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

Quarter Wit, Quarter WisdomHow 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

Thus, our answer is C.

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.

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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!

Quarter Wit, Quarter WisdomTest-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!

Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on FacebookYouTubeGoogle+, and Twitter!

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

Pie ChartIt 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:

DG Circle 1








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:

DG Blog 2









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):

DG Blog 3







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

DG Blog 4







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.)

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTubeGoogle+ and Twitter!

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

stopwatch-620Speed 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?

ACT Triangle 1




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:

ACT Triangle 2




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:

ACT Triangle 3




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) 30ACT Triangle 4
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.

ACT Triangle 5Are 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.

Still need to take the ACT? We run a free online ACT prep seminar every few weeks. And be sure to find us on Facebook, YouTube, Google+ and Twitter!

By Rita Pearson

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.