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Posted on
21
Jul 2021

GMAT 3D Geometry Problem – GMAT Math – Quant Section

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 question

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.

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Posted on
14
Jul 2021

GMAT 680 Level Geometry Problem – No Math Needed!

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.

top level geometry problem

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.

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Posted on
17
Feb 2021

Data Sufficiency: Area of a Triangle Problem

Hey guys! Today we’re checking out a geometry Data Sufficiency problem asking for the area of a triangle, and while the ask might seem straightforward, it’s very easy to get caught up in the introduced information on this problem. And this is a great example of a way that the GMAT can really dictate your thought processes via suggestion if you’re not really really clear on what it is you’re looking for on DS. So here we’re looking for area but area specifically is a discrete measurement; that is we’re going to need some sort of number to anchor this down: whether it’s the length of sides, or the area of a smaller piece, we need some value!

Begin with Statement #2

Jumping into the introduced information, if we look at number 2, because it seems simpler, we have x = 45 degrees. Now you might be jumping in and saying, well, if x = 45 and we got the 90 degree then we have 45, STOP. Because if you’re doing that you missed what I just said. We need a discrete anchor point. The number of degrees is both relative in the sense that the triangle could be really huge or really small, and doesn’t give us what we need. So immediately we want to say: number 2 is insufficient. Rather than dive in deeply and try and figure out how we can use it, let’s begin just by recognizing its insufficiency. Know that we can go deeper if we need to but not get ourselves worked up and not invest the time until it’s appropriate, until number 1 isn’t sufficient and we need to look at them together.

Consider Statement #1

Number 1 gives us this product BD x AC = 20. Well here, we’re given a discrete value, which is a step in the right direction. Now, what do we need for area? You might say we need a base and a height but that’s not entirely accurate. Our equation, area is 1/2 x base x height, means that we don’t need to know the base and the height individually but rather their product. The key to this problem is noticing in number 1 that they give us this B x H product of 20, which means if we want to plug it into our equation, 1/2 x 20 is 10. Area is 10. Number 1 alone is sufficient. Answer choice A.

Don’t Get Caught Up With “X”

If we don’t recognize this then we get caught up with taking a look at x and what that means and trying to slice and dice this, which is complicated to say the least. And I want you to observe that if we get ourselves worked up about x, then immediately when we look at this BD x AC product, our minds are already in the framework of how to incorporate these two together. Whereas, if we dismiss the x is insufficient and look at this solo, the BD times AC, then we’re much more likely to strike upon that identity. Ideally though, of course, before we jump into the introduced information, we want to say, well, the area of a triangle is 1/2 x base x height. So, if I have not B and H individually, although that will be useful, B x H is enough. And then it’s a question of just saying, well, one’s got what we need – check. One is sufficient. Two doesn’t have what we need – isn’t sufficient, and we’re there. So,

I hope this helped. Look for links to other geometry and fun DS problems below and I’ll see you guys soon. Read this article about Data sufficiency problems and triangles next to get more familiar with this type of GMAT question.

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Triangle With Other Shapes on the GMAT
Posted on
16
Feb 2021

Triangles With Other Shapes

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

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:

triangles in other shapes GMAT article

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:

triangles in other shapes gmat article gmat problem

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:

triangles in other shapes gmat article, 45 45 90 degree angle

(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:

  1. Triangles created using the central angle of a circle

triangle in a circle, gmat geometry article

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.

2. Inscribed triangles

triangle inscribed in circle, gmat problem

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.

3. Squares and rectangles inscribed in circles

rectangle in circle, gmat geometry

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:

triangle in other shapes gmat problem

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:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

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Triangle Inequality Rule on the GMAT
Posted on
09
Feb 2021

Triangle Inequality Rule

By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 2021

One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.

Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:

Suppose we have a triangle that has two sides of length 3 and 5:

triangles inequalities 1

What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.

So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]

If you’d like to see that put into words:

**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**

It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)

Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.

triangle inequalities 2

 

If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:

triangle inequalities 3

The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)

Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:

If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five

Strategy: Eliminate Answers

As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).

Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.

Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.

Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.

Note:

It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.

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

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

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similar triangles on the gmat
Posted on
02
Feb 2021

Similar Triangles – GMAT Geometry

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

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:

similar triangles on the GMAT 1

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:

similar triangles on the GMAT 2

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:

similar triangles on the GMAT 3

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:

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

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Isosceles Triangles and Data Sufficiency title
Posted on
26
Jan 2021

Isosceles Triangles and Data Sufficiency

By: Rich Zwelling, Apex GMAT Instructor
Date: 21st January, 2021

Although we’ve already discussed isosceles triangles a bit during our discussion of 45-45-90 (i.e. isosceles right) triangles, it’s worth discussing some other contexts in which you may see isosceles triangles on the GMAT, specifically on Data Sufficiency problems. 

As we discussed before, an isosceles triangle is any triangle that features two equal sides and thus two equal opposite angles:

Isosceles Triangles and Data Sufficiency picture 1

That’s an easy enough definition to remember, but how does the GMAT turn this into more challenging problems? For that, let’s take a look at the following Official Guide problem. Try to solve before reading the explanation below the problem:

Isosceles Triangles and Data Sufficiency picture 2

In the figure above, what is the value of x + y ?
(1) x = 70
(2) ABC and ADC are both isosceles triangles

Explanation

In this case, it’s straightforward enough to determine that each statement alone will be insufficient. Statement (1) gives us a definitive value for x, but no information about y, thus we cannot answer the question (the value of x+y). And although Statement (2) labels each triangle in the diagram as isosceles, we have no way of knowing the specific angles involved nor their relationships. 

However, as with many Data Sufficiency problems, especially those involving Geometry, things can get thorny when we have to combine the statements. The two statements look very complimentary, and that could lead us to prematurely conclude the answer is C (i.e. the two statements are sufficient when combined). But we must do a thorough check. 

Reframing the question

Remember that at any point during a Data Sufficiency problem — beginning, middle, or end — you can reframe the question for simplicity. The question asks for the value of x+y. But now that we are combining the statements, we already know that x=70. In terms of sufficiency, then, what information do we need? The only thing missing is a definitive value of y. The question now might as well be “What is the value of y?”

Now, here’s where the GMAT thinking really comes into play. It’s one thing to understand what an isosceles triangle is. It’s quite another to judge what a diagram of an isosceles triangle does or does not tell you and what you can or cannot extrapolate from it. 

One of my personal favorite things about Geometry Data Sufficiency problems is that they tend to be very intuitive visually. You can often answer them by manipulating figures. 

We know that triangle ADC is isosceles, but is that enough to give us definitive measurements? Visually, which of these does it look like?  

Isosceles Triangles and Data Sufficiency picture 3

Without any numerical evaluations, we can see that we can’t get a definitive measure for the angle at D, which in this case is our y. So even when we combine the statements, we cannot get an answer to our question. The correct answer is E

Here’s another case of a tricky Data Sufficiency problem involving isosceles triangles:

In isosceles triangle RST, what is the measure of angle R?

  • The measure of angle T is 100 degrees
  • The measure of angle S is 40 degrees

Again, give the problem a shot before reading the answer and explanation.

Explanation

This is one for which you can draw a diagram, but it’s not necessary. The trick here is to remember another key property of triangles, namely that all angles in the triangle must sum to 180 degrees.

Since the triangle is isosceles, and since each statement gives you only one angle of three, the temptation can be to say that each statement is insufficient on its own. This is certainly the case for Statement (2), because the 40-degree angle could be one of a pair (in which case we would have a 40-40-100 triangle) or the 40-degree angle could be the odd angle out (in which case we would have a 40-70-70 triangle). 

Because the problem asks for the value of R, and since R could be 40, 70, or 100 depending on the situations outlined above, Statement (2) is INSUFFICIENT.

However, there’s a catch when evaluating Statement (1). Notice that angle T is an obtuse angle, meaning it is greater than 90 degrees. Is it possible that there are two 100-degree angles in a triangle? This would produce a total of 200 degrees, which would exceed the 180-degree total for any triangle. As such, the only possibility is that the 100 degree angle is the odd angle out, and the other two angles are equal acute angles (specifically, we have a 40-40-100 triangle). 

Now we know R must be 40 degrees. Statement (1) is sufficient, and the correct answer is A.

But notice how the GMAT sets the statements up to bait you into thinking that you must combine the two statements to figure out the value of angle R. 

Now that we’ve finished talking about the basic triangle types, we can move on to talking about what happens when triangles are used within different shapes. In the meantime, here are links to our other triangle articles:

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

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The Area of an Equilateral Triangle
Posted on
14
Jan 2021

The Area of an Equilateral Triangle

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

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:

Equilateral triangles GMAT picture 1

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:

Equilateral triangles GMAT picture 2

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

Equilateral triangles GMAT picture 3This 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:

Equilateral triangles GMAT picture 4

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:

Equilateral triangles GMAT picture 5Now, 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?

Equilateral triangles GMAT picture 6

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:

Equilateral triangles GMAT picture 7

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.

Find more articles in our triangle series here:
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

Read more
5-12-13 and 7-24-25 triangles on the gmat
Posted on
12
Jan 2021

The 5-12-13 and 7-24-25 Right Triangles

By: Rich Zwelling, Apex GMAT Instructor
Date: 12th January, 2021

The 5-12-13 and 7-24-25 Right Triangles

Although the 3-4-5 right triangle is by far the most common of the so-called “Pythagorean triples” tested on the GMAT, there are a few others worth knowing. First, a little review: 

You’ll recall that the Pythagorean Theorem ( + b² = c²) holds for any right triangle where a and b are the two legs and c is the hypotenuse, and that the 3-4-5 triangle represents the smallest such triangle with all integer side lengths:

5-12-13 and -7-24-25 Triangle Identities problem 1

This works not only for 3-4-5 but also for 6-8-10, 9-12-15, or any other multiples of each side length.

5-12-13 and -7-24-25 Triangle Identities problem 2No matter what positive integer n you choose for the figure above, you will produce a valid right triangle.

So now we come to the main topic: what are some other common “Pythagorean triples” the GMAT may test? The next base triples that fit the Pythagorean Theorem are 5-12-13 and 7-24-25. These work because if you check the arithmetic, 5² + 12² = 13² and 7² + 24² = 25²:

5-12-13 and -7-24-25 Triangle Identities problem 3

As we’ve continually discussed, however, your success on more difficult GMAT problems will require you to go beyond mere rote memorization. Let’s take a look at an Official Guide Data Sufficiency problem that illustrates how the test can force you to engage some higher-level reasoning skills:

5-12-13 and -7-24-25 Triangle Identities problem 4

If A is the area of a triangle with sides of lengths x, y, and z as shown above, what is the value of A?

(1) z = 13

(2) A = 5y/2

Give it a try on your own before reading any further.

As with any Data Sufficiency question, let’s identify what we’re asked to find. A represents the area of the triangle, which is found by multiplying base by height and dividing by 2. That means A = xy/2, since x and y represent the height and base, respectively. 

Remember, it helps to frame Data Sufficiency questions in terms of what information you need to get to the answer. We need to know the individual values of x and y. Or, as a matter of fact, we could have sufficiency if we knew xy as a product, even if we didn’t know the values of x and y, individually. For example, on a different problem with the same question, if the test had said that the product of the base and height were 30, that would have been sufficient, as that would be enough for us to deduce that the area is 15.  

You can save yourself much time and mental energy by having a solid idea of what information you need from the statements for sufficiency before you actually view the statements. 

Now that we know what information we need for sufficiency, let’s examine each statement on its own. Statement (1) should get you thinking about the 5-12-13 right triangle, as it tells us that the hypotenuse is 13. But be careful: this is where rote memorization only goes so far (and may actually get in the way). 

Does knowing that the hypotenuse is 13 guarantee that the other sides are 5 and 12? For all we know, they could be non-integers that fit + b² = 13². In fact, a and b could be equal — remember that we can’t assume that the figures are drawn to scale. Without a clear idea of what the base and height are, we cannot get a consistent product for xy. Statement (1) is INSUFFICIENT on its own.

Statement (2) is more complicated, as we have two variables, one of which is the area. But we already discussed that A = xy/2, so we can do a substitution:

A = 5y/2
xy/2 = 5y/2

At this point, we can see that the sides are identical, except that the x on the left has been replaced by a 5 on the right. Therefore, x must be 5. Again, this should get us thinking about the 5-12-13 triangle. But we should again remember that this alone does not guarantee that the other sides are 12 and 13. Even though x is 5, there could be multiple values for y, and that means multiple values for the product xy. Statement (2) is also INSUFFICIENT on its own.

This narrows the answer choices down to C (statements sufficient together) and E (statements insufficient together).

This is where previous knowledge of the 5-12-13 triangle helps. Ideally, once you see that the statements together tell you that x=5 and z=13, you will know without much thought that y must be 12. You won’t bother using the Pythagorean theorem and you certainly won’t wonder if y could have multiple values.

Without knowledge of the 5-12-13, one trap a test-taker could possibly fall into is viewing the two statements and noticing that there are 3 variables and only 2 equations. We need a full 3 equations with 3 variables if we’re going to solve for all 3 variables, and that may lead some to prematurely conclude that the answer is E. 

However, why is that a false conclusion?

Well, we’re not trying to solve for all variables. We’re only solving for one. It’s possible to solve for one variable, even if there are fewer equations than variables. 

In this case, now that we know that x=5 and y=12, we have our base and height, and we can solve for A, the area of the triangle. Note that I’m not going to bother solving, because for sufficiency, I don’t need to. I only care that I CAN solve. The final answer is (C).

We’ve now talked about the various Pythagorean triples and special right triangles. Next time, we’ll talk about how triangles can appear within OTHER shapes. And to tide yourself over, you can also link to our other article about triangles:

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

Read more
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. 

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:

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

Read more