Here is a batch of 7 practice QC questions. Explanations will come at the end of the article.

1) In the diagram, AC = 6. CE = 12, DF = 4, and AB is parallel to DE.

2)

3) In the diagram, JL = 4 and JK = 6.

4) In the diagram, O is the center of the circle, and AB is a diameter. Region J is the area between chord AC and the arc of the circle.

5) In the diagram, triangle MNP is equilateral.

6) In the diagram, JKLM is a square. Point S is the midpoint of KL, and point T is the center of the square. Point O is on segment ST, and is the center of the circle, which passes through both K and L.

7)

## Geometry on the GRE QC

As you may appreciate, there’s no guarantee that any geometry diagram on the entire GRE Quant section is drawn to scale. In fact, most diagrams that appear are specifically designed to hoodwink you in the most viciously deceptive way imaginable. For example, if the GRE gives this

with no further explanation or specification, the naïve gullible GRE test taker will think, “OK, an equilateral triangle,” whereas in fact, it could be any of these:

Never be duped by believing a diagram. In particular, one large group of GRE QC Geometry questions gives you diagrams that appear a certain way, but which leaves some ambiguity open, and your job is to spot the ambiguity, the different geometric possibilities, and not be sucked in by trusting the misleading diagram. Some of the questions above are this sort.

Another batch of GRE QC Geometry questions specific everything in detail, and your job is to perform some kind of calculation, a length or angle or area, and compare it to something.

At the very least, it’s important to recognize, when looking at a GRE QC Geometry question, whether it’s the ambiguous-diagram kind or the everything-specified kind. The former case has a high likelihood of having **(D)** as an answer!

## Summary

If the foregoing discussion gave you some insights into the question, you may want to go back and give them a second glance. Please let us know about your experience with GRE Geometry questions in the comment sections below.

## Practice problem explanations

1) The triangles CBA and CDE are similar, so all corresponding lengths are proportional. Each length in CBA is 1/2 as big as the corresponding length in CDE. Think about DF, the altitude in CDE, and the length to which it would correspond in CBA: an altitude from vertex B to the base AC.

Call that **h**. Since DF = 4, h = 2.

Now, for CBA, we have an altitude h = 2 and a base AC = 6.

Answer = **(B)**

2) Everything here is fixed and specified. Triangle ABC is a right triangle, because the longest side is the diameter. See this GMAT blog post for details. In right triangle ABC, AB = 4 is the hypotenuse. Notice, this is NOT a 3-4-5 triangle, because the hypotenuse is 4.

We could take the square root to find the length of AC, but we don’t need to. AC is the side of the square, so AC squared is the area of the square. The square has an area of 7.

Answer = **(C)**

3) This is a deceptive diagram. The angle at L *appears* to be right angle, but absolutely nothing in the diagram or the text guarantees that it is a right angle. The triangle could be:

On the left, the triangle has been “crushed” close to flat, and it could be crushed even further, to an area of almost zero. Certainty the area could be less than 11. By contrast, if JK and JL are perpendicular, then the area would be A = 0.5bh = (0.5)(4)(6) = 12, which is more than 11. Depending on the diagram, it could go either way.

Answer = **(D)**

4) We know AB is a diameter, and we know that angle ACB is 90°. All of that we know. We don’t know the position of point C. In the diagram given, point C is positioned so that the two areas look approximately equal. By moving point C around, we could radically change the situation. For example, we could move C toward B, making Region J much larger.

Or we could move point C toward the other side, closer to point A, in which case the triangle would be much bigger and Region J much smaller.

Since we can vary point C to produce different relationship, we cannot produce a fixed answer.

Answer = **(D)**

5) This is a hard problem, but we can do it entirely without touching a calculator.

The first step is to draw the altitudes from each vertex of the equilateral triangle to the midpoint of the opposite side. This will subdivide the equilateral triangle into six congruent 30-60-90 triangles. For more details on these triangles, see this GMAT blog. It’s very handy to remember that any equilateral triangle can be subdivided this way.

Notice that OP, OM, and ON are all radii of the circle. Let’s just say for convenience that r = 2, so OPO = OM = ON = 2. Look at 30-60-90 triangle TOP. The hypotenuse is OP = 2, so the side opposite the 30° angle is half of this, OT = 1. The remaining side, opposite the 60°, must have a length of:

The area of triangle TOP would be:

Six of these triangles make up the entire equilateral triangle, so that’s

That’s the area of the entire equilateral triangle. Meanwhile, the circle has a radius of r = 2, so

Now the shaded area is the area of the circle minus the area of the equilateral, so the area of the shaded region is:

Now, we have numbers for the areas of both regions:

Whatever the square-root of 3 equals, it is clearly less than 2, so Quantity A is less than 2, and Quantity B is slightly more than 2.

Answer = **(B)**

6) Notice that the one thing not fixed in the problem is the position of point O on the segment ST. It could be toward the top or bottom of that segment. Think about the implications.

If we move point O toward the top of the circle, so that it is very close to S, then we get relative small circles.

In the limit in which O was at S, the circle would be one whose diameter would equal the side of the square: such a circle could fit entirely inside the square, simply touching each side at its midpoint. Such a circle clearly would have much less area than the square.

On the other hand, we could move point O down toward point T, the center of the square. This produces relatively large circles.

In the limit in which O was at T, the circle would neatly contain the square, passing through the four corners. Such a circle clearly has more area than the square.

Because we can choose a position for point O that allows for a relationship either way, we can determine no conclusive relationship between the quantities.

Answer = **(D)**

7) This is a hard question. Let’s approach it this way. Sometimes it is very helpful to introduce a variable into a geometric problem. I am going to say that AC = x and BC = 1. Notice that AB = x + 1. Then

Now, what happens if x = 3? Notice that:

Clearly, they are not equal, as they should be. What happens if we make x bigger? Well, then AC/BC = x simply becomes bigger, but AB/BC becomes smaller, so they wouldn’t get closer to being equal. In fact, they would be getting further away from each other.

What happens if x get smaller? Well, then AC/BC = x clearly gets smaller, and the fraction 1/x gets bigger when its denominator gets smaller, so AB/BC would get bigger. In that case, they would be moving toward each other: the bigger one getting smaller, and the smaller one getting bigger. That indicates they could become equal if we move x in that direction, smaller than 3. In other words, x must be smaller than 3.

Right there, that’s enough to determine that the answer to the QC question is **(B)**.

More information for the curious:

If you are interested, notice that if x = 2, then:

They’re closer, but not equal. If x = 1, then

Here, they have shot past each other. Now, the little one is too big, and the big one is too little! We overshot. This tells us the correct value of x is between 1 and 2.

In fact, a few students may recognize that the diagram & equation above is the definition of the Golden Ratio, and in fact, this is the correct value of x.

If you happened to be familiar with the Golden Ratio, that would have been a shortcut on this particular problem, but in general, you don’t need to know anything about the Golden Ratio for the GRE. The Golden Ratio plays a big role in sacred geometry and all kinds of esoteric wisdom (Egyptian temples, the Parthenon, Leonardo’s paintings, the Gothic Cathedrals, etc.), so you may find it interesting, but all this is well beyond anything relevant to success on the GRE!! 🙂

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I agree with Stephen! In question 3, we have one leg equal to 4 and hypotenuse equal to 6.

Then a triangle has maximal area when it is right-angled. Thus, when its second leg is equal to 36-16 = 2sqrt(5). In this case, the area of the triangle is equal to 4sqrt(5) that is less than 12.

If this is the maximal area, there is no way to get something greater than 12 so that quantity B should be greater.

I am not assuming the triangle is right-angled, it is just the case in which it has the maximal area. While the smallest is nearly zero.

Thanks for creating these questions. For question number two, if OA is the radius and it’s 2, then wouldn’t OB and OC necessarily be 2 as well? if OB and OC are 2, and BC is 3, then wouldn’t AC need to be 3 if OA is 2 and OC is 2? This would make the area of the square 9.

Hi Tates,

Great question! The problem here is that we don’t know the angles of triangle AOC. So while you’re correct that OC is a radius and equals 2, we can’t use that information to find AC because we lack the information necessary to make that determination. The reason that we can conclude that ABC is a right triangle is because of the fact that the longest side of the triangle is the diameter of the circle (read more about that here). Without this vital clue, we can’t determine the rest of the information necessary to find the area of the square! 🙂

In #3 (the triangle with indeterminate third side length and angle): is it not true that the maximum area occurs for a right triangle? For such a case the opposite side length = sqrt(36- 16) = 2sqrt(5). So the area is 0.5 * 4 * 2sqrt(5) ~= 8.9 < 11 . So even the maximum possible value of choice A is < B and the answer should be B.

Hi Stephen,

Please note that you can’t rely on the image. The right angle could potentially be at Angle KJL. If so, the length would be KJ (i.e., 6), and the width would be JL (i.e., 4). Then, the area will be:

= 0.5 * height * base

= 0.5 * 6 * 4

= 12

The diagram for question 3 has the hypotenuse-looking side at 6. In the explanation, the leg-looking side is at 6. It still does not change the answer though. Just wanted to point it out.

Hi Scott,

Thank you for pointing that out! Yes, you are correct that the diagram in the solution looks slightly different from the diagram in the question, but the logic and solution remain the same! Good eye!

In question 2, why are we assuming that AB is a straight Line. Its no where mentioned so why can I assume angle AOB 180.

Hi Mike,

The GRE’s list of mathematical conventions says this- “assume all geometric objects are in the relative positions shown.” Given this, in question 6 wouldn’t it be fair to assume that JM would either be tangential to the circle or lie entirely below it?

This doesn’t change the final answer but I was just wondering 🙂

Do let me know. Thanks.

Hi,

I have read somewhere that the figures are not drawn to scale in GRE, so I was wondering in Question Number 2, is it correct to assume ABC as a triangle? I am saying about the possibility of AC not being the diameter, because AOB may be longer than four. I am confused about when to assume the figures are not drawn to scale, and when they are actually drawn to scale. Any suggestions on this?

Thanks,

Prashanth

Dear Prashanth,

I’m happy to respond. 🙂 It’s true that GRE Geometry Diagrams are not drawn to scale, which means that, in general, any lengths or angles that are not marked could have different sizes than what is indicated in the diagram. Nevertheless, fundamental issues of straightness and what is connected to what is guaranteed. If you see a triangle, you absolutely know that it must be a triangle, although in the absence of further specification, it could be any triangle of any shape.

In fact, in Problem #2, you know that triangle ABC is a right triangle, with a right angle at C, because its hypotenuse is the diameter of the circle. This is a geometry fact discussed in the Magoosh lesson videos. For more on GRE Geometry Assumptions, see:

https://magoosh.com/gre/2013/gre-geometry-diagram-assumptions/

Also, by convention, when a dot is drawn more or less at the middle of the circle, we can assume that this dot is the center. Here the problem explicitly states that, but it would be deeply deceptive for the GRE to put a dot more or less at the center of the circle and not intend for that dot to be the center.

Mike 🙂

Shouldn’t problem #2 explicitly also state that line “AB is a diameter” of the circle? Without this statement, we can easily shift point A to anywhere on the circle, preserve the conditions, and enlarge or shrink the square as a result.

John,

I’m happy to respond. 🙂 We know that O is the center, so any chord that passes through the center is automatically a diameter. You see, the problem *does* tell you that AB is a diameter, indirectly.

Also, in the unadorned diagram, we could slide things around, but when the numerical lengths are applied to the problem, everything becomes locked in place.

Does this make sense?

Mike 🙂

Hi Mike,

Greetings from Chile!

Thanks for this nice set of problems.

I guess there is a problem in question 2) because there is only a graph…

All the best,

A.

Dear Agustin,

Greetings from Berkeley, CA! 🙂 Thank you very much for pointing that out. Somehow, the question part was dropped when we uploaded the blog, but I just fixed that. Best of luck to you and stay warm, my friend.

Mike 🙂

Amazing set of questions. How probable is it to get questions as tough as 6 and 7.

Also, for question 7, once understanding about the substitution of variables fact, can we just solve the algebraic equation and get the roots,

Both the roots are less than 3, so we can say that the answer is B (You have made a typo and written the answer as A).

Thanks!!

Dear Anuj,

I’m happy to respond. 🙂 First of all, thank you very much for pointing out the typo: I just corrected that. I would say #6 is about as hard as you could possibly see. If you perform brilliantly on the first math section, so that the second math section is hard, then #6 could be among the harder questions on that hard section.

In some ways, I would say #7 is perhaps a tad harder than what the GRE would ask. It’s difficult to say whether ETS would pose a question that hard. Yes, absolutely, once we have variables in #7, we could set up the equation and solve. That equation has two irrational roots: if you are particularly skilled with algebra, you can see that both of those roots are less than 3, but there are several steps there that would be hard for many GRE test takers. The fact that you saw the algebraic solution for #7 speaks very highly of your math skills, my friend. BTW, the positive root of that equation is the Golden Ratio, and the negative root is the negative reciprocal of the Golden Ratio.

Best of luck to you!

Mike 🙂