QC سوالات و هندسه

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QC سوالات و هندسه

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QC Questions & Geometry

Quantitative comparison questions and geometry. Here, I will be assuming that you’re familiar with the content of the geometry module.

Now, remember back in the lesson, assumptions and estimations in the geometry module, we established that geometry diagrams in the GRE are inherently unreliable, in other words, every diagram has next to it implicitly the statement, figure not drawn to scale.

And this was really a theme of our discussion of geometry, you can’t really discern anything about these diagrams because they’re not drawn to scale.

They’re not necessarily at all what they appear to be. So for example, you might see a triangle drawn like this. Looks like it’s equilateral or close to equilateral. In fact, that triangle might look like any of those.

Any side could be the biggest side. Any angle could be the biggest angle. We have no idea. So there’s something inherently deceitful about the diagrams, and if you’re naive, and trusting the diagrams, you will go astray 100% of the time. This remains conspicuously true on QC questions.

Unless a property of a figure is explicitly guaranteed, then you cannot assume it is true.

And many people will make the mistakes of believing their eyes, and trusting that something is true simply because it looks true in the diagram, without having an explicit guarantee. Now, some geometry questions on the quantitative comparisons do specify everything that would be doubted.

Essentially they lock the figure into place. And they’re really a tests of your ability to calculate various ge, geometry items. Figure out a length or an area, a diameter, a ratio something like this.

So everything is locked in place, really because everything is very specified and you’re just doing a calculation.

Some specify less, and test your spatial intuition for gif, different geometric possibilities.

So there’s really two kinds of geometry questions here, kind number one, everything is specified and you’re just doing a calculation of this fixed figure, and kind number two, where there’s out there elements in doubt and you can actually change the shape to be several different things.

Now, often, when you can change the shape, the answer is D, not always, but often. .

So here’s a practice problem. Pause the video, and we’ll talk about this. Okay. So this is a vide, this is a problem where everything is locked in place. Everything is specified here.

So we know the radius of the circle. We know length BC. We know that we have a square. We have a right angle.

The angle at C is a right angle on both sides. We aren’t given the side of the square, but we could calculate that.

So everything is fixed here. We know that triangle ABC is a right triangle. We know that AB, the diameter of the circle, happens to be four, because the radius is two, so the diameter is four and AB is the hypotenuse of that triangle. So what we have is a right triangle with a leg of three and a hypotenuse of four.

Now another chance to make a mistake here.

If people think right triangle and they hear three, four, blank, they think oh, three, four, five. The third side must by five. It’s a three, four, five triangle, but that is a mistake.

And it’s a mistake because in a three, four, five triangle, the five would be the hypotenuse.

Here the four is the hypotenuse. Of course, the hypotenuse has to be the longest side so there’s no way that AC could be five, longer than the hypotenuse.

Instead we have to use the Pythagorean Theorem which tells us that three squared, the leg three squared plus the leg AC squared would equal the hypotenuse squared four.

So of course, 3 squared is 9, 4 squared is 16. We get AC equals 16 minus 9, which is 7. Now, we could take a square root, but hold on a second.

Notice that AC is the side of the square. So AC squared is the area of the square. So, the area of the square is seven.

And thus, quantity A and quantity B are equal. So that was an example of a question where everything was fixed, and we simply had to do a calculation. Here’s another practice problem.

Pause the video and then we’ll talk about this. Okay.

Here, we have to be very careful. Think about what we know and what we don’t know. We know the two lengths, we know JL equals four and JK equals six. We know those two.

We know nothing about the angles. And therefore we know nothing about the shape, the true shape of that triangle.

Start by visualizing all the different possibilities for this triangle. So, we could have a triangle where we put these two sides, the four and the six, perpendicular to each other, or you could really crush them very close together.

So the triangle could be crushed down to nearly flat, with almost no area, with an area of almost zero. We could make the area one, or less than one.

We could make the area a thousandth or a millionth depending on how flat we crushed it. So we could get a very very small area, certainly much less than eleven.

Or, if we put those two sides perpendicular to each other, that actually creates the largest possible area.

Then, of course, four is the base and six is the height, so the area of that triangle would be one half base times height, one half four times six, and that equals 12.

So the area could be 0 or it could be 12. Now let’s go back to the comparison. How does the area of that triangle compare to 11?

Well, it could be smaller than 11 or larger than 11. Different relationships, so that means that the answer is D. And not surprisingly, in a geometry question in which there are many different possibilities for the shape, the answer is most likely D.

Sometimes everything in the quantitative comparison geometry question is fixed, and we just have to perform a calculation of some kind. That’s exactly what happened in the first problem. Sometimes, though, as in this problem, the one we just did, not everything that meets the eye is specified.

In other words, it look like the six was the hypotenuse of a right triangle. It looked like that, and many people would make that assumption, and even do calculations based on that.

They would just assume because it looks like a right angle, it is a right angle. And that’s what would lead many people astray.

It pays to explore the range of possibilities before diving into calculations.

It’s always good to have this discernment at the beginning of the geometry problem. How much about the diagram is fixed, and how much is variable? What can we change, and what are we not allowed to change?

You should always be asking yourself that question in a geometry quantitative comparison question. Here’s another problem.

Pause the video and then we’ll talk about this. All right. So this one, if you tried to do a calculation it would really be an elaborate, lengthy calculation. That would be very time consuming.

Instead, think about what could we change?

For example, we have no idea where point C is located. Would changing the position of point C make a difference?

Let’s think about this. Move point C around in your mind. If C is very close to B, then region J is almost the entire semicircle, and the triangle is almost nothing.

So here, region J is gigantic, and that triangle is really, really small. So region J is clearly bigger.

On the other hand, if we put C on the other side of, the other side there, then we make the triangle relatively big, and now region J gets crushed up against that one little side. The region J is now very small.

So here the triangle is clearly much bigger. So we can make the triangle much bigger or much smaller, different possibilities, which allow for different relationships, and as usual, when we can change the geometric diagram in a variety of ways, that often means that D is the answer. Finally, here’s a hard practice problem.

Pause the video and then we’ll talk about this in depth. Okay. So here, everything is locked in place. We have an equilateral triangle that’s part of a circle.

It’s true that we don’t know the, the, the actual size, but, because essentially, we’re looking at a ratio, we’re comparing one area to another.

It doesn’t matter if it were bigger or smaller, the, the, the relationship between those two areas would be the same. So essentially it’s like finding a ratio. The actual lengths don’t really matter. So nothing is variable here.

We have to do some calculations and, most importantly, we’re gonna have to use our estimation skills.

So first of all I’ll say draw the center of the circle, lines to the three vertices, and perpendiculars to the sides of the triangle. So divide it up like this.

And it’s very useful to know that you can divide an equilateral triangle like this. Essentially we’re subdividing the equilateral triangle into six congruent 30-60-90 triangles.

So those six triangles there are equal, and if we could find the area of one, then we could just multiply that by six and we’d get the area of the whole equilateral triangle. And so the, the total lengths don’t matter.

So that’s why we’re free to pick a convenient length, and we’ll just use that length consistently, and of course when we find ratios, that, the ultimate choice of that particular doesn’t matter.

So I’m going to say, let’s just pick OP equals two. So that’s convenient for a few reasons. First of all, means that we know the radius of the circle.

We’ll use that a little later. The radius of the circle is 2, and also 2 is the hypotenuse of the 30-60-90 triangle.

Well that’s very convenient, because you remember the 30-60-90 triangle.

The short side, that would be OT, that’s half the hypotenuse, that’s one. And then the longer leg, PT, would be root three. So those are the three sides of the 30-60-90 triangle, 1, 2, and root 3. All right. So now let’s think about this.

Now if, with that choice, we could find the area of one of those little 30-60-90 triangles, say TOP because we know the height OT is 1 and the base PT is root 3.

So it’s one half times 1 times root 3. Or in other words, root 3 over 2. That’s the area of one of those little 30-60-90 triangles.

We’ll multiple that by 6 now. Multiply it by 6 and we get 3 root 3. That’s the area of all six of those little 30-60-90 triangles. And that equals the area of the big equilateral triangle, MNP. So 3 root 3 is the area of that triangle. That’s very important.

So if the radius of the circle is 2, then of course we find the area by using Archimedes’ amazing formula, pi r squared. And, of course, that’s four pi, so that’s the area of the circle. Now, the shaded region.

That shaded region is the circle minus the triangle. So that would be 4 pi minus 3 root 3.

That’s the area of the shaded region. So now we’re ready to do our comparison. So the area of the triangle we said was 3 root 3, the area of the shaded region is 4 pi minus 3 root 3. Hm, this looks kind of hairy. First thing I’m gonna do is I’m gonna add that radical expression to both quantities, just to get all the radicals in one place.

You’re always allowed to add or subtract anything in a quantitative comparison, so that’s fine. So I’m comparing 6 root 3 to 4 pi. All right, still looks a bit hairy. Now, pi is a little more than 3. We can just approximate pi as 3 with a plus sign, I’ll write it.

3 plus, something slightly bigger than 3. And so this means that 4 times pi would be 4 times something slightly bigger than 3 which is something slightly bigger than 12. So now we’re comparing six root three to something slightly bigger than 12.

Now, simply divide both sides by 6, so now we’ll get root 3 and something slightly bigger than 2.

Well now we can do the comparison very easily. Because whatever value root 3 has, clearly root 3, the square root of 3 is between 1 and 2. So it is less than 2, and quantity, quantity A is less than 2, and quantity B is greater than 2, and so that means that quantity B is definitively bigger.

On the quantitative comparison geometry problems, never be naive about the diagrams. If some property is not explicitly mentioned, then you don’t know it. Do not simply trust what you see.

You have to get, you actually have to see, for example, it has to say that the angle is 90 degrees, or they have to write the little perpendicular circle.

If it’s just drawn to look like a right angle, you have no guarantee. It might not be anything close to a right angle. Very important that you do not believe anything unless it’s specified.

And remember also, there are these two kinds of geometry problems we’ve seen here. One kind, everything is fixed and labeled, and your job is simply to perform some kind of calculation.

And then the other kind, some properties are intentionally left variable.

And your job is to use your imagination and visual reasoning to see the full spectrum of spatial possibilities.

And once again, when there are many, many different spatial possibilities present in the problem, that often means that the answer is D.

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