Posts tagged with: right triangles

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}

Trigonometry does the trick here. Below is that line making a 42° angle with the positive x-axis. I’ve also drawn a dotted segment to make myself a neat little right triangle.

Remember that slope is rise over run—how high the line climbs divided by how far it travels right. In this case, the dotted segment labeled a is the rise and the bottom of the triangle labeled b is the run. And luckily for us, the tangent function calculates that a/b ratio! Remember your SOH-CAH-TOA. Tangent = Opposite/Adjacent.

Just use your calculator to evaluate tan 42°. You’ll get 0.90.

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Think of a 5-12-13 triangle (that’s one of the Pythagorean triples you should know).

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Say angle A measures x°, which would make angle C measure (90 – x)°. (I’m choosing those based on the fact that I already know that the sine of angle C will be 12/13.)

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Now that we’ve got it set up, all we need to do is SOH-CAH-TOA it: sin x° = 5/13.

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Practice question for Circles, Radians, and a Little More Trigonometry, #5, p. 272, 4th Ed.

Here’s the question:

So the first thing you’ll want to do is draw in a strategically placed radius (or two) and label the lengths you know.

Because we know that M is the midpoint of \overline{AB}, we can actually cut that 5\sqrt{3} in half: AM=\dfrac{5\sqrt{3}}{2} andBM=\dfrac{5\sqrt{3}}{2}.

We also know that we can always draw a perpendicular bisector from the center of a circle to the midpoint of a chord, so you know that by drawing in \overline{OM}, you’re creating two right triangles: AOM and BOM.

Let’s just look at triangle AOM.

At this point, we can certainly use the Pythagorean Theorem to solve, but the presence of the \sqrt{3} should have activated our Spidey-senses from the beginning: this is a 30°-60°-90° triangle! So we can also just see how the numbers we know fit into the known 1:\sqrt{3}:2 ratio. The hypotenuse in a 30°-60°-90° is double the length of the short leg. Our hypotenuse is 5, so the short leg we seek must be half that: \dfrac{5}{2}. We know we’re right because the long leg in a30°-60°-90° is the short leg times \sqrt{3}, and sure enough: \dfrac{5}{2}\times\sqrt{3}=\dfrac{5\sqrt{3}}{2}. So we have our answer. OM=\dfrac{5}{2}.

Gettin’ Pythaggy wit it:

    \begin{align*}OM^2+\left(\dfrac{5\sqrt{3}}{2}\right)^2&=5^2\\\\OM^2+\dfrac{25(3)}{4}&=25\\\\OM^2+\dfrac{75}{4}&=\dfrac{100}{4}\\\\OM^2&=\dfrac{25}{4}\\\\OM&=\dfrac{5}{2}\end{align*}

Test 7 Section 4 #36

Two important concepts in this question, closely related: trigonometry and Pythagorean triples.

First, the trig. The fact that \tan B=\dfrac{3}{4} means that the short legs of both triangles in the question are \dfrac{3}{4} the lengths of the longer legs.

Now the Pythagorean triples: 3-4-5 is the most important Pythagorean triple to know! It’s called a Pythagorean triple because it’s a case where three integers work in the Pythagorean theorem: 3^2+4^2=5^2. If you know the legs of a right triangle are in a 3-4 ratio, then you know that you’ve got a 3-4-5 triangle (or one of its bigger cousins)!

The question tells you that BC = 15. That’s the hypotenuse of the longer side, so your larger triangle is a 9-12-15 (AKA a 3-4-5 times 3). So AC = 9 and AB = 12.

If DA = 4, that means DB = 12 – 4 = 8. This means the smaller of the triangles you’re dealing with is a 6-8-10 (AKA a 3-4-5 times 2). You’re asked for the value of DE, which must therefore be 6.

I sorry I am very confused about why the answer question 5 on page 28 is A would you help me please?

Sure. Here’s the question:

This question appears in the plugging in chapter, so the solution (page 301) approaches it that way. I also plug in when I solve this question in the video solutions available to Math Guide Owners. So in this post, I’ll approach the question algebraically.

If you use the Pythagorean theorem, you know that:

    \begin{align*}3^2+a^2&=AC^2\\9+a^2&=AC^2\\\sqrt{9+a^2}&=AC\end{align*}

Since the value of \sin A will be opposite leg over hypotenuse, it will be equal to \dfrac{a}{AC}. Substituting for AC based on the Pythagorean theorem calculation above gives you choice A.

    \begin{align*}\sin A&=\dfrac{a}{AC}\\\\\sin A&=\dfrac{a}{\sqrt{9+a^2}}\end{align*}

Can you please explain Test 3 Section 4 #23?

Yep! The key to getting this one is recognizing that when they tell you that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right), they’re telling you that a+b=90. That’s just a little trig fact you should keep in mind when you’re taking the SAT. It can be derived pretty easily—just picture any right triangle with acute angles measuring a^{\circ} and b^{\circ}:

Of course, since it’s a right triangle, a+b=90. And if you do your SOH-CAH-TOA, you’ll see that \sin \left(a^{\circ}\right)=\cos \left(b^{\circ}\right) and \cos \left(a^{\circ}\right)=\sin \left(b^{\circ}\right).

Anyway, back to the question. Once we recognize that a+b=90, the rest is just substitution.

    \begin{align*}a+b&=90\\(4k-22)+(6k-13)&=90\\10k-35&=90\\10k&=125\\k&=12.5\end{align*}

How do you do #18 in Test 6 Section 3 without a calculator?

The key to this one is recognizing that you’ve got similar triangles. First, both triangle CDB and triangle CEA contain angle C. The question also tells you that \overline{BD}\parallel\overline{AE}, which means that angle CDB is also a right angle. Therefore, you have angle-angle similarity.

It’s also good to recognize a 6-8-10 triangle quickly. Know your Pythagorean triples! 🙂

From there, all you need to do is recognize that the length of \overline{AE} is 3 times the length of \overline{BD}, which means all the sides of triangle CEA are 3 times the length of their corresponding sides in triangle CDB. Therefore, because the length of \overline{CD} is 10, the length of \overline{CE} must be 30.

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.

Subject Test Question:

The area bound by the relationship |x|+|y|=2 is

A) 8
B) 1
C) 2
D) 4
E) there is no finite area.

How do you find this algebraically?

You need to capture all the combinations of negative and positive x and y. There are four equations to consider:

xy = 2
–xy = 2
x + (–y) = 2
x + (–y) = 2

Each of those is a line that can be represented in y = mx + b form. In the same order:

y = –x + 2
yx + 2
y = –x – 2
y = x – 2

Those make a square with vertices at (2, 0), (0, 2), (–2, 0), and (0, –2). The diagonals of the square are 4 long, making the sides of the square 2\sqrt{2} long. The area of a square with those side lengths is \left(2\sqrt{2}\right)^2=8.

A square with an area of 2 is inscribed in a circle. what is the area of the circle?

A) pie

B) Pie^2

C) 2pie

D) 2 radical pie

E) 4pie

It’s “pi,” not “pie.”

A square with an area of 2 will have sides of length \sqrt{2}, and therefore a diagonal of 2. (If you’re wondering how I know that, read up on special right triangles.)

If the square’s diagonal is 2, then the circle’s radius is 1, so the area is \pi r^2 = \pi (1)^2 = \pi.

A right triangle has side lengths of x-1, x+1, and x+3. What is its perimeter?

If you’re really on your right triangles game, you might instantly think of the Pythagorean triples you know and see if any fit into that scheme (hint: one will!). If not, just put those values into the Pythagorean theorem and solve for x:

(x-1)^2 + (x+1)^2=(x+3)^2

(x^2 -2x + 1) + (x^2+2x+1)=x^2+6x+9

2x^2+2=x^2+6x+9

x^2-6x-7=0

(x-7)(x+1)=0

x=7 \text{ or }x=-1

From there, you know x = 7, because the sides of the triangle aren’t going to be negative.

That means your sides are 6, 8, and 10, and your perimeter is 24.

Hi mike! This question is from the May 2015 SAT.

(will post photo)

In the xy plane above, f and g are functions defined by f(x)=abs[x] and g(x)=-abs[x] + 3 for all values x. What is the area of the shaded region bounded by the graphs of the two functions?

You don’t even need to post the picture—I can graph those functions:

First, note that the bounded region is a square. You know this if you remember that the absolute value graphs will create 45º angles with the axes, just like the graph of yx does.

Once you’ve realized that, all you need to do is find the length of one of its sides. Best way to go there is to recognize that the diagonal of the square goes from (0, 0) to (0, 3), so it has a length of 3. A square with a diagonal of length 3 will have sides of length \dfrac{3\sqrt{2}}{2}. (Brush up on your 45º-45º-90º triangles if you’re not clear why.)

So the area of the square is \dfrac{3\sqrt{2}}{2}\times\dfrac{3\sqrt{2}}{2}=\dfrac{9}{2}.

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...

A square and a regular hexagon have the same perimeter. If the area of the square is 2.25, what is the area of the hexagon?

A) 2.250
B) 2.598
C) 2.838
D) 3.464
E) 3.375

The sides of the square are \sqrt{2.25}=1.5, so the perimeter of the square is 4(1.5) = 6.

So the hexagon, which also has perimeter 6, will have sides of length 1.

To find the area of a regular hexagon, break it into 6 equilateral triangles:

Each of those can further be broken into 30º-60º-90º triangles, because all the angles in an equilateral triangle are 60º.

 

It’s easy enough to calculate the area of each of those: \dfrac{1}{2}\left(1\right)\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\sqrt{3}}{4}[latex].  Now if you just remember that there are 6 of those triangles in the hexagon, you get a hexagon area of [latex]6\left(\dfrac{\sqrt{3}}{4}\right)=2.598...