Coordinate geometry problems can throw off the best of test-takers, and with good reason – who among us has worked with a coordinate plane since high school? (I definitely hadn’t when I began studying for the GRE). There’s little reason to fear coordinate geometry, though: Once you become comfortable accurately drawing figures on a place, then you just need to apply the same skills that you use on regular geometry problems. Take this GRE Quantitative Comparison question, for example:
Once we have a clearer idea of what we’re looking at, it can be tempting to throw in the towel and choose answer choice (D). We know that y, the y-coordinate of point B, is negative, but how can we tell whether it’s greater or less than -5, the value in Quantity B? But we don’t have to give up just yet – we have one more piece of information that might be able to help us.
If we go back to the centered information, we see that the total area of the triangle is 12. We can tell from our sketch that the base of the triangle is 4. If we plug our base and area into the formula for the area of a triangle, A = ½*b*h, we get 12 = ½(4)(h). If we continue solving for h, we get:
12 = 2h
6 = h
So we know that the triangle is 6 units “tall”, and we have our value for y: -6. So even though we don’t know exactly where point B is, it doesn’t matter – we were still able to find the y-coordinate’s value. Drawn into our graph, that looks like this:
Now all that’s left is to compare it to the value in Quantity B: -6 is less than -5, so Quantity B is greater and we have our final answer.
The test-writers could have posed a very similar question without testing coordinate geometry: We could just have easily have been told that we had a triangle with a base of 4 and an area of 12, and been asked to determine how the height related to a value in Quantity B. By making us plot the information on a plane instead, the writers are hoping that you’ll either be daunted by the very idea of coordinate geometry, or make a mistake when you draw the figure. Don’t fall for the traps – just sketch carefully, and you’ll be well on your way to gaining points on GRE Test Day!
GRE Geometry problems with multiple figures can often seem intimidating – the components of each figure overlap and intersect in sometimes-unclear ways, and you just know that they’re going to be time-consuming – rarely can GRE multiple figures problems be solved with just a few calculations. Take a look at this problem, for example:
It’s easy to picture the GRE test-maker laughing fiendishly as he inscribed an equilateral triangle into a circle, and then added a square in just to make your head explode. However, that portrait is likely inaccurate – each problem on the GRE is designed to test your critical thinking skills, and multiple figures problems are no exception. Let’s take this one apart: We need to find the area of polygon ABCDE, which is not a regular polygon. The language in the question stem gives us an important clue, though – if we find the areas of triangle ABE and triangle BCDE independently, then we can add them together to get the total area of the larger figure.
The only figure for which we have any real values is the circle, so let’s start there. The area is 36π, and if we plug that into the equation for the area of a circle, we get . Once we divide both sides by π and take the square root of 36, we see that the circle has a radius of 6. It would be very convenient if the radius or diameter of the circle corresponded to any side length of the triangle, but we’re not quite so lucky here. Instead, we’ll have to rely on a lesser-known geometry rule in order to continue solving.
Since we don’t have any radii drawn into the figure, let’s call the center of the circle P and add two radii, as shown:
Now, here’s that not-so-common geometry rule: If an angle is formed by two radii creating an arc (minor arc AE, in this case), and another angle starts on the opposite side of the circle and ends at the same arc-endpoints, then the former angle has twice the measure of the latter angle. If we apply that rule to this figure, we know that angle APE is twice the measure of angle ABE. And since angle ABE is an angle in an equilateral triangle, its measure is 60 degrees. When we double that, we find that angle APE is 120 degrees.
With me so far? I hope so, because we haven’t even found the triangle’s side length yet. We’re getting there, though! Now that we know the angle measure of APE, let’s cut it in half by dropping a height down to line AE, and creating two identical triangles. Since we’ve divided angle APE in half, each of the top angles measures 60 degrees. And since the newly-formed angles along line AE are both right angles, that means that the remaining angles in each triangle must measure 30 degrees each. Do these angles sound familiar? Knowing the ratio of side lengths in a 30:60:90 right triangle will put you far ahead of the pack on GRE Test Day. The ratio of sides in triangles with these angle measures are always x:x√3:2x. Since the known side in this case, the radius of the circle, is opposite the 90 degree angle, we know that 6 = 2x, which means that x = 3 and x√3 = 3√3. Here’s all of the information we currently have, drawn into the figure:
If we double the 3√3, we have the side length of the equilateral triangle: We now know that it’s 6√3. Now that we have the length of the triangle’s base, all we need is its height in order to calculate its area. Let’s take the triangle out of the figure for a moment: If we add a height, we create another pair of 30:60:90 right triangles. The side length opposite the 90 degree angle is 6√3, which means that the base is 3√3 and the height is 3√3 * √3, which equals 9:
Since we finally have the base and height of the equilateral triangle, we can calculate its area:
½ (base)(height) = ½ (6√3)(9), which gives us 27√3.
Looking at the answer choices, which ones now jump out at you? Answers (A) and (E) both have 27√3, and look to be the most promising choices. If you were tackling this problem on Test Day and were running out of time, now would be a perfect time to take a strategic guess and move on.
Calculating the area of square BCDE is now a snap: The equilateral triangle and square share a side, so we know that the square also has a side length of 6√3. Therefore, the square’s area is (6√3)2, or 108. The area of polygon ABCDE, then, is 108 + 27√3, which matches answer choice (A).
That was a marathon of a problem, by anyone’s standards. Let’s review all of the concepts that you needed to know and recognize in order to solve that problem:
- The formulae for areas of circles, triangles, and squares,
- The special geometry rule governing relative angle measures in circles,
- The ratio of sides of a 30:60:90 right triangle, and
- Recognition of the shared sides between the figures that allowed us to transfer information from one figure to another.
Putting all of those pieces together is a test of your critical thinking skills, if I’ve ever seen one! With practice, though, applying all of these geometry concepts to one problem will become easier and faster, and will allow you to rack up points on GRE Test Day.
Sometimes the geometric figures on the GRE can be a little overwhelming, often because of all of the formulas that you need to memorize, including these old favorites:
Granted, you have to remember how to calculate the area of a square/rectangle/quadrilateral (base • height), the area of a triangle (1/2 base • height), and the area of a circle (π • r2). I’m sorry you have to memorize these for the GRE, but you do. Make some flashcards, they will facilitate the process.
If you are not a geometry aficionado like I am, even just that might be enough for you; and expanding into three dimensions may make your head spin. I can hear you exclaiming out there, because I hear you exclaiming in my classes too. “I’ve gotta remember volume formulas now, too?!”
The answer is no, you don’t, not really.
The few of you who were lucky enough to go to a Montessori school will be able to conceptualize this pretty easily. The general rule is that to calculate the volume of any solid, all you have to do is calculate the area of the base and multiply it by the height. If you didn’t go to Montessori school, you can think of it like a deck of cards. Calculate the area of one card; then stacking them gives you the volume of the whole deck.
Therefore, the volume for a regular cylinder (you will only see regular solids on the GRE) is π • r2 • h – that’s the area of the base (a circle) times the height. The volume for a rectangular solid is l • w • h, and the volume for a triangular solid is 1/2 • b • h (of triangle) • h (of solid). You don’t necessarily need flashcards to remember these things, if you already know the foundational formulas and you understand the principle.
Solid geometry problems do not show up often on the GRE, but they are amongst the higher difficulty problems, though they can be very manageable when you have the right foundations. That means understanding this simple principle can help you quickly get those extra couple questions right in the math sections, which can give a healthy boost to your GRE score.
I’d encourage you to think creatively about other geometrical properties on the GRE in order to minimize the number of formulas you actually have to memorize. This is also a particularly good idea because the GRE rewards exactly this kind of understanding. Surface area is another great place to start – share your ideas in the comments!
My GRE students are quite aware of my love of the show 30 Rock. I always find a way to work it into our before-class chat and even into our lessons.
That’s because 30 Rock, if you haven’t seen it, is one of the most brilliantly written shows on television. Tina Fey and her adroit team of writers weave together countless and disparate plotlines into one homogenous, hilarious conclusion.
The reason I bring this up is that the very same thing happens with the quantitative content on the GRE. It’s slightly less hilarious, but it really does all come together.
Case in point: special triangles. Knowing these is a GRE lifesaver, as long as you know how to use them. The problem is, they require some prerequisite math knowledge to be used to their best advantage.
Any Kaplan student with a laminated strategy sheet can tell you that a right triangle with sides 3 and 4 in length will have a hypotenuse with the length of 5. That’s why they’re called 3-4-5 triangles. However, where my students have had trouble in the past is applying that base knowledge to save time on more advanced problems.
Say a question on the GRE asks for the perimeter of a right triangle with known side lengths of 21 and 35. You could use the Pythagorean Theorem to determine that the remaining side is 28. Then, assuming you remember to re-read the question and add all the sides together, you’d have the correct answer. You’d also have about a minute less time for the rest of the section than you would if you’d attacked the problem in a different way. To save time in the long run, sometimes you need to invest a little more time up front.
Anytime you’re working with a figure on the GRE, study it. For this problem, you see that you have a right triangle because the figure either shows that the largest angle is 90 degrees or has a box in the corner where the largest angle is. Determining you have a right triangle is your cue to start examining the side lengths to see if they fit the profile of a special triangle.
Sometimes, you’ll see a 4 and a 5, and your job is easy; the other side is 3. Or sometimes you’ll see a side 5 and a hypotenuse 5√2 and you’ll know that the remaining side is 5 (for a 45:45:90 triangle, one of those other special triangles you’ll be sure to review).
But sometimes, what you’ll see is a special triangle in disguise. In our example, the shortest side of our right triangle was neither 3, 4 nor 5. It was 21. But 21 is a multiple of three. More specifically, it’s 3 x 7. The hypotenuse, or longest side, of the triangle was 35. 35 = 5 x 7. Now we know that we’re dealing with a 3-4-5 triangle. The remaining side has to be a multiple of 4 in the same way the other sides were multiples of 3 and 5. 4 x 7 = 28.
Now we can use the calculator to add it all together and get the correct answer, 84, for the perimeter of the triangle. Or we could save even more time by extending our knowledge of proportions and number properties.
We know that our 21-28-35 triangle was just a 3-4-5 triangle in a “times-seven” disguise. We also know that the distributive property tells us that ab + ac + ad = a(b+c+d). Doesn’t it follow then, that 3(7) + 4(7) + 5(7) = (7)(3+4+5)? On that clunky on-screen GRE calculator, you’ll find that 7 x 12 is much quicker than 21 + 28 + 35.
Now, you can get an idea of how all the content you’re reviewing for the GRE can coalesce to make your life easier. And, as you’re moving onto bigger and better things, you need this part of your life to be as easy as possible.
It’s like Jack Donaghy said to Liz Lemon on 30 Rock: “Lemon, the grown-up dating world is like your haircut. Sometimes, awkward triangles occur.”
To do the truly grown-up thing, we must subvert the awkwardness of such triangles and appreciate them for what they are: masterful amalgams of math magic!