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Special Right Triangles
Special right triangles. Now we can discuss two triangles that the test absolutely loves. Now, why does the test love these two triangles? Both have the property with very little information given, we can figure out all the sides and angles. So, the test loves to put these, these triangles by themselves and in larger figures because just a little bit of information you can figure out so much more.
That’s why the test loves it so much. So, let’s talk about these. The first triangle. This is an isosceles right triangle. Think about this. It’s a right triangle, so it has a right angle.
Obviously, a triangle can’t have two right angles, and then a third angle, because that would be more than 180 degrees. So it has to have a right angle and two equal acute angles. And so it means those two equal angles must be 45 degrees. So the angles are 45, 45, and 90. The two legs must have equal lengths, because if the angles are equal, then the legs have to be equal.
And the hypotenuse is larger than other leg, though less than the sum of the legs. So there we have a picture of the right isosceles triangle. Let’s consider a 45-45-90 triangle with legs of length one. So we’ll pick a very easy leg, length for the legs. So the legs are one, then we can find the hypotenuse with the Pythagorean theorem. So of course that hypotenuse squared equals legs squared plus leg squared, equals 1 plus 1, which is two.
And so the hypotenuse is the square root of 2. Now that was just one 45-45-90 triangle, of one that just happened to have a leg equal to 1. But the big idea is that all other 45-45-90 triangles are similar. They’re all similar to each other because they all have the same angles. That means the sides all have the same proportion.
That is a really big idea. So in any 45-45-90 triangle, it must be true that the two lets are equal and the hypotenuse equals the square root of 2 times the leg. So that is just saying that it has the same proportion as the 45-45-90 triangle that we just looked at on the previous slide. We could think of the sides as leg equals s and hypotenuse equals s times square root of 2.
We can refer to this triangle in three different ways. A more formal way to refer to it is the isosceles right triangle. That actually implies all the properties right there. We can refer to it as the 45-45-90 triangle, or the 1-1- root 2 triangle. And if you remember those three different ways to refer to the triangle, you automatically remember all of its properties.
So you have to know this triangle very well, but don’t simply memorize it. Instead make sure you understand why each set of properties implies all the others. So in other words, if you just start out with those sides, it must have those angles. If you just start out with those angles, it must have that ratio of sides. Make sure you understand why that is true.
Don’t simply memorize it. Notice that we divide a square in half along the diagonal, we get two of these triangles. So that interesting. This is one place where this triangle so up frequently. It make it very easy to see that the diagonal of the square equals the side times the square root of 2.
We can also divide a square into four of these triangles, and the hypotenuse of each of those triangles is on the side of the square. So that’s another way that this triangle can show up inside a square. Here’s a practice problem. Pause the video, and then we’ll talk about this. Okay, from the fact that KLMN is a square, we know we have a right angle at N, KN is perpendicular to JM.
And so we need, in the triangle, we have a right angle, we have a 45 degree angle, means the other angle has to be 45 degrees. So that’s a 45-45-90 triangle. So that means that the hypotenuse KN, sorry, the leg KN times root 2 equals the hypotenuse JK. Well, what’s tricky here is you don’t know the length of the leg, but we know the length of the hypotenuse.
So, we divide by root 2, 6 divided by root 2. And remember here we have to simplify, we have to rationalize, so we multiply by root 2 over root 2. We get 6 root 2 over 2, we can cancel the factors of 2 and we get 3 root 2. Now, this is the length of KN.
And because it’s an isosceles triangle, it’s also the, the length of JN. That’s also the side of the square. So that means the other three sides of the square also have to equal root 3 over 2. So there are actually five segments in this diagram that have that same length. So now, we can find the area of the square. The area of the square, all we have to do is square 3 root 2.
And of course that is what we do there, is we square 3 and we square 2. If this is something unfamiliar to you, I would suggest going back and watching the video Operations with Roots, in the Power and Roots module. So, 3 squared is 9, Root 2 squared is two, 9 times 2 is 18. That’s the area of the square. Now, we don’t even have to do that many calculations for the area of the triangle.
All we have to do is notice, that triangle has exactly half the area of the square. So that means the triangle has an area of nine. Well, the whole figure 18 plus 9 is 27, and that’s the area of the figure. The second triangle. For this we will begin with an equilateral triangle in which each side has a length of 2.
So there’s our equilateral triangle. This triangle, while very symmetrical, is not a right triangle of course. But we can create a right triangle by drawing an altitude from point A. So we draw this altitude. And remember, as in any isosceles triangle, this altitude will bisect the angle at A and also bisect the opposite.
Remember who we were talking a previous video, that line down the middle of a isosceles triangle plays the role of a perpendicular bisector. A line to the midpoint, an angle bisector, and an altitude, it plays all those roles all at once. So we’ll draw that line, and there we get our right triangle. The angle at B is one of the original angles, 60 degree angles of the equilateral triangle.
And the hypotenuse AB is one of the original sides. So we actually have a, a piece of the whole equilateral triangle there. The 30-degree angle at A is the bisected angle. So we know that angle has to be 30 because the 60-degree angle was bisected. Well, we also because we have a right angle. Three angles have to add up to 180.
And we also know that DB. We know D is the midpoint of the base. And so DB equals 1. So we know two of the sides. AB is 2 and DB is 1. Call the the third side AD, call that H.
We can find this with the Pythagorean theorem. So it must be true that hypotenuse squared equals leg squared plus leg squared. So h squared equals 4 minus 3. 4 minus 1, which is 3. And we take the, the square root, h equal the square root of 3. Sometimes this is called the 30-60-90 Triangle.
Sometimes people also call it the 1-2-root 3 triangle. We can call it that. But make sure you understand that the largest side, the longest side is actually root, is actually 2. 2 is actually a larger number than the square root of 3. And finally, it’s very helpful to remember the relationship this triangle has with the whole equilateral triangle from which it comes.
In fact, sometimes, it’s helpful, if you get confused at all. Just take your 30-60-90 triangle and draw the missing part so that you can see the full equilateral triangle. And that makes it very easy to see where the 60 degree has to be, what side is bisected, and all that. As with the 45-45-90 triangle, any 30-60-90 degree triangle is similar to this one, and the sides have the same proportion.
So let’s think about these proportions. Hypotenuse to short leg, 2 over 1 is 2. And so that means that any hyp, for any of these triangles, the hypotenuse has to be twice the length of the short leg. It also means that the long leg over the short leg equals root 3. Or in other words, long leg equals root 3 times the short leg.
And so the, those relationships, those proportions are always true for this triangle because they’re all similar to each other. Now that we know the 30-60-90 triangle, we can create a general formula for the area of an equilateral triangle with side s. This is an unlikely thing, but it is possible that the test could expect us to know the area of an equilateral triangle, now it would just give us one side.
And so it’s a handy formula to know. I’ll say this again when we get to the end of this process, I do not want you to memorize this formula. What I want you to do is understand where this formula comes from. If you understand it, then you really own it. So let’s think about this carefully.
We have an equilateral triangle with side S. We’re gonna draw an altitude, and of course that divides the whole triangle in two 30-60-90 triangles. The hypotenuse of the triangle is S and ML, the bisected side, that has be S over 2. And we know that the length of KM has to be root 3 times the short leg.
The long leg equals root three times the short leg. So the long leg, KM has to equal S root 3 over 2. But once we have that. KM is the altitude. JL is the base, and of course that has the length of S. All three sides are S.
So one half base times height. We’re multiplying the base, S times S root 3 over 2 and multiply by one half. When we simplify, we get, S squared root 3 over 4. This is a handy formula to know but do not memorize it. Once again, make sure you understand the argument and you can reproduce that result on your own.
When you can really derive it on your own then you really understand it. Find the perimeter of triangle ABC. This is a practice problem. Pause the video and then we’ll talk about this. Okay, so the short leg has to be half the hypotenuse.
So that’s gonna be 4 root 3. The long leg is gonna be the short leg times the square root of 3. So we’re gonna multiply square root of 3 times 4 root 3. Again, operations with roots, you’ll remember all we do is just multiply the roots. So we get 4 times root 3 times root 3.
Well root 3 times root 3 is 3. So this four times three which is 12. And so that’s the length of the long leg. So now we simply add things up. We have a hypotenuse 8 root 3, the short leg 4 root 3, and the long leg 12. We can combine the two like terms to get 12 root 3 plus 12.
We can’t simplify that any further. That is the perimeter. Here’s another practice problem. Pause the video, and then we’ll talk about this. Okay, in the diagram, we have a square in the middle, NGH equals 10.
Well, because that’s a square, we know that HM and JL are both perpendicular to the base GK. So we have right angles at the base. And so what that means, on each side what we have is a 30-60-90 triangle. And so both of those slanted sides have a length of 10, those are two congruent triangles on each side.
And let’s see. The short leg from GM has to be half the length of the hypotenuse. And the long leg, HM, has to be root 3 times the short leg. So the short leg is 5 and HM is 5 root 3. So once we have that from the properties of the 30-60-90 triangle. All right.
Now we can start figuring out areas. So one triangle will be one-half base times height, but we have two identical triangles. So, we might as well just figure out both triangles together. That’s base times height of the triangles. The base is 5.
The height is 5 root 3, and so that gives us 25 root 3. So that’s the combined area of the two triangles. Now put that aside for a moment. Now the square, the side of the square HM is the side of the square, that’s 5 root 3. So the area is 5 root 3 squared.
Square the five, that’s 25. Square the root 3, that’s just 3, and 3 times 25 is 75. That’s the area of the square. So now we just add everything together. The total area is the area of the square plus the area of the two triangles. 75 plus 25 root 3.
In summary, one special triangle, the isosceles right triangle has angles of 45-45-90. And sides of 1-1-root 2. So that’s one of our special triangles. The other results from dividing an equilateral triangle in half, and has angles of 30-60-90.
And if we put the sides in order of length, it would be 1-root 3-2. And again, it’s very important to go back and think about the relationship that the 30-60-90 triangle has to the equilateral triangle. If you understand that, you really understand this triangle. We can use these patterns and proportional reasoning to solve a variety of problems.
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