## Posts tagged with: area

An 84 meter length of fencing is attached to the side of a barn in order to fence in a rectangular area, as shown in the figure above. If the length of the side of the fence running perpendicular to the barn is half the length of the side of the fence that is running parallel to the barn, what is the area of the fenced off land?

You know that 84 meters of fencing is used to make 3 sides of a rectangle (the barn forms the fourth side). You also know that one side of the rectangle is twice as long as the other. Let’s call the short side x and the long side 2x.

You know that x + x + 2x = 84, so you can solve for x. 4x = 84; x = 21.

That means the rectangle’s width and length will be 21 meters and 42 meters. Its area will therefore be (21 meters)(42 meters) = 882 square meters.

Hello, can you please explain number 27 section 4 test 1?

Yeah, this is a really tricky one. The key to getting it right is recognizing just how much of the field the students are NOT covering! In a 10 ft by 10 ft field, there will be 100 1 ft by 1 ft squares. Look, below is a 10 by 10 grid with X’s marking places where the 10 students in this question might be.

The data in the table tells us that, on average, there are about 150 worms in each of the small boxes. Since there are 100 small boxes in the big field, we need to multiply 150 by 100 to get our answer for the approximate number of earthworms in the whole field.

150(100) = 15,000

Will you please answer question # 27, Test 3, section 4. Is it possible to use real numbers in your example? Thanks!

Sure! The thing you want to remember when you’re plugging in numbers is that you’re doing it to make your life easier, so pick numbers that will really make your life easier. In this case, that means I’m going to say that the original rectangle is a 10 by 10 square, so its area is 100. Obviously, it’s very easy to work with 100 when dealing with percents, and it’s also easy to take different percents of 10 to increase/decrease side lengths.

So, yeah. Here’s our original rectangle:

Now we need to increase its length by 10 percent (10 percent of 10 is 1, which means the length increases from 10 to 11), and decrease its width by p percent. What should we do for p? Well, why not backsolve?

Say p = 20, like answer choice C says. (I’m picking that because it’s in the middle, and also because 20 is the easiest choice to work with so why not try it first.) If p = 20, then we decrease the width by 2, taking it from 10 to 8. So here’s the new rectangle:

What’s the difference between the original area of 100 and the new area of 88? 12. And here’s where we pat ourselves on the back for picking numbers that gave us 100 as our starting area: 12 is what percent of 100? 12 percent! So the first choice we tried, C, is right. Awesome!

(Whenever I illustrate backsolving and the first choice I try is the right choice, I feel compelled to show what a wrong choice would look like. So let’s quickly look at choice B. We’re still increasing the length by 10 percent, so the length is still 11. The width is now decreased by 15 percent, making it 8.5. The new area is . Is that a 12 percent decrease from the original area of 100? No, 93.5 is too big! So if we didn’t already know the answer, we’d know that the next choice we should try should be smaller.)

Hi Mike,
question on Math Quiz 4 #9: I get the answer and the proposed solution (area of large triangle minus area of small triangle). But my daughter worked the problem differently and I think what she did was correct, but she got a slightly different answer and I can’t figure out exactly why.
Her solution: she used Pythag Theorem to get the hypot of small triangle: so 4^2 + 1^2 = c^2…. so c=Sqrt(17). This is base of large triangle. So A= 1/2 bXh, or 1/2 (sqrt 17)(8), she got 16.492.

When you’re finding the area of a triangle, the height you use needs to be perpendicular to the base you choose. In this case, drawing a segment from (2, 6) that is perpendicular to that base would require drawing it outside of the triangle.

If you’re going to use as your base (segment BC), then you need to use segment AD in the figure above as your height. This can be done easily with software, but I wouldn’t advise trying to do it by hand on test day.

If points (6,0), (0,0), (0,2), and (a,2) are consecutive vertices of a trapezoid of area 7.5, what is the value of a?

A) 1.5
B) 2
C) 2.5
D) 5
E) 9

Backsolve this. If a is 2.5, like choice C says, then your trapezoid looks like this:

I put the dashed line in there to make it easier to see how you can calculate the area without remembering any trapezoid area formula: just break it into a triangle and a rectangle! You’ll find that the area of that trapezoid is 8.5, which is too big. So go smaller.

Because I know I want a .5 at the end of my area, I decided to skip 2 and go straight to 1.5 for a. Lo and behold, that works!

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:

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?

Since you guys tore up my last challenge question so quickly, I figured I’d make this one a tiny bit harder. Let’s see if I can’t stump you for more than 8 minutes.

First correct response in the comments gets a free Math Guide. Usual contest rules apply. Ready? Here we go!

In the figure above, triangle ABC is inscribed in a circle with center P. If the area of triangle ABC is equal to $\frac{x^2-y^2}{2}$, and AC = xy, what is the area of the circle in terms of x and y? (Be careful typing your answer—parentheses matter!)

MUAHAHAHA. Good luck!

UPDATE: Tuấn got it first, although a others came VERY close (careful with parentheses!!!). Solution below the cut.

It might not be immediately clear what to do here, but as is true of most geometry questions, it’s a good idea to start by listing everything you know. First, you know that triangle ABC is a right triangle, because BC is a diameter, and point A is on the circle. You might recall from geometry class that if an angle lies on a circle, its measure is half of the measure of the arc it corresponds to. Since A is on the circle and corresponds to a 180º arc, the measure of angle A is 90º. You probably won’t see that rule tested on the SAT, but it’s not outside the realm of possibility. So here’s what we know:

Of course, we also know that the formula for the area of a triangle is ½bh. So let’s figure out AB:

\begin{align*}\frac{1}{2}(AB)(x-y)&=\frac{x^2-y^2}{2}\\(AB)(x-y)&=x^2-y^2\\(AB)(x-y)&=(x-y)(x+y)\\AB&=x+y\end{align*}
Oh look. Difference of two squares. Crazy how often that appears!
Now we can solve for BC, using the Pythagorean Theorem:
\begin{align*}(x+y)^2(x-y)^2&=(BC)^2\\(x^2+2xy+y^2)+(x^2-2xy+y^2)&=(BC)^2\\2x^2+2y^2&=(BC)^2\\\sqrt{2(x^2+y^2)}&=BC\end{align*}
Nice. Here’s what we’ve got now:
Of course, BC is the diameter of the circle, and to find the circle’s area we want the radius, which is half of BC:
$r=\frac{\sqrt{2(x^2+y^2)}}{2}$
\begin{align*}A&=\pi\left(\frac{\sqrt{2(x^2+y^2)}}{2}\right)^2\\A&=\pi\left(\frac{2(x^2+y^2)}{4}\right)\\A&=\frac{\pi(x^2+y^2)}{2}\end{align*}

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!

In honor of Father’s Day, this weekend challenge is inspired by my dad’s hobby business. He sells wooden flags, like the one above, on Etsy. Basically, he finds discarded pallets (this is a pallet), chops ’em up into pieces of the size he needs, and mounts them.

As usual, these challenge questions are not really SAT questions, they just require you to work the same muscles that you need to exercise to do well on the SAT. Consider them cross-training. Here we go.

The creation of one of these flags requires 13 solid strips of wood that are 1 inch thick, 2 inches wide, and 36 inches long. If my dad makes as many flags as possible out of a pallet with 10 usable slats that are one inch thick, 7 inches wide, and 40 inches long, what is the volume of the leftover scrap in cubic inches? Disregard sawdust loss and any part of the pallet that was not originally deemed usable.

Put your answers in the comments. First correct (non-anonymous) answer gets a Math Guide. Good luck, and don’t forget to wish the fathers in your life a happy Father’s Day!

UPDATE: Congratulations to Shahriar, who got it first. I’ll be in touch shortly to get your shipping information, Shahriar, so I can send you your Math Guide.

Solution below the cut.
I realized after I posted this that my measurements weren’t very realistic at all, but whatever. It’s the math that’s important, not whether my fictional pallet measures up to industry standards.

The first step to solving this problem is to recognize that by making the thickness of the slats 1 inch, I went easy on you: the value of the volume of wood left over will be the same as the value of the area of its front-facing face. In other words, you can basically treat this as a two-dimensional problem even though I’m asking for volume. You’re welcome.

How many usable strips of wood can be made from one pallet slat? 3 of them. And once those 3 usable strips are cut away, each slat produces 36+28=64 in3 of scrap.

Of course, there are 10 such slats, so you’re looking at something like this:
Oh but wait. That produces 30 usable slats. Which is only enough to make 2 flags, since each flag requires 13 strips. I guess 4 of those strips become unusable scrap, too. 🙁

So really, the situation is that each slat produces at least 64 in3 of scrap. We have to add the volume of the 4 unnecessary strips (32×2=72 in3 each) to our final volume, since they’re not actually used in the making of flags.

(10 slats × 64 in3 scrap) + (4 strips × 72 in3 scrap) = 928 in3 of scrap

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:

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 =

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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This is a little harder than a typical SAT question, and obviously formatted with a bit more flair and color than the College Board would use, but it deals with the same concepts you’ll need to master to kick the SAT where it hurts most, so have a go at it:

I looked out my window the other day and realized that the shadow that my apartment building was casting was a perfect square. If the distance from the edge of the shadow to the top of the building (the dotted line on the diagram) was 100 feet, and my building (which has a square base) is 7 times as tall as it is wide, what was the area of the shadow?

I’ll refrain from answering this myself just yet. Put your answer in the comments, and check back in a few days to see if you were right!

Update: This post had LOTS of views and very few folks attempting an answer, so either people were finding this post by mistake when they Googled for something else, people are shy, or this question was harder than I thought. Answer and explanation below the cut.

Basically, this problem requires three insights to solve:

1. If the shadow is a perfect square, then it has to be exactly as long as it is wide. That means it has to be as long as the building is wide. Once you’ve got that figured out…
2. This is a right triangle problem. The dotted line in the diagram is the hypotenuse of the triangle formed by the height of the building and the length of the shadow (which is the same as the width of the building).
3. The area of the shadow will be the width of the building squared.
So let’s call the width of the building x and get to work:

Pythagorize it:

x2 + (7x)2 = 1002
50x2 = 10000
x2 = 200

There you go. The area of the shadow is 200 ft2. Nice job to the one kid who got it right, and all the others who got it right and didn’t have the confidence to post it. Maybe next time I’ll offer a prize or something.
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:

$\large&space;\inline&space;\dpi{300}&space;Area_{shaded}=Area_{whole}-Area_{unshaded}$

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):