3 Ways to Solve a 750+ Level GMAT Question About Irregular Polygons

We have examined how to deal with polygons when you encounter them on a GMAT question in a previous post. Today, we will look at a relatively difficult polygon question, however we would like to remind you here that the concepts being tested in this question are still very simple (although we won’t give away exactly which concepts they are yet). First, take a look at the question itself:

The hexagon above has interior angles whose measures are all equal. As shown, only five of the six side lengths are known: 10, 15, 4, 18, and 7. What is the unknown side length?

(A) 7
(B)10
(C) 12
(D) 15
(E) 16

There are various ways to solve this question, but each takes a bit of effort. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. The angles, however, are all equal. Let’s first find the measure of each one of those angles using the formula discussed in this previous post.

(n – 2)*180 = sum of all interior angles
(6 – 2)*180 = 720
Each of the 6 angles = 720/6 = 120 degrees

Though we would like to point out here that if you see a question such as this one on the actual GMAT exam, you should already know that if each angle of a hexagon is equal, each angle must be 120 degrees, so performing the above calculation would not be necessary.

Method 1: Visualization
This is a very valid approach to obtaining the correct answer on this GMAT question since we don’t need to explain the reasoning or show our steps, however it may be hard to comprehend for the beginners. We will try to explain it anyway, since it requires virtually no work and will help build your math instinct.

Note that in the given hexagon, each angle is 120 degrees – this means that each pair of opposite sides are parallel. Think of it this way: Side 4 turns on Side 18 by 120 degrees. Then Side 15 turns on Side 4 by another 120 degrees. And finally, Side 10 turns on Side 15 by another 120 degrees. So Side 10 has, in effect, turned by 360 degrees on Side 18.

This means Side 10 is parallel to Side 18.

Now, think of the 120 degree angle between Side 4 and Side 15 – it has to be kept constant. Plus, the angles of the legs must also stay constant at 120 degrees with Sides 10 and 18. Since the slopes of each leg of that angle are negatives of each other (√3 and -√3), when one leg gets shorter, the other gets longer by the same length (use the image below as a visual of what we’re talking about).

Hence, the sum of the sides will always be 15 + 4 = 19. This means 7 + Unknown = 19, so Unknown = 12. Our answer is C.

If you struggled to understand the approach above, you’re not alone. This method involves a lot of intuition, and struggling to figure it out may not be the best use of your time on the GMAT, so let’s examine a couple of more tangible solutions!

Method 2: Using Right Triangles
As we saw in Method 1 above, AB and DE are parallel lines. Since each of the angles A, B, C, D, E and F are 120 degrees, the four triangles we have made are all 30-60-90 triangles. The sides of a 30-60-90 triangle can be written using the ratio 1:√(3):2.

AT = 7.5*√3 and ME = 2*√3, so the distance between the sides of length 10 and 18 is 9.5*√3. We know that DN = 3.5*√3, so BP = (9.5*√3) – (3.5*√3) = 6*√3.

Since the ratios of our sides should be 1:√(3):2, side BC = 2*6 = 12. Again, the answer is C. Let’s look at our third and final method for solving this problem:

Method 3: Using Equilateral Triangles
First, extend the sides of the hexagon as shown to form a triangle:

Since each internal angle of the hexagon is 120 degrees, each external angle will be 60 degrees. In that case, each angle between the dotted lines will become 60 degrees too, and hence, triangle PAB becomes an equilateral triangle. This means PA = PB = 10. Triangle QFE  and triangle RDC also become equilateral triangles, so QF = QE = 4, and RD = RC = 7.

Now note that since angles P, Q, and R are all 60 degrees, triangle PQR is also equilateral, and hence, PQ = PR.

PQ = 10 + 15 + 4 = 29
PR = 10 + BC + 7 = 29
BC = 12 (again, answer choice C)

Note the geometry concepts that we used to solve this problem: regular polygon, parallel lines, angles, 30-60-90 right triangles, and equilateral triangles. We know all of these concepts very well individually, but applying them to a GMAT question can take some ingenuity!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Today, we will look at a Geometry concept involving parallel lines and transversals (a line that cuts through two parallel lines). This is the property:

The ratios of the intercepts of two transversals on parallel lines is the same.

Consider the diagram below:

Here, we can see that:

• “a” is the intercept of the first transversal between L1 and L2.
• “b” is the intercept of the first transversal between L2 and L3.
• “c” is the intercept of the second transversal between L1 and L2.
• “d” is the intercept of the second transversal between L2 and L3.

Therefore, the ratios of a/b = c/d. Let’s see how knowing this property could be useful to us on a GMAT question. Take a look at the following example problem:

In triangle ABC below, D is the mid-point of BC and E is the mid-point of AD. BF passes through E. What is the ratio of AF:FC ?

(A) 1:1
(B) 1:2
(C) 1:3
(D) 2:3
(E) 3:4

Here, the given triangle is neither a right triangle, nor is it an equilateral triangle. We don’t really know many properties of such triangles, so that will probably not help us. We do know, however, that AD is the median and E is its mid-point, but again, we don’t know any properties of mid-points of medians.

Instead, we need to think outside the box – parallel lines will come to our rescue. Let’s draw lines parallel to BF passing through the points A, D, and C, as shown in the diagram below:

Now we have four lines parallel to each other and two transversals, AD and AC, passing through them.

Consider the three parallel lines, “line passing through A”, “BF”, and “line passing through D”. The ratio of the intercepts of the two transversals on them will be the same.

AE/ED = AF/FP

We know that AE = ED since E is the mid point of AD. Hence, AE/ED = 1/1. This means we can say:

AE/ED = 1/1 = AF/FP
AF = FP

Now consider these three parallel lines: “BF”, “line passing through D”, and “line passing through C”. The ratio of the intercepts of the two transversals on them will also be the same.

BD/DC = FP/PC

We know that BD = DC since D is the mid point of BC. Hence, BD/DC = 1/1. This means we can also say:

BD/DC = 1/1 = FP/PC
FP = PC

From these two calculations, we will get AF = FP = PC, and hence, AF:FC = 1:(1+1) = 1:2.

Therefore, the answer is B. We hope you see that Geometry questions on the GMAT can be easily resolved once we bring in parallel lines.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The Pythagorean Triples Properties You’ll See on the GMAT

Today, let’s discuss a few useful properties of primitive Pythagorean triples. A primitive Pythagorean triple is one in which a, b and c (the length of the two legs and the hypotenuse, respectively) are co-prime. So, for example, (3, 4, 5) is a primitive Pythagorean triple while its multiple, (6, 8, 10), is not.

Now, without further ado, here are the properties of primitive Pythagorean triples that you’ll probably encounter on the GMAT:

I. One of a and b is odd and the other is even.
II. From property I, we can then say that c is odd.
III. Exactly one of a, b is divisible by 3.
IV. Exactly one of a, b is divisible by 4.
V. Exactly one of a, b, c is divisible by 5.

If you keep in mind the first primitive Pythagorean triple that we used as an example (3, 4, 5), it is very easy to remember all these properties.

If we look at some other examples:

(3, 4, 5), (5, 12, 13), (8, 15, 17) (7, 24, 25) (20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53) (11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73), etc.

we will see that these properties hold for all primitive Pythagorean triples.

Now, let’s take a look at an example GMAT question which can be easily solved if we know these properties:

The three sides of a triangle have lengths p, q and r, each an integer. Is this triangle a right triangle?

Statement 1: The perimeter of the triangle is an odd integer.
Statement 2: If the triangle’s area is doubled, the result is not an integer.

We know that the three sides of the triangle are all integers. So if the triangle is a right triangle, the three sides will represent a Pythagorean triple. Given that p, q and r are all integers, let’s use the properties of primitive Pythagorean triples to break down each of the statements.

Statement 1: The perimeter of the triangle is an odd integer.

Looking at the properties above, we know that a primitive Pythagorean triple can be represented as:

(Odd, Even, Odd) (The first two are interchangeable.)

Non-primitive triples are made by multiplying each member of the primitive triple by an integer n greater than 1. Depending on whether n is odd or even, the three sides can be represented as:

(Odd*Odd, Even*Odd, Odd*Odd) = (Odd, Even, Odd)
or
(Odd*Even, Even*Even, Odd*Even) = (Even, Even, Even)

However, the perimeter of a right triangle can never be odd because:

Odd + Even + Odd = Even
Even + Even + Even = Even

Hence, the perimeter will be even in all cases. (If the perimeter of the given triangle is odd, we can say for sure that it is not a right triangle.) This statement alone is sufficient.

Statement 2: If the triangle’s area is doubled, the result is not an integer.

If p, q and r are the sides of a right triangle such that r is the hypotenuse (the hypotenuse could actually be either p, q, or r but for the sake of this example, let’s say it’s r), we can say that:

The area of this triangle = (1/2)*p*q
and
Double of area of this triangle = p*q

Double the area of the triangle has to be an integer because we are given that both p and q are integers, but this statement tells us that this is not an integer. In that case, this triangle cannot be a right triangle. If the triangle is not a right triangle, double the area would be the base * the altitude, and the altitude would not be an integer in this case.

This statement alone is sufficient, too. Therefore, our answer is D.

As you can see, understanding the special properties of primitive Pythagorean triples can come in handy on the GMAT – especially in tackling complicated geometry questions.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Find the Maximum Distance Between Points on a 3D Object

How do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal?

A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)

There are various different diagonals in a rectangular solid. Look at the given figure:

BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H?

The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH.

The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH.

Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult.

In our question we know that:

l = 10 inches
w = 10 inches
h = 5 inches

Let’s consider the right triangle DHB. DH is the length, so it is 10 inches.

DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem:

DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125

Going back to triangle DHB, we can now say that:

BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15

Thus, our answer to this question is A.

Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out:

The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C?

(A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15

Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:

The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points:

Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5

In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Quarter Wit, Quarter Wisdom: Beware of Sneaky Answer Choices on the GMAT!

Test-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2)
(B) 24√(3)
(C) 72√(2)
(D) 144√(2)
(E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that  there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

How to Use the Pythagorean Theorem With a Circle

It does not surprise anyone when they learn that the properties of circles are tested on the GMAT. Most test-takers will nod and rattle off the relevant equations by rote: Area = Π*radius ^2; Circumference  = 2Π* radius; etc. However, many of my students are caught off guard to learn that the equation for a circle on the coordinate plane is our good friend the Pythagorean theorem. Why on earth would an equation for a right triangle describe a circle?

Take a look at the following diagram in which a circle is centered on the origin (0,0) in the coordinate plane:

Designate a random point on the circle (x,y.) If we draw a line from the center of the circle to x,y, that line is a radius of the circle. Call it r. If we drop a line down from (x,y) to the x-axis, we’ll have a right triangle:

Note that the base of the triangle is x, and the height of the triangle is y. So now we have our Pythagorean theorem: x^2 + y^2 = r^2. This is also the equation for a circle centered on the origin on the coordinate plane. [The more general equation for a circle with a center (a,b) is (x-a)^2 + (y-b)^2 = r^2. When a circle is centered on the origin, (a,b) is simply (0,0.)]

This ends up being an immensely useful tool to use on the GMAT. Take the following question, for example:

A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?

(1) The radius of the circle is 4
(2) The sum of the coordinates of P is 0

So let’s draw this, designating P as (x,y):

Now we draw our trust right triangle by dropping a line down from P to the x-axis, which will give us this:

We’re looking for x^2 + y^2. Hopefully, at this point, you notice what the question is going for – because we have a right triangle, x^2 + y^2 = r^2, meaning that all we need is the radius!

Statement 1 is pretty straightforward – if r = 4, we can insert this into our equation of x^2 + y^2 = r^2 to get x^2 + y^2 = 4^2. So x^2 + y^2 = 16. Clearly, this is sufficient.

Now look at Statement 2. If the sum of x and y is 0, we can say x = 1 and y = -1 or x = 2 and y = -2 or x = 100 and y = -100, etc. Each of these will yield a different value for x^2 + y^2, so this statement alone is clearly not sufficient. Our answer is A.

Takeaway: any shape can appear on the coordinate plane. If the shape in question is a circle, remember to use the Pythagorean theorem as your equation for the circle, and what would have been a challenging question becomes a tasty piece of baklava. (We are talking about principles elucidated by the ancient Greeks, after all.)

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

Improve Your Speed on the ACT Math Section Using Math Fluidity

Speed is key on the Math Section of the ACT – you have only 60 minutes to complete 60 questions. However, this doesn’t mean you should spend one minute on each question, as not every question on in this section is created equal. Many questions (particularly Questions 1-30) are problems that you can solve in under one minute. In fact, you should aim to solve Questions 1-30 in less than 30 minutes – around 25 minutes is the goal.

That’s because some of the later questions, particularly the questions from Questions 40-60, will require more than a minute. Basically, you want to put aside extra time for the tricky questions at end of the section by completing the easier, earlier questions as quickly as possible. If you do Questions 1-30 in 25 minutes, then you have 35 minutes to do Questions 31-60.

One way to improve your speed on the Math Section is to develop what I call “math fluidity.” That means recognizing how common patterns, formulas and special rules can help you solve any particular problem. To illustrate, take a look at the following triangle problem:

Triangle ABC (below) is an equilateral triangle with side of length 4. What is the area of triangle ABC?

The first step to any geometry problem is writing down what relevant common formula you’ll need to solve the problem; i.e. whenever I’m asked the area of a triangle, at the top of my work space I’ll write:

A = (b*h)/2

Having the formula in front of you will be helpful because right away, it’s clear that although we have some information, we don’t have all the information we need to solve this problem – we have the base of the triangle (4), but not the height. Since the height of an equilateral triangle always goes from one angle to the opposite side, where it forms two 90-degree angles, drawing the height of an equilateral triangle creates two identical triangles, as shown below:

Many students would now conclude that they need the Pythagorean theorem to solve for the height (that line bisecting the equilateral triangle). This is where math fluidity comes in. Although you could use the Pythagorean theorem, it’s much faster to instead recognize what type of triangle you are dealing with.

Whenever you split an equilateral triangle in half, you create two 30-60-90 triangles. These are also called “special right triangles” because they always follow the rule that the shortest side is always “x,” the side opposite the 60-degree angle is always x√3, and the hypotenuse is always 2x. See the triangle below:

So, rather than spend any time solving for the height of the our triangle by using the Pythagorean Theorem, recognize that because the hypotenuse is 4 and the base is 2 (of either of the smaller triangles), and because the triangle is a right triangle, the height must be 2√3. Therefore, the area of the larger triangle is  (2√3)(4)(1/2), which equals 4√3.

Instantly recognizing that the two smaller triangles are 30-60-90 triangles only saves a little bit of time – if you can regularly shave off 20 seconds on question after question by recognizing special rules or how best to apply formulas, you’ll accrue saved time that can later be spent on harder math questions. Speaking of which, math fluidity also applies to tricky questions – similar to what we previously saw, recognition will break down hard questions into easier, faster steps.

So, let’s take a look at a more difficult question. Note, this next example is especially relevant for students shooting for 99th percentile or perfect scores. Although many students can solve the following question if given enough time, few students can solve it quickly enough to get it correct on the ACT. Here’s the problem:

In triangle ABC below, angle BAE measures 30 degrees. What is the value of angle AED minus angle ABE?

A) 30
B) 60
C) 90
D) 120
E) 150

Although there are several ways to solve this problem, math fluidity will help with whatever approach you choose. As I mentioned earlier, it is always best to start by writing down a relevant formula, as it will include what information you have and what information you need. In this case, I’m looking for AED-ABE. Because I’ve also been given the measure of angle BAE, I’ll write down:

BAE = 30 and BAE + ABE = AED

Here’s where math fluidity comes in; the second formula is based off a theorem that you probably learned (and then forgot!) in your geometry class. I do recommend (re)memorizing it for the ACT as follows: a measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

Are you drawing a blank? If so, take a moment to think about why that statement is true. If the smaller two angles of a right-angle triangle, as shown at left, are 40 and 50, then if we extend a line as shown to form the adjacent exterior angle x, then x + 50 = 180, so x = 130.

Also, 40 + 50 + 90 = 180, since the sum of interior angles of a triangle always add up to 180. So, if x + 50 = 180, and  40 + 50 + 90 = 180, then x+ 50 = 40 + 50 + 90.

Removing the 50 from both sides, we can conclude that x = 40 + 90, or x (the adjacent  exterior angle of one interior angle) is equal to the sum of the other two interior angles.

Now, returning to our original problem:

If BAE = 30 and BAE + ABE = AED, then:

30 + ABE = AED

AED – ABE = 30

Therefore, our answer is A, 30.

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GMAT Tip of the Week

Approach Geometry Questions from the Right (tri)Angle

(This is one of a series of GMAT tips that we offer on our blog.)

As the writers of the GMAT create difficult quantitative questions, one favorite technique is to require test-takers to use skills that would not, on the surface, seem relevant. As examinees struggle to find a foothold on the question with the tools that seem more obvious by comparison, they lose time, make calculation errors, and suffer not only from an incorrect answer but also from a decreased level of confidence that carries on to future questions.

One such seemingly-irrelevant skill that appears on a majority of geometry-based questions is the right triangle, a shape for which you should train yourself to look whenever a geometry question appears. Every shape that the GMAT tests could contain a right triangle:

• The area of a non-right triangle is created by taking a perpendicular line from the base to the angle opposite it; this process divides the original triangle in to two right triangles.
• The diagonal of a square or rectangle is a right triangle.
• The area of a parallelogram or trapezoid is found by taking a perpendicular line between the bases; doing so can create a right triangle.
• Even in a circle, connecting the diameter to a point on the edge of the circle, using two chords to do so, will create a right triangle with the diameter as the hypotenuse.
• In 3-D figures, the longest difference between two points on the item is always a hypotenuse of a right triangle that must be drawn in space.

By training yourself to look for opportunities to use your knowledge of right triangles, even if it doesn’t appear to be the obvious skill to apply, you will save time and build confidence on most geometry-based questions.

For more help on the GMAT, take a look and see what Veritas Prep has to offer.