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Special Right TrianglesPowerPoint Presentation

Special Right Triangles

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## PowerPoint Slideshow about ' Special Right Triangles' - chelsea-roth

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### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

### Special Right Triangles

One of the good things about math is that you can recreate it yourself, if you can remember the basics.

So let’s pretend you suddenly have a Special Right Triangles test, but only vaguely remember anything about them.

Looking carefully, I see there are only two kinds of right triangles here….

Let’s deal with this one first...

1

1

And instead of dealing with x, let’s make it easier and have the length of the legs be 1.

Ok… so, it’s a right triangle… and the first thing I think of when I see a right triangle is…..

1

1

THE PYTHAGOREAN THEOREM!

1 + 1=c2

2=c2

c = sqr root 2

…and if I want the hypotenuse, all I have to do is solve 12 + 12 =c2.

1

1

• r

sqr root 2

…and every triangle that has the same angles as this one will be similar to it…

1 • r

1

1

1 • r

…which means that they will all be dilations of this one… with some zoom factor/ratio that I can call r.

• r

sqr root 2

This one is 45-45-90.

The length of one leg is 18… which means 18 = 1 • r.

1 • r

1

1

1 • r

So it’s easy enough to figure out that 18 = r.

And since the hypotenuse is r • sqr root 2…

• r

sqr root 2

NEXT!

This one is also 45-45-90.

1 • r

1

1

1 • r

In fact, the only difference is that r = 3 • sqr root 2

And since the hypotenuse is r • sqr root 2…

• r

sqr root 2

1 • r

1

1

1 • r

x = (3 • sqr root 2) • sqr root 2

x = 3 • (sqr root 2 • sqr root 2)

x = 3 • 2

x = 6

• r

sqr root 2

NEXT!

This one is also 45-45-90.

1 • r

1

1

1 • r

But we’re given the hypotenuse, instead of a leg!

We know the hypotenuse is r • sqr root 2…

• r

sqr root 2

1 • r

1

1

1 • r

18 = r • sqr root 2

18• sqr root 2 = r • sqr root 2• sqr root 2

18• sqr root 2= r • 2

9• sqr root 2= r … and so does x

600

600

600

Let’s take on the 30-60-90 now.

This one starts off as an equilateral triangle… with all sides equal… and all angles equal to 60 degrees.

Then, we cut it in half.

300

300

2

2

600

600

1

1

So now, the two angles at the top are 30 degrees each.

And if the original sides of the equilateral triangle had a length of two, the bottom is cut in half, too!

300

2

600

1

Now, let’s just look at the half we care about… the 30-60-90 triangle.

Notice that the hypotenuse is twice as long as the side opposite the 300 angle.

That’s always going to be true!

12 + h2 =22

1 + h2 =4

300

2

h2 =3

h

h = sqr root 3

600

1

What about the height?

This is a job for…..

THE PYTHAGOREAN THEOREM!

a2 + b2 =c2

300

2

• r

r •

sqr root 3

600

1

• r

Because every 30-60-90 triangle will be similar to this one…

The sides will always be proportional to these sides!

So we are all set to get started.

300

300

2

• r

r •

sqr root 3

600

1

• r

The missing angle is 300.

We are given the length of the side opposite that angle, so r = 8.

The hypotenuse, y, is equal to 2r… or 16.

The side across from the 600 angle has to be r • sqr root 3…

so x = 8 • sqr root 3

300

2

• r

r •

sqr root 3

600

1

• r

The hypotenuse, which has to be 2 • r, is equal to 11.

That means r, the side opposite the 300 angle,has to be 5.5….

and so x = 5.5.

The side across from the 600 angle has to be r • sqr root 3…

so y = 5.5 • sqr root 3

300

2

• r

r •

sqr root 3

300

600

600

1

• r

Since this is an isoceles triangle, the other base angle is also 600.

And the half-angle on the right is 300.

And we can focus on just the part we care about!

300

2

• r

r •

sqr root 3

300

600

600

1

• r

The hypotenuse, which has to be 2 • r, is equal to 20.

That means r, the side opposite the 300 angle,has to be 10….

and so y = 10.

The side across from the 600 angle has to be r • sqr root 3…

so x = 10 • sqr root 3

300

2

• r

r •

sqr root 3

600

1

• r

This time, we are given the length of the side opposite the 600 angle, which has to be r • sqr root 3.

If 12 = r • sqr root 3…

12 • sqr root 3 = (r • sqr root 3)• sqr root 3

12 • sqr root 3 = r • (sqr root 3 • sqr root 3)

12 • sqr root 3 = r • 3

4 • sqr root 3 = r

300

2

• r

r •

sqr root 3

600

1

• r

Since r = 4 • sqr root 3…

and that is the side opposite the 300 angle…

x = 4 • sqr root 3

300

2

• r

r •

sqr root 3

600

1

• r

And, again, since r = 4 • sqr root 3…

and the hypotenuse (y) has to be twice as long…

y = 8 • sqr root 3

300

2

• r

r •

sqr root 3

1 • r

1 • r

600

1

• r

So, if you ever have to answer questions about Special Right Triangles,

now you know that you can create the “formulas” from scratch,

just by using the Pythagorean Theorem.

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