Browsing articles in "GRE Geometry"
Nov
7
2012

# Coordinate geometry – think regular geometry, with an extra step or two

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:

We’re given two of the triangle’s three vertices, and we know that point B is in quadrant III since both of its coordinates are negative. If we draw that out, it looks like this:

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!

Sep
7
2012

# GRE Geometry with Multiple Figures

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:

1. The formulae for areas of circles, triangles, and squares,
2. The special geometry rule governing relative angle measures in circles,
3. The ratio of sides of a 30:60:90 right triangle, and
4. 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.

Sep
3
2012

# GRE Math in Real Life: Return of Mega Carrot Cake

In my last post, I discussed how GRE math shows up in real life situations.  My wife and I, while baking a carrot cake, decided to increase the recipe by adding more of each ingredient.  This would, we surmised, create a Mega Carrot Cake (MCC).  The problem we faced, though, was that we didn’t quite have enough flour to double the recipe.  So instead of making a cake twice as big, we were forced instead to increase the amount of each ingredient proportionally, thus creating an Almost Twice As Big As Normal But Not Quite Twice As Big Mega Carrot Cake (ATABANBNQTABMCC).

Actually, you know what?  That acronym is much too long.  Let’s just stick with MCC.

Once we determined the amounts for each ingredient, we threw everything together in a bowl and began mixing.  But then another thought hit me.  Will this MCC fit in our cake carrier?  We were making this cake for a friend’s housewarming party, and it was essential that when all was said and done, we be able to transport the MCC from our house to theirs.  The cake pan we bought was deeper than our other pans, and the diameter was an inch larger.  So not only would each layer of the MCC be thicker and wider, but stacking the layers would make the cake, as a whole, higher than a normal carrot cake.  Our cake carrier was about 12″ tall, and previous iterations of this particular carrot cake stood about 9″.  Would there be enough room in the cake carrier for our MCC?

If you remember back to my last post, you’ll remember that we increased the ingredients by 60% (going from 2.5 to 4 cups of flour is an increase of 1.5 cups, and 1.5/2.5 is 60%).  That means the volume of our cake should increase by 60%, as well.  So if I could find the volume of the old carrot cake, increase it by 60%, then divide by the area of the new pan, I should have the height of my new cake.  Whew.  I’m getting tired just thinking about it.  All I wanted was to make a Giant Carrot Cake.  Is that too much to ask?

Our old cake pan was advertised as having a diameter of 8 inches.  If the height of the normal carrot cake was also 8 inches, what was the volume of the older cake?

If you’re having trouble finding that value, remember the formula for the volume of a cylinder:

p * r2 * h

In other words, if we can find the area of the round cake pan, then multiply it by the height of the cake, we should have our volume.  The cake pan has a radius of 4 inches, so its area is 16p inches2.  Multiply that by 9 and we end up with 144p inches2.  Now, in order to determine the volume of the larger cake, we have to increase it by 60%.  We can do that by multiplying by 1.6.  When we do that, we find that our Mega Carrot Cake has a volume of 230.4p inches3.

That’s a big cake.

But we’re not done, are we?  We still need to find the height of the MCC.  (Please be less than 12″. Please be less than 12″.)

• Remember that the new cake pan actually has a larger diameter – this one is 9 inches across.
• If the diameter is 9 then the radius must be 4.5 inches, which means that the area of the base of the cake will be (4.5in)2p, or 20.25p in2.
• Divide 20.25p in2 inches into 230.4p inches3 and…YES!!  11.38 inches is the height of the MCC!

The giant cake will live!  Well, live long enough to make it to the housewarming party, that is.  After that, all bets are off.

As you can see, math in real life can be just as complicated as math on the GRE, and vice-versa.  The concepts and principles that are being tested by the GRE test-makers are not relevant simply in the abstract. Ratios, proportions, percents, and volumes are mathematical concepts that have daily, real-world applications.  And the more you practice your GRE math skills, the better you’ll get at navigating the waters of a math-filled world.

Now, where’s my fork?  This MCC isn’t going to eat itself.

Aug
3
2012

# GRE Math: Volume Transfer and Phelpses in the Pool

Ah, the Summer Games.  A time to sit back, relax, watch the world’s best athletes compete, and contemplate nonsensical GRE-related questions like, “If you dumped all of the water from the diving pool into the swimming pool, how many Michael Phelpses would be needed to completely fill up the rest of the swimming pool?”

What?  You don’t sit around wondering about such things?  Okay, well, maybe I’m just weird then.

As strange as this little math riddle may sound, it’s actually not all that different from a question type you could see on the GRE – the volume transfer question.  In these questions, the GRE test-makers will ask you to evaluate the volume of two separate shapes, then determine the change of volume when the contents of one shape are transferred to the other shape.  In our example, we’re looking at the diving pool and the swimming pool at the Summer Games.  First, let’s talk about the sizes of the pools: the swimming pool has dimensions of 164 ft. long, 82 ft. wide, and 6.5 feet deep, while the diving pool is only 100 ft. long and 50 ft. wide, but 14 ft. deep.  Now, let’s say that for some unknown reason, London suffers a major water shortage during the Summer Games.  In this hypothetical situation, the games organizers only have enough water to fill up the diving pool.  After the divers compete, the organizers must pump all the water over to the swimming pool.  So, here’s my first question:

Once all of the water has been pumped from the diving pool to the swimming pool, how much of the swimming pool is filled with water, as a percent of total possible volume?

First you need to determine the volume of each pool.  To do that, simply multiply the length, width, and height of each pool.  When you do that, what do you get?

The volume of the diving pool is 100 x 50 x 14, or 70,000 cubic feet.

The volume of the swimming pool is 164 x 82 x 6.5, or 87,412 cubic feet.

Now, imagine all of the water from the diving pool is pumped into the swimming pool.  How can you determine the percentage of the swimming pool that is now filled with water?  Percent can be found by taking the partial value and dividing it by the whole value.  In this case, there are 70,000 cubic feet of water in a pool that has a total possible volume of 87,412 cubic feet.  70,000 divided by 87,412 becomes, roughly, 80%.  So there you go.  The swimming pool is 80% full.

While the question above does mimic a common GRE question type (volume transfer), this next question does not.  Regardless, it’s still fun to occasionally ponder such things.  Here’s what I want to know now:

How many Michael Phelpses are needed to fill up the rest of the swimming pool?

To answer this random question, what’s the first thing we need to figure out?  We need to know how much volume Michael Phelps takes up.  To do this, we can approximate by finding his weight (about 186 pounds), and dividing by 62 pounds, which is equivalent to one cubic foot of water.  Because the human body and water have an almost equal density, we can essentially equate 62 pounds of human mass with one cubic foot of water.  If Michael Phelps weighs 186 pounds, then he must be taking up 3 cubic feet of water.

Okay, so, here we go.  I’ve got an endless row of completely identical, totally dominating, 100% gold-medal-winning Michael Phelpses ready to jump into our pool until it’s completely full.  ”Michaels, are you ready?” (Silence, of course, as the fictional Michaels are very confused as to why there are so many of them, and why they’re being asked to jump into a pool that is only partially filled.)  ”Okay, Michaels, go, go, go!”  And I send them off, one at a time, jumping into the pool.

Let’s go back to the question.  How many Michael Phelpses are we going to need to fill up the rest of that swimming pool?  Well, let’s think about this.  If we’ve already dumped 70,000 cubic feet of water into a pool with capacity 87,412 cubic feet, then there must be 17,412 cubic feet left to go to reach full capacity.  Each Michael that jumps into the water will add an average of 3 cubic feet to the volume.  One by one they go in, slowly filling up the pool.  When do I say stop?

Right!  17,412 divided by 3 comes out to 5,804.  So, as the Michaels go jumping in, one after the other, I’ll need to stop the 5,805th Michael and softly, kindly let him know that we have all the Michaels we need.  Meanwhile, 5,084 Michael Phelpses are now sitting in the completely full pool, wondering just what exactly it is they’re supposed to do now.

So, first and foremost, be on the lookout for volume transfer questions on the GRE.  These are relatively straightforward questions that may appear difficult but can be mastered with a little bit of memorization. If we know the formula for the volume of a rectangular solid (length x width x height), then these problems can be made quite simple.

In addition, as you sit on your couch enjoying the Summer Games with friends, don’t be afraid to drop some of the knowledge you’ve gained by reading through this post.  ”Hey, did you know that if you dumped all of the water from the diving pool into the swimming pool, you’d still need over 5,000 Michael Phelpses to completely fill it up?”

If they give you any funny looks, just tell them that knowing such vital facts is simply part of your preparation for the GRE.

Jun
25
2012

# GRE Geometry: Volume of Solids

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!

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