special right triangles

Special Right TrianglesOne 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.

Special Right Triangles

YIKES!Don’t Panic.


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

Special Right TrianglesThere is the 30-

60-90….

… and the 45-45-90.


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!


…and if I want the hypotenuse, all I have to do is solve 12 + 12 = c2. 1 1

1 + 1 = c2

2 = c2

c = sqr root 2


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

1 1

sqr root 2

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

1 • r 1 • r

• r


OK… Now I’m ready… bring on the problems.

1 1

sqr root 2

1 • r 1 • r

• r


This one is 45-45-90.1 1

sqr root 2

1 • r 1 • r

• r

The length of one leg is 18… which means 18 = 1 • r.So it’s easy enough to figure out that 18 = r.And since the hypotenuse is r • sqr root 2…


x = 18 • sqr root 21 1

sqr root 2

1 • r 1 • r

• r


NEXT!1 1

sqr root 2

1 • r 1 • r

• r

This one is also 45-45-90.In fact, the only difference is that r = 3 • sqr root 2And since the hypotenuse is r • sqr root 2…


x = (3 • sqr root 2) • sqr root 2

1 1

sqr root 2

1 • r 1 • r

• r

x = 3 • (sqr root 2 • sqr root 2)x = 3 • 2x = 6


NEXT!1 1

sqr root 2

1 • r 1 • r

• r

This one is also 45-45-90.But we’re given the hypotenuse, instead of a leg!

We know the hypotenuse is r • sqr root 2…


18 = r • sqr root 21 1

sqr root 2

1 • r 1 • r

• r

18 • sqr root 2 = r • sqr root 2 • sqr root 218 • sqr root 2 = r • 2 9 • sqr root 2 = r … and so does x


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.

600 600

600


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!

600 600

300300

2 2

11


600

300

2

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.

1

That’s always going to be true!


600

300

2

What about the height?

This is a job for…..

1

THE PYTHAGOREAN THEOREM!

h

a2 + b2 = c2

12 + h2 = 22

h = sqr root 3

1 + h2 = 4 h2 = 3


600

300

Because every 30-60-90 triangle will be similar to this one…The sides will always be proportional to these sides!

1

• rsqr root 32

• r

r •

So we are all set to get started.


600

300

The missing angle is 300.We are given the length of the side opposite that angle, so r = 8.

1

• rsqr root 32

• r

r •

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


600

300

Let’s do another.1

• rsqr root 32

• r

r •


600

300

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….

1

• rsqr root 32

• r

r •

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


600

300

Bring on the next one!1

• rsqr root 32

• r

r •


600

300

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

And the half-angle on the right is 300.

1

• rsqr root 32

• r

r •

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

600

300


600

300

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….

1

• rsqr root 32

• r

r •

and so y = 10.

600

300

The side across from the 600 angle has to be r • sqr root 3… so x = 10 • sqr root 3


600

300

And, finally….1

• rsqr root 32

• r

r •


600

300

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

1

• rsqr root 32

• r

r •

If 12 = r • sqr root 3… 12 • sqr root 3 = (r • sqr root 3) • sqr root 312 • sqr root 3 = r • (sqr root 3 • sqr root 3) 12 • sqr root 3 = r • 3

4 • sqr root 3 = r


600

300

Since r = 4 • sqr root 3…1

• rsqr root 32

• r

r •

and that is the side opposite the 300 angle…x = 4 • sqr root 3


600

300

And, again, since r = 4 • sqr root 3…1

• rsqr root 32

• r

r •

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

y = 8 • sqr root 3


600

300

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

1

• rsqr root 32

• r

r •

now you know that you can create the “formulas” from scratch,just by using the Pythagorean Theorem.

sqr root 2

1 • r 1 • r

special right triangles

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