GMAT Geometry Practice Problems

gmat geometry tools placed on top of graph paper, including protractor, compass with lead pencil attached, two triange rulers, standard ruler, and eraser -image by magoosh

GMAT Geometry problems test your spatial reasoning ability. Can you look at a diagram of points, lines, and/or circles and extract the essential details that lead to a correct answer?

If you answered no, well, have no fear! After reading this post, learning the fundamental geometry formulas, and working through these practice geometry questions, you will have the tools you need to succeed on test day!

Table of Contents

How to Use Geometry Formulas

It’s really important to understand that geometry formulas are useful tools, NOT magic wands. Geometry formulas are definitely important! But it could very tempting to think that all you need to do is memorize a bunch of formulas. By themselves, formulas cannot guarantee you a great score on the GMAT Quant section. You also need to know when and how to apply the formulas.

Furthermore, it’s rare that a problem would require only a single formula to solve it. Most often, you’ll need to put a few different formulas together like pieces of a puzzle. The best problem-solvers take a goal-oriented approach. In other words, start with what you need to solve. Then work backward, identifying what info would be useful to get to that goal. In addition, you need to keep in mind the given info, both from the diagram and also from the question statement. Use that to build a bridge to your goal.

This post walks you through the most important formulas for GMAT Geometry. The purpose here is just to help you review—so click on the links to learn more about the material.

Then, you can try out your skills on a set of geometry practice questions. Detailed solutions are given at the very end.

Ready? Let’s go!

Lines and Angles

First of all, know your terms: parallel (same direction) vs. perpendicular (crossing at right angles) lines, interior angles vs. exterior angles, supplementary (angles adding to 180°) vs. complementary (angles adding to 90°).

You should review the basic geometry formulas. For example, this diagram shows all of the possibilities involving a line crossing two perpendicular lines.

Two parallell lines cut by a transveral
 

magoosh video lesson icon with human figure For more review on lines and angles, check out our post on Angles and Parallel Lines on the GMAT and our video lesson Geometry: Lines and Angles.
 

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Triangles

There are tons of formulas and a huge amount of terminology related to triangles! We can only scratch the surface in this article.

Terminology Related to Sides

  • Equilateral – All three sides are equal. All angles are equal to 60°.
  • Isosceles – Two equal sides and corresponding equal angles.
  • Scalene – None of the sides or angles are equal to one another.

Terminology Related to Angles

  • Acute – All three angles measure less than 90°
  • Right – One angle is 90° (right)
  • Obtuse – One angle is greater than 90°

Sum of angles = 180° (for any triangle)

Pythagorean Theorem: \( a^2 + b^2 = c^2\), where \(a, b\) are the legs and \(c\) is the hypotenuse of a right triangle. (But also try to have memorized the most common Pythagorean Triples: 3-4-5, 5-12-13, 8-15-17, and 7-24-25.)

Area: \(A = \frac{1}{2}bh\), where \(b\) is the base, and \(h\) is the height.

Area of an equilateral triangle with side \(s\): \(A = \frac{3}{2} \cdot \sqrt{3} \cdot s\)

magoosh video lesson icon with human figure
You can learn more by watching our video lessons, Triangles – Part I and Right Triangles.
 

And even more resources can be found here:

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Quadrilaterals and Other Polygons

The basic area formula for rectangles and parallelograms is: \(A = bh\) (base times height). That’s all you really need for GMAT geometry because more complicated shapes can usually be broken down.

It’s useful to know the following angle formulas:

Sum of interior angles of an \(n\)-sided polygon = \(180(n – 2)\) degrees.

If the polygon is regular (all sides and angles are equal), then any angle has measure \(\frac{180(n – 2)}{n}\) degrees.

magoosh video lesson icon with human figure
For additional review, check out this video lesson about Regular Polygons.
 

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Circles

Circles show up in many problems on the GMAT test.

Area: \(A = \pi r^2\)

Circumference: \(A = 2\pi r\)

The majority of problems involving circles can be reasoned out without relying on a lot of fancy geometric formulas. You just have to use your mathematical common sense. Need to know the area of a sector? Well just find out what fraction of the whole circle it represents!

Additional resources can be found here:

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Solids

There are typically only a couple questions about solid geometry on any given GMAT test. So we won’t delve into that topic here, but you can check out the following links for more.

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GMAT Geometry Practice (Problem Solving Questions)

Problem 1

An equilateral triangle intersects a circle (not shown). The possible number of distinct intersection points could be the following:
 
I. 3
II. 4
III. 5

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

Click here for the answer

 
E.

All three are possible. (In fact, if you think about it, the number of intersections points could have been any of 0, 1, 2, 3, 4, 5, or 6!)

Three diagrams of circle and triangle intersection.

Problem 2

Ten circles arranged in a pattern with some parts shaded

Ten circles of radius \(r = 6\) are equally spaced in a regular pattern as suggested by the diagram. What is the total area of the shaded regions?

A. \(34\pi\)
B. \(36\pi\)
C. \(39\pi\)
D. \(40\pi\)
E. \(42\pi\)

Click here for the answer

 
B.

It would be a huge mistake to attempt any advanced calculations to solve this. By symmetry, the three smaller shapes can be inserted into the three hollow spaces adjacent to the upper shaded region. Then together, the four shapes will neatly form a single circle. The circle has radius 6, so the area is \(\pi r^2 = \pi 6^2 = 36\pi\).

Problem 3

Triangle STV (not shown) has sides ST = TV = 17, and SV = 16. What is the area?

A. 85
B. 100
C. 120
D. 136
E. 165

Click here for the answer

 
C.

To find the area, we need to know the base and height. Triangle STV is isosceles, so we know that SV = 16 is the base but we don’t know the height.

The height would be represented by a perpendicular line segment from vertex T is bisects base SV, at a point we’ll call W.

Isosceles Triangle with altitude

Thus, SW = 8. Now, look at right triangle STW: it has leg = 8 and hypotenuse = 17. It will save you a tremendous amount of calculations here if you already have memorized the 8-15-17 Pythagorean Triple. Thus, TW = 15, and that’s the height. This allows you to find the area: \(\frac{1}{2}\)\(b\)\(h\)\( = \frac{1}{2}\)\((16)\)\((15)\)\( = 120\)

Problem 4

Pentagon with two diagonals

Given that ABCDE is a regular pentagon, what is the measure of ∠ACE?

A. 24°
B. 30°
C. 36°
D. 40°
E. 45°

Click here for the answer

 
Use the formula for the angle of a regular polygon (with \(n=5\)):

\(\frac{180(5 – 2)}{5} = 108\) degrees.

Now, look at isosceles triangle ABC, with an angle of 108° at B. The other two angles are equal: call each \(x\).

Knowing that the angles in a triangle add up to 180, we know that \(108 + x + x = 180\), which leads to \(x =\) 36°.

Finally, ∠BCA = ∠ECD. Given that ∠BCA = \(x\) = 36°, then ∠ECD = 36°. This means that ∠ACE = 108° − 36° − 36° = 36°

Problem 5

Parallel lines, triangles, and a circle

In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?

A. 240
B. 270
C. 300
D. 330
E. 360

Click here for the answer

 
B.

Since ED is parallel to GH, triangles FED and FHG are similar. Why? Well, the vertical angles are equal: ∠GFH = ∠DFE, and the pairs of alternate interior angles are also equal: ∠G = ∠D and ∠H = ∠E.

Let’s begin by focusing on triangle FED. The angle ∠E spans a diameter, so ∠E = 90°. Thus, triangle FED is right with hypotenuse FD = 13 and leg ED = 5. That means that FE = 12 (just remember the 5-12-13 Pythagorean Triple).

Next, since GH = 15 is three times ED, the scaling factor is 3. Scale up FE by 3 to get FH = 36. Finally, find the area using the familiar formula for triangles: \(A = \frac{1}{2}(36)(15) = 270\).

Below is the image again with the length of each side labeled (the givens are in yellow):

Problem 6

Two intersecting circles with triangles

In the diagram above, A and B are the centers of the two circles, each with radius 6, and ∠A = ∠B = 60°. What is the area of the shaded region?

A. \(12\pi – 18\sqrt{3}\)
B. \(18\sqrt{3} – 6\pi\)
C. \(24\pi – 36\sqrt{3}\)
D. \(36\sqrt{3} – 12\pi\)
E. \(36\pi – 72\sqrt{3}\)

Click here for the answer

 
D.

This one is definitely tricky! I’ll break up the solution into 5 steps.

Step #1: Each circle has an area of \(\pi(6)^2 = 36\pi\).

Step #2: One sector (“slice of pie”) occupies 60°, which is one sixth of the circle.

Sector of a circle

Therefore, the area of the sector is: \(\frac{1}{6}(36\pi) = 6\pi\).

Step #3: Now look at the equilateral triangle.

Equilateral triangle within a circle

Its side length is \(s = 6\), so using the shortcut formula, its area of the equilateral triangle is \(9\sqrt{3}\).

Step #4: Find the area of the circular segment, which is the name for that little leftover piece, the part of the sector that’s beyond the triangle.

Circle Segment

Area of segment = (Area of sector)(Area of triangle) = \(6\pi – 9\sqrt{3}\).

Step #5: Now notice that the shaded region in the diagram is just the two equilateral triangles minus two circular segments.

\(2(9\sqrt{3}) \)\( 2(6\pi – 9\sqrt{3})\)\( = 18\sqrt{3} – 12\pi + 18\sqrt{3} = 36\sqrt{3} – 12\pi\)

Problem 7

Diagram with many line segments

In the diagram above, AB is parallel to EH, and BD is parallel to FH. Also, AB = BC, and EF = FH. If ∠EGC = 70°, then ∠D =

A. 65°
B. 70°
C. 75°
D. 80°
E. 85°

Click here for the answer

 
D.

Since ∠EGC = 70°, we know that ∠A = 70° (alternate interior angles).

Next, because AB = BC, we see that triangle ABC is isosceles, which means that ∠ACB = 70°. The three angles have to add up to 180°, so this tells us that ∠B = 40°.

At this point, we reach a very tricky move: both ∠B and ∠H are angles formed by the pairs of parallel lines — the sides of each one are parallel to the corresponding sides of the others. This means ∠B = ∠H = 40°.

Next, because EF = FH, triangle AFH is also isosceles, which means ∠GEF = 40°. Again, the angles of a triangle have to add up to 180°, so this tells us that ∠F = 100°.

Finally, ∠F and ∠D are two angles on the same side of the same line between two parallel lines (same side interior angles). These angles must be supplementary, which means that ∠D = 180° − 100° = 80°.

 

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Additional GMAT Geoemtry Practice (Data Sufficiency Questions)

The 7 problems above are all of the Problem Solving variety. You can also practice a few Data Sufficiency GMAT geometry questions by following these Magoosh links:

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Conclusion

GMAT geometry doesn’t require a huge number of sophisticated formulas. If anything, you should focus more on improving your geometry strategies, particularly how to use diagrams to your benefit.

What is the diagram is telling you: What assumptions can you make? What shouldn’t be assumed? Can you use estimation?

magoosh video lesson icon with human figure Our video lessons on geometry strategies and estimation will help you build those skills!
 

If you made it to the end of this post, then kudos! Hopefully, you can take what you’ve learned here and apply it to ace the GMAT Quantitative section!

Ready to get an awesome GMAT score? Start here.

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20 Responses to GMAT Geometry Practice Problems

  1. Nimmy July 18, 2019 at 4:59 am #

    Hi,
    Just wanted to point out an error

    In the solution to the 8th problem, the line “Now, we know that, because EF = EH, triangle AFH is also isosceles, which means ∠GEF = 40°. The three angles have to add up to 180°, so this tells us that ∠F = 100°.”

    It says since EF= EH, when actually the problem stated that, EF= FH. I just broke my head trying to figure out how EF=EH leads to angle GEF to be 40degrees when EF=EH haha. Thought I just forgot all my basics.

    Otherwise, great explanations!

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert July 19, 2019 at 4:04 pm #

      Thanks for pointing that out, Nimmy! I’ve fixed the error 🙂

  2. Bishal Shah December 23, 2017 at 7:01 am #

    its became easy too solve.. Thank u mike too help me.

  3. Franco November 28, 2015 at 10:27 pm #

    In problem number 5, how do you know that the diameter is actually FD? Since it was not given that the center of the circle is at Q?

  4. Franco November 25, 2015 at 2:51 am #

    Hi Mike,

    I would like to ask you for a general “speed/efficient” strategy for any geometry problem that would involve drawing the figure.

    For example, for problem #7, it takes me about 50s to 1min to read the problem and draw the figure. This still excludes understanding the problem and knowing how to attack it.

    (Even by re-doing the problem, it still takes about 50s-1min to read + draw the figure).

    I would like to hear your thoughts on this.

    Thanks.

    Cheers,
    Franco

    • Claud February 29, 2020 at 7:59 pm #

      Hi Mike, Franco’s question is a question I have as well. Any input would be appreciated it. Thanks!

  5. Srini June 5, 2014 at 6:41 am #

    Mike:

    In problem 6, step 3, based on the given data, how do you determine that triangle is equilateral? In other words, how do you determine CD=6?

    • Mike MᶜGarry
      Mike June 5, 2014 at 10:27 am #

      Dear Srini,
      I’m happy to respond. 🙂 Look at triangle ACD. We know A is the center of the circle, so AC = AD are radii, and therefore of equal length. This means that triangle ACD is isosceles, with at least two equal sides, AC & AD. Now, angle A = 60, so the other two angles have a sum of 120 degrees. Well, by the Isosceles Angle Theorem, those other two angles must be equal, 60 degrees each. Three 60 degree angles: that’s an equilateral triangle.
      This is definitely an example of the kind of geometric reasoning the GMAT could expect you to do in a problem.
      Does all this make sense?
      Mike 🙂

      • Srini June 5, 2014 at 4:18 pm #

        Yes, it does Mike. Thank you!

        • Mike MᶜGarry
          Mike June 5, 2014 at 5:13 pm #

          Dear Srini,
          You are more than welcome, my friend. Best of luck to you in the future!
          Mike 🙂

  6. Sads May 12, 2014 at 8:17 am #

    Hi Mike,

    I am confused as to why triangle ABC and EFG in the 8th problem not similar triangles?

    AB and BH are parallel aren’t they then by AA rule aren’t the two triangles Similar?

    Thanks

    • Mike MᶜGarry
      Mike May 12, 2014 at 10:36 am #

      Dear Sads,
      I’m happy to help. 🙂
      It’s true that AB and EH are parallel, but one set of parallel lines is not enough to determine more than one angle if the triangles are not between the two parallel lines. It’s true that angle B = angle H, because of the double parallel lines. Because EF = FH, by the Isosceles Triangle theorem, we know that angle H = angle EFG, Then, by transitivity, we know angle B = angle EFG. That’s one equal angle the two triangles have in common. In triangle ABC, we know AB = BC, so the angles at B & C would have to be equal. We have no guarantee that EF and EG are the same length — in fact, in this “drawn to scale” diagram, it very much looks as if EF is considerable longer than EG. Therefore, ABC is an isosceles triangle, and EFG is not, so they can’t possible be similar.
      Does all this make sense?
      Mike 🙂

  7. tensor May 7, 2014 at 10:59 pm #

    Hi mangoosh,

    I am stuck @ 6th question. I am able to follow till the step where we subtract the area of triangle from the area of the sector.

    Area of circular segment=6*Pi*- 9 *sqrt{3}.

    We have two circular segments so shouldn’t the answer be 2*(6*Pi*- 9 *sqrt{3})

    = 12*Pi*- 18 *sqrt{3}.

    Can you please elaborate step 5 too??

    thanks

    Tensor

    • Mike MᶜGarry
      Mike May 8, 2014 at 11:43 am #

      Dear Tensor,
      I save all the “doubing” for the last step, step #5. In step #4, I simply calculate the area of ONE circular segment. That’s precisely what I intend to do there, because I wait until step #5 to put everything together and double — TWO triangles minus TWO circular segments. Those are the factors of 2 in front of the two parentheses in that step.
      Does all this make sense?
      Mike 🙂

      • tensor May 8, 2014 at 10:25 pm #

        Thanks mike. My bad, I thought question asked us to find the area of two circular segments. Now it makes sense.

        In question number 7. can you elaborate how angle B and angle H are corresponding angles? I don’t see any traversal between line segment AB and GH.

        Thanks

        tensor

        • Mike MᶜGarry
          Mike May 9, 2014 at 1:00 pm #

          Tensor,
          That’s a really trick one — both of those angles are formed by two pairs of parallel lines, so they have to be congruent. One way to think about this — suppose segment DEF were extended to the left of D, and suppose AB were extended above A, so that it intersected the extension of DEF — call that point Q. Then, we would have two triangles, HFE and BDQ — the pairs of angles at F & D and at E & Q would be equal, because the extended version of QDEF would a common transversal crossing both sets of parallel lines. Then, the two triangles would be similar, HFE ~ BDQ, by AA Similarity, and that would mean the angles at B & H would have to be congruent.
          Does all this make sense?
          Mike 🙂

          • Sads May 13, 2014 at 11:04 pm #

            At this point, we reach a very tricky move: both ∠B and ∠H are angles formed by the pairs of parallel lines — the sides of each one are parallel to the corresponding sides of the others. These means ∠B = ∠H = 40°.

            Sorry Mike Im still not able to grasp this point! 🙁

            Pls help

            Thanks

            • Mike MᶜGarry
              Mike May 14, 2014 at 2:20 pm #

              Dear Sads,
              I explained one way to think about this in this response to Tensor, and because you asked, I went ahead and copied that text and added it to the text explanation above. If you cannot visualize all that in your head, I suggest that you write it out on paper and look at it. Often, the best way to understand geometry is visually. Draw out the diagram, add the extended lines as I suggest, and that should make everything clear.
              Mike 🙂

              • Sads May 14, 2014 at 8:48 pm #

                Hi Mike,

                Yes now I understand that the two angles are corresponding angles between parallel lines BD and FH!

                Thanks a lot 🙂

                • Mike MᶜGarry
                  Mike May 15, 2014 at 12:18 pm #

                  Dear Sads,
                  You are quite welcome. Best of luck to you!
                  Mike 🙂


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