Posts tagged with: geometry

Thomas is making a sign in the shape of a regular hexagon with 4-inch sides, which he will cut out from a rectangular sheet of metal. What is the sum of the areas of the four triangles that will be removed from the rectangle?

So it’ll look like this:

It’s helpful just to know that a regular hexagon’s interior angles all measure 120°, but you can also calculate that using (n-2)\times 180^\circ:

\dfrac{(6-2)\times 180^\circ}{6}=120^\circ

That means that the four triangles you’re cutting off the rectangle are each 30°-60°-90° triangles with 4-inch hypotenuses.

Those will have legs of 2 and 2\sqrt{3}, and therefore areas of \dfrac{1}{2}(2)(2\sqrt{3})=2\sqrt{3}. Since there are four such rectangles, the total area you’re cutting off is 8\sqrt{3}

A semicircle is exactly half a circle, so take the formula for circumference (C = 2πr) and divide by 2. Since r = 2, you end up with (2π(2))/2 = 2π. That covers the curved part.

You know the straight part has a length of 4 because the radius of the semicircle is 2.

4 + 2π is your answer.

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First, a circle inscribed in a square looks like this:

image

If that square has an area of 2, that means each of its sides has a length of √2. And THAT means that the radius of the circle is √2/2.

image

The area of a circle is πr^2. Plug √2/2 in for r and you’ve got your answer:

image

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PSAT/NMSQT practice test #1 section 4 question 18

This is a great one to plug in on. Say the height of the trapezoid is 2, the lower width is 5, and the upper one is 3. That way, assuming you don’t know the formula for the area of a trapezoid (which exists—see below—but which you don’t need to memorize for SAT purposes) you can break the trapezoid neatly into two right triangles and a rectangle thusly (note that my drawing is not to scale):

The rectangle has an area of 3\times 2=6, and the triangles each have an area of \dfrac{1}{2}(1)(2)=1. Total area of the trapezoid: 1+6+1=8.

Now if you do the manipulations the question asks you to do, the height gets cut in half and the bases are doubled:

The areas of the triangle are still \dfrac{1}{2}(2)(1)=1. The area of the rectangle is also unchanged: 6\times 1 = 6. Therefore, the area of the trapezoid is still 1+6+1=8.

The answer is C: the area does not change.

Since this post will be a reference point now, you can also do this using the area of a trapezoid formula:

A=\dfrac{b_1+b_2}{2}h

If you double both bases and cut the height in half, you get:

A=\dfrac{2b_1+2b_2}{2}\dfrac{h}{2}

A=\left(b_1+b_2\right)\dfrac{h}{2}

Of course, that’s equivalent to the original formula. Therefore, doubling the bases and halving the height won’t change the area of any trapezoid.

I want to stress, though, that the lesson you should take from this question is not that you need to memorize the trapezoid area formula. Rather, the lesson should be that you can 1) plug in, and 2) break more complex shapes into things like triangles and rectangles that you already know how to work with.

Can you please do #20 from the no calculator section in test 5?

Sure. Here’s (basically) the image you’re given:

To find the value of x, you need to remember that the angles that make up straight lines and the angles that make up triangles will both sum to 180°. You’ve got a straight line (marked in red below) that helps you calculate that the upper left angle in the triangle is 180° – 106° = 74°.

From there, you can calculate the measure of the third angle in the triangle: 180° – 23° – 74° = 83°. Just one more 180° straight line to find the value of x:

= 180º – 83°
= 97°

A note since I’m sure someone would bring it up in the comments if I didn’t: another way to go here is the Exterior Angle Theorem, which states that the measure of an exterior angle of a triangle (the 106° angle in this case) equals the sum of the measures of the nonadjacent interior angles (the 23° and the 83°). You can use that theorem as a “shortcut” to find the 83° angle in one fewer step if you like. 🙂

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.

 

Can you please explain College Board Test 5 Section 4 #36?

I understand that angle A is half of x- the only part of the college board explanation I’m not understanding is why we do (360-x) while setting up the equation. Thanks!

Sure. Without looking at CB’s solution, here’s how I do it.

First, as you pointed out, the measure of angle A is half of . That’ll be useful. The other thing I think is useful is to draw segment BC and mark the angles you create (I’ll say their measures are  and ). Now you’ve got two triangles, ABC and PBC.

Each triangle, of course, has angles that add up to 180°. You can write two equations that show this:

Triangle ABC: \dfrac{1}{2}x+20+20+b+c=180

Triangle PBC: x+b+c=180

All you need to do to solve for x is set those left sides equal to each other since they both equal 180 (b and c will cancel right out!).

    \begin{align*}\dfrac{1}{2}x+20+20+b+c&=x+b+c\\\dfrac{1}{2}x+40&=x\\40&=\dfrac{1}{2}x\\80&=x\end{align*}

In the picture, Lines L1 and L2 are parallel, Lines L3 and L4 are parallel, L1 and L3 intersect on the circle, and L2 passes through the center of the circle. The angle theta equals pi/3, and radius R = 6. Find the area of the region A, which lies outside the circle and is bounded within the three line segments. Round your answer to the nearest hundredth. The answer is 29.72. Help!!

Can you upload a picture? I can’t draw this from your description alone. Put it in the comments below this post, and then I’m happy to help.

Update: thanks for uploading a pic. Here’s a version I whipped up that’s a little easier to read—hopefully I got all the important bits right. Note that I just put 60° for the angle instead of \dfrac{\pi}{3} just to keep things a little simpler.

Here’s the first thing you want to recognize: the quadrilateral we have here is a parallelogram with an angle of 60°. That means its other three angles will be 60°, 120°, and 120°.

We can calculate the area of a parallelogram by multiplying its base and its height. We know the base is the radius, 6. We need to calculate the height, but that’s easy because if we draw in the height in the right place, we make a 30°-60°-90° triangle! Here’s where I make my usual reminders that the special right triangle ratios should always be at the front of our minds.

So the height of the parallelogram is 3\sqrt{3}. Let’s calculate the area of the parallelogram:

    \begin{align*}A&=bh\\A&=(6)(3\sqrt{3})\\A&=18\sqrt{3}\end{align*}

Now we just need to subtract the area of the sector (AKA pizza slice) from that.

Since the circle’s radius is 6, its area is 36\pi. 60° is one sixth of the circle, so the sector we need to pull out is one sixth of the circle’s full area.

    \begin{align*}\dfrac{60}{360}&=\dfrac{A_\text{sector}}{36\pi}\\6\pi&=A_\text{sector}\end{align*}

Now we have what we need to solve! The answer will be 18\sqrt{3}-6\pi. Put that in your calculator to get a decimal approximation and you’ve got your answer–I’m getting 12.33. Is that what you were getting?

The book you’re working from is wrong. Here’s a sanity check: the area of A should be a bit less than half the area of the parallelogram. We already know the whole parallelogram has an area of 18\sqrt{3}, so half that will be 9\sqrt{3}\approx 15.59.

 

Test 3 Section 3 #20

To get this one, first draw triangle ABC with the information given: angle B is a right angle,
BC = 16, and AC = 20.

t3s3-20-1

Because you know your Pythagorean triples, you know that this is a big cousin of the 3-4-5 triangle—it’s a 12-16-20!

Now, let me point out something that you may already have realized: if triangle DEF is similar to triangle ABC, that means it has the same angles! Since the sine of an angle is always the same ratio regardless of the lengths of the sides in any particular triangle, we don’t need to draw DEF or calculate its side lengths to know what the sine of angle F is! The question tells us that angle F corresponds to angle C, so sin F = sin C. All we need to do is calculate the sine of C. So, SOH-CAH-TOA that bad boy.

\sin F=\sin C =\dfrac{12}{20}=\dfrac{3}{5}

Note that you can grid the fraction, 3/5, or the decimal equivalent .6 and be correct.

Test 3 Section 3 #18

First, here’s a mockup of the figure (not to scale):

You’re told that 180 – z = 2y, and then you’re told that y = 75. (This is typical of the SAT: give you the information in the reverse order in which you should use it.) So let’s solve for z, and then mark up the figure a bit.

180 – z = 2(75)
180 – z = 150
z = –30
z = 30

To move forward from here, you need to remember the base angle theorem: in isosceles triangles, the angles across from the congruent sides are also congruent. So the blue angles and the red angles below are congruent.

We want x, so let’s figure out what those red angles are (I’ll use a for them):

30 + 2a = 180
2a = 150
a = 75

If the red angles each measure 75°, then remembering that a straight line measures 180°, we can solve for x:

75 + x = 180
x = 105

Test 2 Section 4 #36

OK, so here’s my version of the diagram:

The first thing you need to recognize is that because segments LM and MN are tangent to the circle, and segments LO and NO are radii, you have two right angles:

Furthermore, with those two right angles and the 60º angle you’re given in the question, you have three out of the four angles in a quadrilateral:

The angles in all quadrilaterals add up to 360º, so you can solve for the measure of angle LON.

360° = 90° + 60° + 90° + m∠LON
120° = m∠LON

So far so good? Now we just need to use the fact that a circle has 360° of arc, and we’re dealing with 120°, or \frac{1}{3} of that. If the whole circumference of the circle is 96, then…

\dfrac{120}{360}=\dfrac{x}{96}

\dfrac{120}{360}\times 96 = x

32=x

Test 1 Sec4 #24 help!

For this one, you’re looking for the equation of a circle with a center of (0,4) that passes through \left(\dfrac{4}{3},5\right). The best way to go here is just to build the equation.

For this test, you should know the standard circle equation: A circle with center (h, k) and radius r has the equation (x-h)^2+(y-k)^2=r^2. You can fill in the left side immediately: The center of this circle is (0, 4), so the left side should be (x-0)^2+(y-4)^2=x^2+(y-4)^2.

To fill in the right side, you need to find r (or r^2). Easy enough:

\left(\dfrac{4}{3}\right)^2+1^2=r^2\\ \dfrac{16}{9}+1=r^2\\ \dfrac{16}{9}+\dfrac{9}{9}=r^2\\ \dfrac{25}{9}=r^2

So the answer you want is A.

A note for those who like quick and dirty solutions: Once you’ve got the left side of the equation, you know the answer must be either A or C. If you’re paying close attention, you’ll notice that the only difference between A and C is on the right side of the equation: A says r^2=\dfrac{25}{9} and C says r^2=\dfrac{5}{3}. If you’re in a hurry, that’s probably good enough—you know you’re looking for a squared value, and A provides it.

Another note: if you can’t remember the circle equation, you can still get this question right! Use your calculator to plug x=\dfrac{4}{3} and y=5 into each answer choice. Only one will work. 🙂

If Paul is using a piece of fencing 80 meters long to build a rectangular enclosure for his dog, what is the greatest possible area that can be enclosed?

Long way:

Say the sides are x and y long. The perimeter of the fence must be 80, so 2x + 2y = 80. The area of the fence will be xy.

We can use the first equation to get the second equation in terms of only one variable!

2x + 2y = 80
y = 40 – x

Now you can say that the area of the rectangle will be x(40 – x). You can graph that if you want, or you can just recognize that it will be a downward facing parabola with roots at 0 and 40. That means it’ll have its maximum when x = 20. That means the greatest possible area will be when x = 20, when the area will equal 400 square meters.

Shortcut:

Just remember that in a case like this, the biggest rectangular area you can enclose is with a square. If you know the perimeter needs to be 80, then make a 20×20 square.

In the figure above, point R lies on segment OP. The area of the circle with center O is 4(pi), and the area of the circle with center P is 100(pi). What is the length of segment RP?

Put the figure in the comments.

Points A and B lie in the same plane. How many isosceles right triangles in the plane have A and B as vertices?

A) Three
B) Four
C) Five
D) Six
E) Infinitely many

That side can be the hypotenuse of such a triangle on either side of it, and it can also be one leg of such a triangle in 4 orientations. Look:

Untitled_1

The smaller, darker triangles are where segment AB is the hypotenuse. The larger  triangles are the four other orientations.