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GRE Geometry Diagram Assumptions

What can you assume on GRE geometry diagrams, and what can’t you assume?  First, here are a couple of practice questions.


Quadrilateral ABCD is a rectangle.



What you can’t assume on GRE geometry problems

The biggie is: no diagram is drawn to scale.  That means — if nothing is specified about the lengths, then any lengths may be the longest or shortest lengths of the diagram.  It also means that any angle that is not marked could be acute, right, or obtuse.   For example, if this diagram appears without further comment —

…. then it is cruelly and deceitfully trying to tempt you into believing that it is really a square.  If this were a square, then you would know a whole boatload of things (four congruent sides, four right angles, congruent & perpendicular diagonals, etc.)  The great unwashed masses taking the GRE will fall into this erroneous assumption, and all its implications, like lemmings running to the sea.  In fact, because the diagram specifies no lengths or angles, it could be any one of the following:

You always have to have your visual imagination warmed up for possible alternatives, with different lengths and different angles.  Any angle could be the largest or smallest angle.  Any side could be the largest or smallest side.  The shape could look not even vaguely like the explicit diagram that appears.  That’s all fair game in the GRE’s blanket statement: “no diagram drawn to scale.”

If angles appear to be right angles, you can’t assume they are right angles unless the test says so, or unless a little “perpendicular square” appears in the diagram.  For example, in #1 above, the text specifies that the figure is a rectangle, so this means you can assume it has all the properties of rectangles (four right angles, congruent opposite sides, etc.)  In #2, ∠EGF is marked with the little perpendicular square, so we are guaranteed that ∠EGF = 90°.  If nothing is marked or specified, you are falling into a trap to assume that an angle that looks right in the diagram truly is right.

If lengths appear equal, they may not be.  If JK appears longer than WX, then the relationship may be that way, or they may be equal, or it may be that WX  is in fact longer that JK.  If lines appear parallel or perpendicular, you cannot assume either if it’s not specified.  You should always be suspicious about a GRE geometry diagram, and the less that is specified in words, the more suspicious you should be.


What you can assume on GRE geometry problems

The most fundamental thing you can assume about any geometric diagram: any line or line segment that looks straight, is straight.   For example, in problem #2 above, we are absolutely guaranteed that D-E-F are collinear and that C-E-G are collinear — that those two are straight lines with no “hidden bend” at point E.  If it looks straight, it is straight.

BTW, don’t confuse “straight” (meaning, “lying in a line, collinear”) with “horizontal.”  Many people say “straight” when they mean “horizontal”, and this is a 100% wrong mistake that leads to a great deal of confusion.  For example, you must assume a line on the GRE is “straight”, but you absolutely cannot assume it is “horizontal” if that is not stated.

If you are told a shape in a GRE diagram is a polygon in general or a particular kind of polygon (triangle, rectangle, etc.), then you can assume

(a) it is a closed shape:

(b) that the sides don’t “cross” each other and that the figure is “convex” (all vertices pointing outward), not “concave” (some vertices pointing inward).

Finally, it’s important to know the theorems and the defined properties of shapes.  For example, if you are told that two angles in a triangle are equal, then by that extraordinary theorem, the Isosceles Triangle theorem, you also know the two opposite sides are congruent.  If you are told that a figure is, say, a rhombus, then you automatically know it has all the defined properties of a rhombus (four congruent sides, congruent opposite angles, perpendicular diagonals).   Technically, if you know something as the result of a definition or a theorem, then you are deducing it, not assuming it, but this is still in the broad categories of things you need to know about your diagram that were not explicitly stated.



Be very clear on what you can assume and what you can’t assume — I would guess that over 80% of all the mistakes that folks make on GRE Geometry problems involving diagrams result from improper assumptions.   It may pay to go back to the practice questions above and re-evaluate them in terms of what you have learned here.  Here’s a further practice question:




Practice question explanations

1) In this question, we are guaranteed that it’s a rectangle, which substantially reduces the range of possible shapes.  Notice, though, we know absolutely nothing about the relative lengths of the two sides — either one could be much larger than the other.  It’s good to visualize two versions, one with each of the sides exaggeratedly longer than the other:

In the version on the left (AD > AB), it’s clear that AC/AB is much much bigger than 1, may be equal to 5 or 6, while AB/AD is small, only say about 1/5.  In that version, column A definitely would be bigger.

Meanwhile, in the version on the right (AB > AD), now AC/AB is still larger than 1, but here just larger than 1 — maybe 6/5 or 7/6, much less than 2 — and now AB/AD is very large, maybe 4 or 5.  In this version, column B would be bigger.

Because the diagram can be changed to make either column bigger, we cannot draw any conclusions.  Answer = D


2) It’s true, we know ∠EGF = 90°, but we don’t know whether EG > FG or EG < FG.  Furthermore, we know nothing about the slant of segment CD — for the angles at C & D, either one could be acute or right or obtuse.  There is a lot we don’t know.

We do know, though, that D-E-F are collinear and that C-E-G are collinear — that those two are straight lines.  This means, the two angles at point E, ∠CED and ∠FEG, have to be the angles formed by intersecting lines —- known in Geometry as “vertical angles” — and according to a basic theorem, they must be congruent.

Think about it — D-E-F is a straight line, so ∠CED + ∠DEG = 180°; C-E-G is a straight line, so  ∠FEG + ∠DEG = 180°.   Rearranging these two equations, we know that ∠CED = 180° – ∠DEG and that ∠FEG = 180° – ∠DEG.  The right side of those two equation are identical, so that means the left sides must be equal to each other — equal to the same thing means equal to each other!  Therefore, ∠CED =  ∠FEG.

Answer = C


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17 Responses to GRE Geometry Diagram Assumptions

  1. Mahmud Ashraf shamim July 17, 2015 at 12:56 am #

    Mike you a great math teacher . Thanks for your crystal clear presentation of the GRE math.

  2. Meh September 15, 2014 at 6:58 am #

    Hi Mike

    You are very smart and unique person. I like the way you explain the math. You make math easy to understand in a nice way. Thank you

    • Mike MᶜGarry
      Mike September 15, 2014 at 9:45 am #

      Dear Meh,
      You are quite welcome, my friend. 🙂 It gives me immense satisfaction to know that you found my explanations helpful. I wish you the very best of luck in your future.
      Mike 🙂

  3. Pia August 4, 2014 at 9:15 pm #

    Hi Mike,

    I am confused between example 1 and 3; aren’t they kind of the same? When I looked at example 1, I thought the AC/AB would be greater than than AB/AD because AC is a hypotenuse, kind of how it works in the third example. I read the explanation behind example 1, and the figures drawn do allow for the provided conclusion, but I am still confused.

    Thank you!

    • Mike MᶜGarry
      Mike August 5, 2014 at 2:43 pm #

      Dear Pia,
      In #1, it is definitely true that hypotenuse AC is longer than either AB or AD, and therefore, AC/AB is definitely a fraction greater than one. The trouble is, depending on the shape of the rectangle, we don’t know whether the ratio of sides, AB/AD, is much smaller than 1 or much larger than one. It could be either.
      In #3, be careful: the order of the fractions is switched. In the triangle on the left, triangle ABD, AB is the hypotenuse and longest side, so BD/AB is a fraction less than one. In the triangle on the right, triangle BCD, BC is the hypotenuse and longest side, so BC/DC is greater than one. Anything greater than one is always bigger than anything less than one. In one fraction of #3, the hypotenuse is in the denominator, and in the other, it’s in the numerator. This is in contrast to #1, in which the hypotenuse appears in only one of the two fractions.
      Does this make sense?
      Mike 🙂

  4. Hashir October 8, 2013 at 6:33 pm #

    Hi Mike !

    In diagram 1, AD is horizontal besides straight. I can conclude because we have to keep angle between AD and AB 90. Right?

    In diagram 2, CEG is straight but not horizontal. We can play with all angles and sides in this diagram except angle G. So i can keep this angle 90 and come up with other possible shapes. Like making CEG downward and DEF more downward inclined to the extent that angle G is 90. Right Mike ?

    • Mike MᶜGarry
      Mike October 9, 2013 at 10:10 am #

      Dear Hashir,
      Because you are told explicitly that the figure in diagram #1 is a rectangle, you know all four angles are 90 degrees. If the word “rectangle” did not appear, you could assume absolutely nothing about the angle.
      In diagram #2, CEG is straight. Whether it is “horizontal” has absolutely no relevance to this problem. Yes, you could rearrange the lengths and angles other than G to make a variety of shapes. In all of them, it would turn out that angle CED equals angle FEG.
      Does all this make sense?
      Mike 🙂

      • Hashir October 9, 2013 at 10:27 am #

        Hi Mike !

        Yeah it all makes sense.

        Soory i didn’t clear the purpose of my post which was to check wether i have understood the assumptions which i can make regarding diagrams. There was no query regarding given questions. and yeah i do know that whatever shape i come up with in diagram#2, angle CED will be equal to angle FEG. I just wanted to make sure that i can reshape diagram #2. So thanks for the clarification Mike. You have been a great help.

        • Mike MᶜGarry
          Mike October 9, 2013 at 10:41 am #

          I’m glad you found my explanation helpful. Best of luck to you!
          Mike 🙂

  5. Glory July 18, 2013 at 12:39 pm #

    Small error********right/left

    “Meanwhile, in the version on the left “

    • Mike MᶜGarry
      Mike July 18, 2013 at 1:02 pm #

      Dear Glory,
      Thank you very much. I fixed that typo.
      Mike 🙂

  6. abc April 15, 2013 at 1:20 pm #

    Hi Mike,
    The ratio of sides of a 30:60:90 triangle is x:3^.5 x :2x.

    However, the Pythagorean triplets do not satist this sides formulae…

    for ex, 3,4,5 form pythagorean triplet, but do not satisfy above sides ratio rule ..

    This has been having me all mixed up

    • Mike MᶜGarry
      Mike April 15, 2013 at 1:33 pm #

      The 30-60-90 right triangle is a very special right triangle — the angles are very special, and as it happens, the ratios of its sides are irrational, so it’s impossible to have all three sides of a 30-60-90 triangle be integers at the same time. You can have at least two sides be integers at once, so that’s also a special property. They always will satisfy the Pythagorean Theorem, but they can’t always be integers. That’s just one special right triangle.
      There’s a separate group of special right triangles that form the Pythagorean Triplets — these are right triangles in which all three sides can be integers. All of the angles in these triangles are irrational — for example, the smallest angle in a 3-4-5 triangle is approx 36.86989765… (an irrational number of degrees). These also satisfy the Pythagorean Theorem. You can read more about these here:
      Then, there are other right triangles that have integer angles (e.g. a 20-70-90 triangle) but all the sides are irrational ratios. These also satisfy the Pythagorean Theorem.
      Then, there’s the general population of right triangles with both irrational sides and irrational angles — there’s more of these than any others. While this is by far the most common type of right triangle, this is the type you will see the least on the GRE, because there are no special properties at all. All of these also satisfy the Pythagorean Theorem.
      Does all this make sense?
      Mike 🙂

  7. Cindy January 31, 2013 at 9:07 am #

    Do you have a list of properties/rules that may be helpful to memorize? For example, special triangles like the 45,45,90 right triangle have a ratio of 1:1:√2 so if you are given the length of one side of such a triangle you can easily fill in the missing info. If so, can you please provide a link? Thanks in advance. 🙂

  8. Ryan January 22, 2013 at 1:02 pm #

    Hi Mike,

    Would it be safe to reason that the figure in example 1 could be considered a square, due to all squares being a type of rectangle but not all rectangles being squares?

    • Mike MᶜGarry
      Mike January 22, 2013 at 2:29 pm #

      Ryan: yes, if you know it’s a rectangle, one possibility among many is that it is the most elite of all rectangles — the perfect square. It’s good to remember that is *one* of many possibilities, but beware of falling into the trap of assuming it *has to* be a square — allowing your mind to be hypnotized by that single magically symmetrical possibility to the exclusion of all other. That is an all-too-typical mistake where squares are concerned.
      Mike 🙂

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