ve talked about the basic triangles, we can start looking at how the GMAT can make problems difficult by embedding triangles in other figures, or vice versa.

Here are just a few examples, which include triangles within and outside of squares, rectangles, and circles:

Today, we’ll talk about some crucial connections that are often made between triangles and other figures, starting with the 45-45-90 triangle, also known as the isosceles right triangle.

You’ve probably seen a rectangle split in two along one of its diagonals to produce two right triangles:

But one of the oft-overlooked basic geometric truths is that when that rectangle is a square (and yes, remember a square is a type of rectangle), the diagonal splits the square into two isosceles right triangles. This makes sense when you think about it, because the diagonal bisects two 90-degree angles to give you two 45-degree angles:

(For clarification, the diagonal of a rectangle is a bisector when the rectangle is a square, but it is not a bisector in any other case.)

Another very common combination of shapes in more difficult GMAT Geometry problems is triangles with circles. This can manifest itself in three common ways:

## Triangles Created Using the Central Angle of a Circle

In this case, notice that two of the sides of the triangle are radii (remember, a radius is any line segment from the center of the circle to its circumference). What does that guarantee about the triangle?

Since two side are of equal length, the triangle is automatically isosceles. Remember that the two angles opposite those two sides are also of equal measure. So any triangle with the center of the circle as one vertex and points along the circumference as the other two vertices will automatically be an isosceles triangle.

## Inscribed Triangles

An inscribed triangle is any triangle with a circle’s diameter as one of its sides and a vertex along the circumference. And a key thing to note: an inscribed triangle will ALWAYS be a right triangle. So even if you don’t see the right angle marked, you can rest assured the inscribed angle at that third vertex is 90 degrees.

## Squares and Rectangles Inscribed in Circles

What’s important to note here is that the diagonal of the rectangle (or square) is equivalent to the diameter of the circle.

Now that we’ve seen a few common relationships between triangles and other figures, let’s take a look at an example Official Guide problem:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A) 19,200
B) 19,600
C) 20,000
D) 20,400
E) 20,800

## Explanation

The diagonal splits the rectangular park into two similar triangles:

## Use Signals to Avoid Algebra

It can be tempting to then jump straight into algebra. The formulas for perimeter and diagonal are P = 2L + 2W an D2 = L2 + W2, respectively, where L and W are the length and width of the rectangle. The second formula, you’ll notice, arises out of the Pythagorean Theorem, since we now have two right triangles. We are trying to find area, which is LW, so we could set out on a cumbersome algebraic journey.

However, let’s try to use some SIGNALS the problem gives us and our knowledge of how the GMAT operates to see if we can short-circuit this problem.

We know the GMAT is fond of both clean numerical solutions and common Pythagorean triples. The large numbers of 200 for the diagonal and 560 for the perimeter don’t change that we now have a very specific rectangle (and pair of triangles). Thus, we should suspect that one of our basic Pythagorean triples (3-4-5, 5-12-13, 7-24-25) is involved.

Could it be that our diagonal of 200 is the hypotenuse of a 3-4-5 triangle multiple? If so, the 200 would correspond to the 5, and the multiplying factor would be 40. That would also mean that the legs would be 3*40 and 4*40, or 120 and 160.

Does this check out? Well, we’re already told the perimeter is 560. Adding 160 and 120 gives us 280, which is one length and one width, or half the perimeter of the rectangle. We can then just double the 280 to get 560 and confirm that we do indeed have the correct numbers. The length and width of the park must be 120 and 160. No algebra necessary.

Now, to get the area, we just multiply 120 by 160 to get 19,200 and the final answer of A.

Check out the following links for our other articles on triangles and their properties:

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 2021

Posted on
02
Feb 2021

## Similar Triangles – GMAT Geometry

Triangles, and Geometry in general, are something you need to master as much as you can before your GMAT exam. It can seem complicated in the beginning, but once you start working with different geometric objects and its shapes, thing will definitely get easier. The topic of this article is similar triangles. One of the most important things to highlight here is that “similar” does not necessarily mean “identical.” Two triangles can be similar without being the same size. For example, take the following:

Even though the triangles are of different size, notice that the angles remain the same. This is what really defines the triangles as similar.

Now, what makes this interesting is that the measurements associated with the triangle increase proportionally. For example, if we were to present a triangle with lengths 3, 5, and 7, and we were to then tell you that a similar triangle existed that was twice as large, the corresponding side lengths of that similar triangle would have to be 6, 10, and 14. (This should be no surprise considering our lesson on multiples of Pythagorean triples, such as 3-4-5 leading to 6-8-10, 9-12-15, etc.)

You can also extend this to Perimeter, as perimeter is another one-dimensional measurement. So, if for example we ask:

A triangle has line segments XY = 6, YZ = 7, and XZ = 9. If Triangle PQR is similar to Triangle XYZ, and PQ = 18, as shown, then what is the perimeter of Triangle PQR?

Answer: Perimeter is a one-dimensional measurement, just as line segments are. As such, since PQ is three times the length of XY, that means the perimeter of Triangle PQR will be three times the perimeter of Triangle XYZ as well. The perimeter of Triangle XYZ is 6+7+9 = 22. We simply multiply that by 3 to get the perimeter of Triangle PQR, which is 66.

Things can get a little more difficult with area, however, as area is a two-dimensional measurement. If I double the length of each side of a triangle, for example, how does this affect the area? Think about it before reading on…

## Scenario

Suppose we had a triangle that had a base of 20 and a height of 10:

The area would be 20*10 / 2 = 100.

Now, if we double each side of the triangle, what effect does that have on the height? Well, the height is still a one-dimensional measurement (i.e. a line segment), so it also doubles. So the new triangle would have a base of 40 and a height of 20. That would make the area 40*20 / 2 = 400.

Notice that since the original area was 100 and the new area is 400, the area actually quadrupled, even though each side doubled. If the base and height are each multiplied by 2, the area is multiplied by 22. (There’s a connection here to units, since units of area are in square measurements, such as square inches, square meters, or square feet.)

Now, let’s take a look at the following original problem:

Triangle ABC and Triangle DEF are two triangular pens enclosing two separate terrariums. Triangle ABC has side lengths 7 inches, 8 inches, and 10 inches. A beetle is placed along the outer edge of the other terrarium at point D and traverses the entire perimeter once without retracing its path. When finished, it was discovered that the beetle took three times as long as it did traversing the first terrarium traveling at the same average speed in the same manner. What is the total distance, in inches, that the beetle covered between the two terrariums?

A) 25
B) 50
C) 75
D) 100
E) 125

## Explanation

This one has a few traps in store. Hopefully you figured out the significance of the beetle taking three times as long to traverse the second terrarium at the same average speed: it’s confirmation that the second terrarium has three times the perimeter of the first. At that point, you can deduce that, since the first terrarium has perimeter 7+8+10 = 25, the second one must have perimeter 25*3 = 75. However, it can be tempting to then choose C, if you don’t read the question closely. Notice the question effectively asks for the perimeters of BOTH terrariums. The correct answer is D.

## GMAT Triangle Series Articles

By: Rich Zwelling, Apex GMAT Instructor
Date: 2nd February, 2021

Posted on
14
Jan 2021

## The Area of an Equilateral Triangle

As promised, we will now connect the 30-60-90 triangle to the equilateral triangle, specifically its area. There is a formula for the area of an equilateral triangle as it relates to the length of its side s, and it is as follows:

But more likely than not for the GMAT, you’ll need to understand how this formula is derived. And the √3 term in the area is a big clue.

First, it helps to remember that an equilateral triangle has all equal angles as well as all equal sides. And given that the angles in a triangle must sum to 180 degrees, each angle must be 60 degrees:

Now, what happens when we take such a triangle and split it down the middle?

This should look familiar. Because the line segment down the middle acts as an angle bisector, the 60 degree angle at the top vertex becomes two 30-degree angles. Take a moment to consider what this produces and what the implications are.

As you might have guessed, this line segment produces two 30-60-90 right triangles:

Not only that, but we can then use s to denote the side length of the equilateral triangle and map out each segment of the 30-60-90 right triangles. Before viewing the diagram below, take a moment to consider what the height of the triangle would be.

Remember that the ratio of side lengths is 1 : √3 : 2. If we fill in all of the appropriate lengths, we would get the following:

Now, we’re very close to deriving the area of the triangle, which is simply base*height/2. In this case, the base is s, while the height is s√3/2.

This is how we finally get the universal formula for an equilateral triangle:

Area = base * height / 2
Area = (s) * (s√3/2) / 2
Area = (s) * (s√3/4)
Area = (s2√3) / 4.

Now that we’ve seen the relationship between equilateral and 30-60-90 triangles, let’s see how it plays out in an official GMAT problem:

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

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

## Using signals

This is a complex problem that seems intimidating at first. However, if we use signals the problem is giving us, we can get to the answer more quickly than we might initially think. What signals does the area of the shaded region give us? Think about it before reading on…

If we look closely at the diagram, we see that an equilateral triangle is involved. We know this because each side of the triangle consists of two radii of each circle (i.e. the distance from the center to the outer edge of the circle), thus each side of the triangle must be equal. That’s a big hint that the √3 term is linked to the area formula we’ve been talking about.

Likewise, although it is not the subject of this post, the term using π is associated with circles in this case, the areas of the identical circles. (For reference, the area of a circle is πr2, and the circumference of a circle is 2πr.)

Conceptually, we should be able to see that 64√3 – 32π represents the area of the equilateral triangle minus the area of the three small sectors from the circles

Now, rather than do any unnecessarily complicated math, we should take notice that the question asks for the radius of each circle, and each side of the equilateral triangle is 2r:

We already know that the area of the equilateral triangle is 64√3, and we have the formula for that area, so we are just a few steps away from solving for the radius.

Remember the formula, where s is the length of the side of the equilateral triangle:
Area = (s2√3) / 4

Substitute:
64√3 = (s2√3) / 4

Since √3 is common to both sides, you can divide it out:

64 = s2 / 4
256 = s2

Now, normally, you would say that s could be 16 or -16, but since this is a geometric quantity, we only deal in nonnegative quantities. Therefore:

s = 16, giving us the length of each side of the equilateral triangle.

Be careful, however. This could trap you into picking answer choice C. Remember to check exactly what the question asks for. We were asked for the radius of the circle, which as we see in the above diagram is half the length of s. The correct answer is B.

Again, it’s very important to notice that we didn’t do anything with the circles. The 64√3 term and the equilateral triangle were enough to get us the length of each side and thus the radius. Look for signals to help short-circuit problems and avoid lengthy solution paths.

Now that we’ve reviewed all of the basic triangles, we’ll do a little more next time on how triangles can appear in other shapes, such as circles and rectangles. We got a little taste today, so hopefully that will give you a good idea.

By: Rich Zwelling, Apex GMAT Instructor
Date: 14th January, 2021

Posted on
08
Jan 2021

## The 30-60-90 Right Triangle

By: Rich Zwelling, Apex GMAT Instructor
Date: 7th January, 2021

## 30-60-90 Right Triangle

In a previous piece, we covered the 45-45-90 right triangle, also known as the isosceles right triangle. There is another so-called “special right triangle” commonly tested on the GMAT, namely the 30-60-90 right triangle.

Like the isosceles right, its sides always fit a specific ratio, as seen in the above diagram (1 : √3 : 2). And it’s worth noting, as with all triangles, that the shortest side is opposite the smallest angle, while the longest side is opposite the largest angle, etc.

Now, it’s easy enough to memorize this ratio and deduce what each side length will be, given that we are dealing with a 30-60-90 triangle. For example, Suppose we are given the following information:

This is low-level memorization, and we can deduce that the side opposite the 60-degree angle will be length 5√3, while the hypotenuse will be length 10.

But let’s look to this GMAT Official Guide problem to see something a little more high-level. Give it a shot before reading further:

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object?

(A) 10−53
(B) 10−52
(C) 2
(D) 2 1/2
(E) 4

(For starters, notice that the question they’re asking for — the distance between the actual position and the perceived position — is just line segment RS. Remember that the GMAT is very good at using complicated wording to frame a simple concept. Always simplify the question as quickly as possible.)

To understand this problem, let’s first talk about one of the higher-level ways the GMAT could test 30-60-90 triangles. Take this example:

Notice we are given no angles except the right angle. But we do have 2 sides and 1 angle in total, which is sufficient to form a unique triangle. Furthermore, did you identify anything that gives this away as a 30-60-90?

The hypotenuse is twice the length of one of the sides, giving them a 2:1 ratio. That guarantees that the third side fits the √3 component of our ratio, giving that side a length of 5√3. So even without labeled angles:

A right triangle with a hypotenuse twice the length of one of its legs must be a 30-60-90 triangle.

That’s much more the kind of critical thinking the GMAT is interested in testing.

Similarly, in this Official Guide problem, we are told that VR is length 10:

Notice that at this point, it’s up to you to make the deduction that we have a 30-60-90 triangle, and thus the distance from the right angle marker to point R must be 5√3:

From there, it’s straightforward to see that RS is simply the marked length of 10 minus the length of 5√3 we just deduced, thus leading us to answer choice A.

In terms of strategy, another point: a brief look at the answer choices at the start of the problem gives a strong hint that either a 30-60-90 or 45-45-90 triangle is involved. Notice that the first two answers feature a √3 and a √2 term, and this is clearly a geometry question. This gives you the opportunity to be preemptive and use the test’s patterns against itself.

It goes without saying that these kinds of problems get easier to be solved with practice. Solve as many problems as you can, compare and contrast the strategies you used, and make sure you understand the process properly. You can also consider having a GMAT tutor to help you during your GMAT prep and get a stellar score.

In our next post, we’ll talk about how 30-60-90 triangles can be used directly to calculate the area of equilateral triangles. You can also link to our other article about triangles:

Posted on
06
Jan 2021

## 45-45-90 Right Triangle – GMAT Geometry Guide

By: Rich Zwelling, Apex GMAT Instructor
Date: 6th January, 2021

## 45-45-90 Right Triangle

Another of the commonly tested triangles on the GMAT is the 45-45-90, also known as the isosceles right triangle. Know that term, as it could appear by name in a question.

As shown in the above diagram, the side lengths of this triangle always fit the same ratio (1 : 1 : √2) , where the legs are the same length and the hypotenuse length is √2 times the leg length. For example, if the leg lengths were 3 instead of 1, then the hypotenuse would be 3√2 instead of simply √2.

But likewise, don’t forget that you can go backwards and divide the hypotenuse length by √2 to get to the leg length. It may seem obvious, but it presents an important point: what’s more important than simply memorizing the ratio is understanding the mathematical relationship between the side lengths. This will help you avoid trouble if the GMAT happens to give you a problem that doesn’t conform to expectations.

For example, the following problem fits expectations quite nicely:

A yard in the shape of an isosceles right triangle has a hypotenuse of length 10√2. What is the area of this yard?

From this information, it’s easy enough to deduce that the leg length is 10, and we can draw a diagram that looks roughly like this:

From there, we can easily calculate the area, which is base*height / 2, or in this case 10*10/2 = 50.

But what happens if we give the problem a little twist:

A yard in the shape of an isosceles right triangle has a hypotenuse of length 10. What is the area of this yard?

Did you catch the twist? We’re used to the hypotenuse including a √2. This is what the GMAT will do. They’ll throw you off-center, and you’ll have to adjust. But this is also why we said earlier that what matters more than memorizing the ratio of sides is understanding the relationships between the sides of an isosceles right triangle…

Remember we said that, just as we multiply the leg length by √2 to get to the hypotenuse length, so we must divide the hypotenuse length by √2 to get to the leg length. That must mean each leg has length 10/√2.

You can then take 10/√2 and multiply it by √2/√2 to de-radicalize the denominator and get (10√2) / 2, or a leg length of 5√2:

Notice again that we have a more unfamiliar form, with the √2 terms in the legs and an integer in the hypotenuse. We can’t count on the GMAT to give us what we’re used to.

Now we can calculate the area:

Area = (base*height)/2 = (5√2)(5√2)/2 = (5*5)(√2*√2)/2 = (25)*(2) / 2 = 25

### Problem #1

Now, to try this on your own, take a look at this Official Guide problem:

If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?

(A)   40
(B)   60
(C)   80
(D)   100
(E)   120

Explanation:

This is a nice change-up, because it involves another shape. Did you notice that splitting a square along its diagonal creates two isosceles right triangles

Once you realize this, you can divide 20 by √2 to get 20/√2, then multiply top and bottom by √2 to get x=10√2.

Since the question asks for perimeter, we can multiply this by four to get 40√2.

The final step is to realize that √2 is approximately 1.4. If we multiply 40 by 1.4, the only answer choice that possibly makes sense is 60, and thus the correct answer is B

Obviously, practice is always the key for problems like this. All you need to do is remember the formulas we used above and try to tackle different kinds of problems that are related to this topic. In addition, working with a GMAT tutor can be a great addition to your GMAT prep. There are many strategies and techniques that they will provide you with which will make your GMAT journey smoother and more productive.

After reviewing the 45-45-90 triangle identity, these further articles in the triangle geometry series will take you through more identities, each of the specific triangles and how the GMAT uses them to test your critical and creative solving skills:

Posted on
15
Dec 2020

## The 3-4-5 Right Triangle – GMAT Geometry Guide

By: Rich Zwelling, Apex GMAT Instructor
Date: 17th December, 2020

## Right Triangle Identities: 3-4-5

Right triangles always adhere to the same basic relationship, reflected by the Pythagorean Theorem, or + b² = c², where a, b, and c match the triangle sides as pictured above. c always represents the longest side, called the hypotenuse.

But rather than use the formula directly, the most common way the GMAT will test knowledge of the formula is through the simplest integer values that fit this relationship. The most common is + 4² = 5² → 9 + 16 = 25, as pictured below:

What’s important to remember is that this relationship works not only for 3-4-5, but also for any corresponding multiples, such as 6-8-10 or 9-12-15 or any other multiples of the original numbers.

#### GMAT Triangle Problem #1

If you rely solely on the formula, you could certainly get the job done, but it will take you a lot longer. Here’s an Official Guide problem that drives this point home:

The figure above shows a path around a triangular piece of land. Mary walked the distance of 8 miles from P to Q and then walked the distance of 6 miles from Q to R. If Ted walked directly from P to R, by what percent did the distance that Mary walked exceed the distance that Ted walked?

(A)   30%
(B)   40%
(C)   50%
(D)   60%
(E)   80%

If you know your so-called “Pythagorean Triples” from memory (e.g. 3-4-5, 6-8-10), this problem moves along much faster. By test day, you should know within seconds that segment PR is length 10, no calculations involved.

After that, the bulk of your time should be spent calculating the percent difference between Mary’s total distance (14) and Ted’s total distance (10). (Answer: Since Mary walked 4 more miles more than Ted’s original 10, and 4 is 40% of 10, this makes B the correct answer.)

#### GMAT Triangle Problem #2

Also, it’s much more likely that the GMAT will test your knowledge of completeness of information with respect to Right Triangles, especially on Data Sufficiency. Give this problem a shot before reading on:

What is the area of triangle ABC pictured above?

1. The length of segment AB is 5
2. The perimeter of triangle ABC is 12

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

If you know your 3-4-5 Pythagorean triple really well, you may prematurely view Statement (1) as sufficient, since you may believe the hypotenuse of 5 automatically guarantees the legs of the triangle must be 3 and 4. But this assumes the legs must have integer lengths. In reality, the legs could be any non-integer lengths that satisfy + b² = 5²

This is a classic way the GMAT could throw you off your guard. And it’s another way to test that only one side of a triangle is not enough to give you complete information about the entire triangle. Statement (1) is actually INSUFFICIENT, because we do not have information about a unique triangle, and thus could not possibly know the area.

Likewise, Statement (2) is INSUFFICIENT, because there are many ways to generate a right triangle with a perimeter of 12.

When we combine the statements, however, it’s interesting to note that, as a rule, we know we have a unique triangle if we’re given both the perimeter and the hypotenuse. As such, we would be able to find the area (even though we don’t have to calculate it), and thus the answer is C. Don’t do any math!

#### Takeaways

The big takeaway here is that, rather than have you use the Pythagorean theorem directly, the GMAT will try to force you into false conclusions, such as believing a hypotenuse of 5 gives you all the information you need. Be on your toes! Make sure to thoroughly examine all information given to you!

The 3-4-5 triangle is not the only identity to review in this triangle geometry section, here are some other identities and triangle related topics to review:
A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles
Triangle Inequality Rule

Posted on
15
Dec 2020

## A short meditation on GMAT triangles – Read this if you struggle with triangle problems on the GMAT!

GMAT Triangles

Which of the following did you picture? Something different?

Well, every one of the above is tested on the GMAT. And you can’t stick to just one conception of a single shape. You have to be familiar with different types of triangles and all of their characteristics.

For example, suppose you need to find the two-dimensional area within a triangle. What’s the proper way to do this for a right triangle (2), which contains a 90-degree angle? Even if I know the proper formula, would applying it prove different if I needed to use it on triangle (4), an obtuse triangle, which has one angle greater than 90 degrees?

What about the perimeter, or the sum of the lengths of the three sides of the triangle? Are there simple, straightforward ways to calculate this in the case of the equilateral triangle (1), where each side is the same length. Or the isosceles triangle (3), where two of the three sides are the same length?

What about that isosceles triangle? Did you notice that the two sides that are the same length are opposite two angles of equal measure? What are the implications of this?

How about the fact that all the angles in any triangle add up to the same degree measure, no matter how the triangle looks?

And then there’s something as simple as the base of the triangle. Take triangle (4):

As shown, the shortest side is the base, or bottom side, of the triangle. But in truth, any side of a triangle could be considered the base. What if we took the same triangle and rotated it clockwise:

Now suddenly, the same triangle has the longest side as its base. This could completely change how you calculate, for example, the area of the triangle.

What about the relationships among the sides of a right triangle, for example?

It turns out that the sides a, b, and c, of all right triangles conform to a special mathematical relationship, no matter their lengths. And in its most elementary forms as well. For example, did you know that if a right triangle has short sides (legs) of lengths 3 and 4, the longest side (hypotenuse) will always be 5?

The GMAT is less interested in very complex math and much more interested in variations on basic facts like this one.

Also, what happens when triangle types are combined? What happens when a triangle is both isosceles AND right? How does this affect the relationships among the sides?

Another way the test can give triangles (and other shapes) a twist is to combine them with other shapes. You may see questions that involve, for example, a triangle within a circles (inscribed):

This presents all sorts of opportunities for testing not only the measurements of triangles and of circles individually but also the relationships among those measurements and the ways in which they might overlap

This is of course meant to be a brief overview of the basic forms GMAT triangles can take. But the above poses critical thinking questions you should examine as you move forward with your preparation. Remember: don’t just think in terms of memorizing facts! There will be a handful of formulas to know, to be sure, but once those are out of the way, it will be up to you to be on the lookout for new ways that knowledge could be applied and tested.

As the next step in your discovery and practice of GMAT triangles visit these posts:

By: Rich Zwelling, Apex GMAT Instructor
Date: 15th December, 2020