Posts tagged with: shaded region

In figure 5, rectangle ABCD is inscribed in a circle. If the radius of the circle is 1 and AB = 1, what is the area of the shaded region?

A) 0.091
B) 0.285
C) 0.614
D) 0.705
E) 0.732

Draw a diagonal of the rectangle. That’s going to be the diameter of the circle, too, so you know it’ll have a length of 2. That makes a right triangle with a leg of 1 and a hypotenuse of 2. Do you know your special right triangles? If you do, you recognize that that’s a 30º-60º-90º triangle, so you can save yourself a tiny bit of time: the long side of the rectangle is \sqrt{3}. Nice, right?

OK, now draw the other diagonal.

What you’ve got there, now, is a central angle, which tells you that a 120º angle cuts out the only part of the circle you care about. If you don’t know why that’s important, read this.

The area of the whole circle is π, which you know because the radius is 1. The sector you care about is \dfrac{120}{360} = \dfrac{1}{3} of that, so \dfrac{\pi}{3}. The triangle you’re taking away from it has an area of \dfrac{1}{2}(\sqrt{3})\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{4}.

To find the area of the shaded region, subtract!

\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4}=0.614...

Yesterday I went to a museum and waited in line for a very long time for my turn to spend 5 minutes staring at a piece of art that completely mystified me. I guess it’s good art if I’m still thinking about it a day later, even if the thoughts I’m having mostly revolve around my own resignation that I will never really understand art. But it wasn’t a complete loss! There was a circular pattern on the floor that I spent a lot of time staring at while in line, and it ended up inspiring this challenge question.

As always, first correct answer in the comments will win a Math Guide. All the usual contest rules apply: previous winners can’t win; if you live outside the US you have to pay for shipping; etc.

In the figure above, two congruent circles are tangent at point D. Points D, E, and F are the midpoints of AB, AC, and BC, respectively. If AB = 12, what is the area of the shaded region?

Good luck!

UPDATE: Congratulations to John, who got it first. Solution below the cut…


When we’re asked to solve for the areas of weirdly shaped shaded regions, we’re almost always going to find the area of a larger thing that we know how to calculate, and then subtract small things we know how to calculate until we’re left with the weird shaded bit:

AwholeAunshaded = Ashaded

The first thing we should do is mark this bad boy up. We know AB = 12, and D is the midpoint of AB and also the endpoint of two radii. We also know E and F are endpoints of two radii, and midpoints of AC and BC, respectively.

At this point, we actually know a great deal. First, we know the radius of each circle is 6. That means each circle has an area of π(6)2 = 36π. We’ll come back to this in a minute.

It should also be obvious that ABC is an equilateral triangle. This is awesome, because equilateral triangles are easily broken into 30º-60º-90º triangles, which is what we’ll do to find the triangle’s area.

So triangle ABC has a base of 12 and a height of 6√3.

Now that we have that, all we need to do is subtract the areas of the circle sectors (in green below) that aren’t included in the shaded region.

Areas of sectors are easy to calculate. All we do is figure out what fraction of the whole circle the sector covers by using the central angle. In this case, the angles are 60º, so we’re dealing with 60/360 = 1/6 of each circle.

We need to subtract two sectors from the area of triangle ABC to find our shaded region:

And there you have it! Cool, right?

I’m heading out this morning to go camping for the weekend (Woohoo! Bug bites!) and I won’t have any way to access the Internet reliably, so if you enter this contest please be patient. I probably won’t be able to declare a winner or post a solution until Monday evening.

I’m told the place we’ll be camping is in a wireless dead zone, which inspired this question. Note that I’m pretty sure this isn’t really how cell carriers set up their networks, but it’s a fun problem. Because it’s a bit more involved than normal, I’m going to say this challenge is worth TWO Math Guides—one for you and one for a friend. They’ll both get shipped to you, and then if you don’t have any friends you can sell it on eBay or something. I dunno. I’m sure you have friends.

Annnnd here we go!

A wireless phone company’s cell towers are arranged in regular hexagonal structure, as shown in the figure above. If each tower’s coverage radius is exactly half the length of a side of the regular hexagon, what percentage of the hexagon’s area will not be covered by the towers?

Put your answers in the comments! The usual rules apply: you must not be anonymous to win, and although you may win if you’re not in the US, you will have to help me out with shipping costs via PayPal.

Good luck, and have a great weekend.

UPDATE: Nice work, Rex T95! Two books are on their way to you! Solution below the cut.


So you’ve got a couple of tricky things going on here, but this question is not NEARLY as gnarly as it looks at first.

We’re going to treat it like a shaded region question (because that’s what it is). First, we’ll find the area of the hexagon. Then, we’ll find the areas that are covered in the hexagon by the towers. Do a little subtraction, and voila! We’ll have our answer.

Step 1: Find the area of the hexagon
There are a lot of ways to do this, but the way I find most intuitive is to remember that all regular polygons are made up of congruent isosceles triangles that meet in the center, like so:

Since the center is 360º, you can figure out the measures of all the center-touching angles by dividing 360º by the number of triangles (which, of course, is the same as the number of sides):

360º ÷ 6 = 60º

So in a regular hexagon, the central angles of the isosceles triangles are all 60º, which means these triangles are not only isosceles, but equilateral. See how I did nice to you? You’re welcome.

So we need to find the area of an equilateral triangle, and multiply it by 6 to get the area of our hexagon.

Step 1a: Find the area of an equilateral triangle
At this point, I’m going to make an executive decision and plug in 2 for the length of the sides of my hexagon. We could do this without plugging in, of course, but it’d be more of a pain and I don’t subscribe to the school of thought that math should be difficult if it doesn’t have to be. So we’re dealing with an equilateral triangle with sides of length 2:

Surely, you know by heart that the area of a triangle is (½)bh. When we drop an altitude to find the height of an equilateral triangle, we create 2 30º-60º-90º triangles, so we don’t even need to Pythagorize since we know our special rights so well

The base of the above triangle is 2, and its height is √3.

A = (½)2√3 = √3
Ahexagon = 6×A
Ahexagon = 6√3

Step 2: Find the area that is covered by the towers
This is actually a bit easier. We’ve already decided to make the sides of the hexagon equal 2, so the radii of all the circles in the diagram are 1, and therefore each has an area of π.

Furthermore, the 6 circles that are only partially included in the hexagon are all sliced the exact same way: in 120º chunks.

So let’s do a little part-whole ratio calculating to figure out the sector areas.

Of course, there are 6 of those sectors. So the total area covered by the cell towers is 2π (for the 6 sectors) + π (for the center circle) = .

Step 3: Find the uncovered area
This is the easy part! The area that’s not covered is simply AhexagonAcovered.

Auncovered = 6√3 – 3π

Step 4: Find the percentage of the hexagon that’s uncovered
This last step, where we convert to a percentage, is why it doesn’t matter that I chose to plug in a value instead of working through this whole problem a bit more algebraically.

Throw that in your calculator, and you’ll get 9.31%.

Phew!

I didn’t do a Weekend Challenge last weekend. It’s not like I didn’t want to, y’all. Things just got totally cray-cray. My use of “cray-cray” in the previous sentence should be enough of an indication to you that things remain squarely thus.

I actually wrote two questions for this weekend, but I’m only going to use one of them. The other one is going into the strategic reserve. For emergencies.

If you’re the first to answer this correctly (and not anonymously), I’ll bestow upon you coveted access to the PWN the SAT Math Guide Beta.

Other ways you can gain access to my Magnum Opus:

  • Buy it. It’s $5. You get about 300 pages of useful SAT math help. I get a footlong meatball sub at Subway.
  • Send me a question of your own making. Added bonus: this is a fantastic way to solidify your knowledge of the test.
On to the question:

In the figure above A is the center of the circle, A and D lie on BF and CE, respectively, and B, D, F, and G lie on the circle. If BC = 3, and DG (not shown) bisects BF, what is the total area of the shaded regions?

Good luck, and have a great weekend. I’ll post the solution early next week.

UPDATE: Commenter Katieluvgold got it first. Nice work Katie!  Solution posted below the cut.

This is a shaded region question, and the usual shaded region technique applies, but with a twist. Since only parts of the top half of the circle are shaded, let’s just look use the top semicircle as Awhole.

We know AD is perpendicular to CE because it’s a radius connecting to a tangent line. Those are always perpendicular. It’s a rule.  And since A and D are both on rectangle BCEF, and BC = 3, AD must also equal 3.

So the radius of the circle is 3, so the area of the circle is 9π. That means the area of the semicircle is half that. Awhole = 4.5π.

Point A is the center of the circle, so it’s obviously the center of diameter BF. When the question tells us that DG bisects BF, it’s telling us that DG is also a diameter because it also passes through the center of the circle. That means ΔCEG has both a base and a height of 6.

Because the rectangle’s top and bottom are parallel, that means the top triangle (which is our Aunshaded) is similar to the large triangle, because all the angles of the two triangles are the same (the top one is shared, and the bottom ones are corresponding angles across parallel lines). We don’t have a way of knowing what the base is automatically, but we DO know that the height of it is 3, because the height of the little triangle is a radius. So the small triangle has a base of 3 and a height of 3.

The area of a triangle is (1/2)bh, so Aunshaded = 4.5.

To calculate Ashaded, we just subtract Aunshaded from Awhole:

AwholeAunshaded = Ashaded
4.5π – 4.5 = Ashaded

So the question writing contest I proposed a few weeks back didn’t exactly explode onto the scene like I thought it would, but I still think it’s a fantastic way for you to improve your skills, so just FYI: it’s still open.

I got the question above in an email the other day. The writer prefers to remain anonymous, but I’ve given him Beta access to the Math Guide because it’s an awesome question. It incorporates circle properties, shaded regions, and has great, well-planned incorrect choices. That’s good hustle.

Here’s an important thing to remember: all figures on the SAT are drawn to scale unless indicated otherwise. In other words, if it doesn’t say “Note: figure not drawn to scale,” underneath it, it is drawn to scale. Most figures on the SAT are drawn to scale, which means it’s a good idea to guesstimate whenever possible.

Guesstimating could mean actively trying to eyeball relative angle measures, areas, or segment lengths, or it could mean sliding pieces of the diagram around in your mind. You might still end up doing some math because guesstimating doesn’t lead you all the way to an answer. But it’s important that you not waste the opportunity when a diagram is drawn to scale. Let’s dig right into an example:

  1. The figure above depicts two intersecting diameters of two concentric circles of radius 6 and 10. If the diameters are perpendicular, what is the area of the shaded regions?
     
    (A) 32π
    (B) 50π
    (C) 58π
    (D) 64π
    (E) 74π

STOP DOING MATH! You need almost none of it to solve this problem. How can you guesstimate this?

Well, what’s the area of the large circle? 100π. What’s half of that? 50π. Good, the answer is (B). Done.

Why? Glad you asked. What happens if I rearrange the pieces of the puzzle?

OH HELL YES. Look at that. Doesn’t that excite you? I love guesstimating so friggin’ much.

Note that guesstimating doesn’t just apply to shaded regions. You can use it to solve all kinds of geometry questions. As long as a figure is drawn to scale, you should ponder the implications of guesstimate for a few seconds before you start doing any math. This takes practice, but I promise you it’s worth it when you become proficient.

The questions below can be solved with math, but your mission is to solve them with guesstimate. Make me proud.

Don’t stop! Don’t ever stop!

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If you’ve ever sat down and taken a practice (or real) SAT, you’ve come across shaded region questions. They’re among the most iconic question types on the test, so much so that you may find that the memory of them remains with you long after your SAT taking days have passed. True story: I had a roommate in college that used to talk in his sleep sometimes, and one time I woke up in the middle of the night to hear him plaintively moaning about shaded regions.

Should you let yourself get intimidated by a shaded region questions? ABSO-EFFING-LUTELY NOT.

Say I told you that the area of the entire blob shape in the figure above was 15, and then asked you for the area of the shaded region. It’d be cake, right?

If I know you like I think I do, you’d probably say something like: “Thank you for insulting my intelligence with this asinine question; it’s 5.”

All you did was recognize that since areas just add up, and you know that the unshaded areas add up to 10, the shaded region has to make up the rest of the total area. If the total area is 15, and the unshaded part is 10, then the shaded one has to be 5. Easy, yes?

So it is with all shaded region questions:

Awhole – Aunshaded = Ashaded

Let’s try an example, shall we?

  1. In the figure above, P is the center of the circle and also the intersection of the two right triangles. If the radius of the circle is 3, what is the area of the shaded region?
     
    (A) π
    (B) 6π
    (C) 9π – 9
    (D) 9π – 6
    (E) 9π – 3

In order to solve for the shaded region, we need to find Awhole and Aunshaded.

The area of a circle is πr2, so Awhole = π32 = 9π.

What’s the area of the unshaded bits? Note that they’re both right triangles, and that each leg is a radius. In other words, we know the base and height of both triangles are 3. The area of one of the triangles is ½bh = ½(3)(3) = 4.5 Since there are two of the triangles, Aunshaded = 9.

So far so good? Now we can solve:

Awhole = 
– Aunshaded = 9

Ashaded  = 9π – 9

That’s choice (C). Easy, right? Just remember Awhole – Aunshaded = Ashaded and you’ll be fine.

I know this seems painfully obvious, but it’s important to remember that the SAT specializes in making it tricky to deal with concepts that are, on the surface, obvious.

Think about Rubin’s Vase (pictured at the top of the post) for a minute. Are you familiar with this image (or this kind of image)? If I told you I was about to show you a picture of a vase and showed you the black and white version, you’d see a vase. But if I told you I was about to show you a picture of two faces and showed you the same image, you’d see the faces.

The writers of the SAT, of course, know this. So when they ask you to find the area of a shaded region, they’re usually trying to make you focus on the most difficult thing to find directly (and sometimes, an impossible thing to find directly). That’s why it’s important to always keep the formula above at the front of your mind. If it’s not, you’ll likely find yourself choking down a dish of geometrical futility, drizzled with a balsamic disgruntlement reduction and served with a side of anguish fries. It can happen.

Just remember, the SAT would LOVE to misdirect you—to make you focus on the part that’s difficult to solve for, instead of the part that isn’t. But since you’re an expert pattern recognizer, you’re not going to fall for it. You know that the SAT likes to make it difficult to solve directly for the shaded regions, so you’re going to solve for everything else instead. Now go forth, intrepid one. Slay shaded regions, one and all.

Practice makes perfect, you know.

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Also try this previously posted question (post contains an explanation).

Before we get into triangles, we need to take a very quick look at the ingredients of a triangle: line segments and angles. Please tell me you already know this stuff:
 

We good? Cool. Prove it:

  1. In the figure above, AE, BS, CG, DS, and FS intersect at point S. Which of the following pairs of angles must be congruent?
     
    (A) ∠ASF and ∠BSF
    (B) ∠ASG and ∠CSE
    (C) ∠ASG and ∠FSG
    (D) ∠ASB and ∠ASG
    (E) ∠ASC and ∠BSE

What to do, what to do? It’s often possible to guesstimate on a question like this: they’re asking you which angles are congruent, and the diagram is drawn to scale, so look at the damn thing. Enough choices look plausible here that we can’t take that shortcut, although we can use it to eliminate choice (C) if we like.

Which angles are vertical angles? We know that all these segments meet at point S, but which ones actually go through it? AE and CG both go all the way through, so they’ll create a set of vertical angles: ∠ASG and ∠CSE. We know vertical angles are always congruent, so (B) is the answer. You’re really going to want to make sure you’re solid on this kind of question.

OK, that was fun, right? Now let’s talk triangles. There are only a few things you need to know about angles and triangles for the SAT, many of which are given to you at the beginning of each math section. This post is going to be long (I know what you’re thinking: it’s already long) because there are a lot of different kinds of questions you might be asked and I want you to see them, but don’t be daunted by its length; I’m willing to bet you already know pretty much everything you need to know.

Now, as always, you just need to study the scouting report: know what the SAT will throw at you, and you’ll have a better chance of knocking it out of the park. Note also that there is a separate post for the special case of Right Triangles, just to keep the length of this post somewhat manageable. Lastly, please forgive me if my drawings look janky.

Triangle Facts:

The sum of the measures of the angles in a triangle is 180°:

The area of a triangle can be found with Abh:

In an isosceles triangle, the angles across from the equal sides are also equal:

In an equilateral triangle, all the angles are 60°, and all the sides are of equal length:



The bigger the angle, the bigger its opposite side:



No side of a triangle can be as long as or longer than the sum of the other two sides:



This is called the Triangle Inequality Theorem and there’s a cool demonstration of it here. The basic thrust is this: if one side were longer than the other two, then how would those two shorter ones connect to form the triangle? They couldn’t. And if one side was equal to the sum of the other two, would you have a triangle? No, you’d just have a straight line.

To drive this home: imagine your forearms (apologies to my armless friends) are two sides of a triangle, and the imaginary line that connects your elbows is the third side. If you touch your fingertips together and pull your elbows apart, eventually your fingertips have to disconnect…that’s when the length between your elbows is longer than the sum of the lengths of your forearms. Neat, huh?

Everything else:

There’s an awful lot to know about non-right triangles if you’re doing math, but on the SAT, the preceding basically covers it. You will never need trigonometry of any kind, nor will you need any knowledge about things like circumcenters or orthocenters of triangles (Google them if you’re curious). Occasionally it’ll be nice to know triangle congruence/similarity rules, but I’m not going to recreate them here because honestly, it’s incredibly rare that you’ll actually need one. There are some who think you need to know things like the external angle rule, or rules about alternate interior angles and opposite exterior angles in transversals. I say: if you know a straight line is 180° and you know a triangle’s angles add up to 180°, you already know them! Don’t overcomplicate your life with overlapping rules.

I’m in the business of preparing you for a very specific test. The above is, in my expert opinion, the salient stuff. Isn’t it awesome how little of it there is?

Let’s try one more example together:
Note: Figure not drawn to scale.
  1. If y = 180 – x and all lengths are integers, which of the following could be the perimeter of the triangle in the figure above?
     
    (A) 19
    (B) 22
    (C) 23
    (D) 24
    (E) 26

Wait, what? We don’t know anything about any of the lengths, how are we supposed to figure out anything about the perimeter? Relax, friend. We’ll get through this.

We know y = 180 – x. How can we use that? Well, note that the red arc in the image to our right here is a straight line, which means it’s 180°. Therefore, the unlabeled angle in our original figure must be equal to 180 – y. Do a little algebra now:

[unmarked angle measure] = 180 – y

[unmarked angle measure] = 180 – (180 – x)

[unmarked angle measure] = x

WTF. It doesn’t look like it, but we’ve got an equilateral triangle on our hands: all the angles inside it are equal so they must all be 60°! If all our lengths are integers, then, the perimeter must be a multiple of 3. Only one of our choices is: (D) 24.

 

Is it over yet? Nope. Drills.

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  1. In the figure above, AB is the diameter of the circle, and AC = BC. What is the area of the shaded region?
     
    (A) 4π – 2
    (B) 2π – 1
    (C) π
    (D) π – 1
    (E) π – 2

Answer and explanation after the jump…

As is usually the case with shaded region problems, the easiest way (and in this case, the only way) to solve for the shaded region is to solve for the regions around it first. The relationship we want to keep in mind is:

In this particular case, we’re going to have to do a little legwork to figure out what out “whole” is before we get down to business.

Let’s start by dropping a vertical from the top of our isosceles triangle (and noting that in doing so, we’re drawing a radius, so it’s got a length of 2):

That vertical is of course perpendicular to AB, and creates a right angle that nicely frames the area we’re looking to solve for. So the Areawhole we’re looking for here is actually only the part of the circle marked off by that right angle. Since a circle has 360 degrees of arc and we’re only dealing with 90 of them, we’re dealing with one fourth of the circle.

Areacircleπr2
Areacircle = π(22)
Areacircle = 4π

So the area of the sector we care about is simply one fourth of that, or π:

Areawholeπ

Now we just need to find the area of the unshaded part (the right triangle we created, in red):

Areaunshaded = 1/2 (2)(2)
Areaunshaded = 2

So the area of our shaded region must be…

Areashadedπ2

That’s answer choice (E). 

One more note about this one: since the diagram is drawn to scale, it’s possible (and wise) to use your guesstimation skills once you’ve found an answer (or even to eliminate answers before you do much calculating). Which is to say: when you know the area of the triangle is 2, does it make sense that the shaded region is about 1.14159? I’d say, by eyeballing it, that yeah, it does. It wouldn’t have made sense, though, had we made a math mistake somehow and ended up with a different choice, like (D) π – 1. To pick that answer would be to say (insanely) that the shaded region is as big as the right triangle we made by dropping that vertical. That’s crazy talk.