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### Familiar Triangles

```
Date: 01/19/99 at 21:56:41
From: Noel
Subject: Integrated math 2b

I need help figuring out special right triangles. I've tried to start
with a and b but I get confused. I don't know the next step.

For example one problem is:

Find a and b. Simplify radicals whenever possible.

| .
|60 .
|     .
|       .
|         .14
|           .
a |             .
|               .
|____________________.

b

I need help with 45-45-90 triangles and 30-60-90 triangles. If you have
any available notes to show me step by step, I'll be very grateful.
Thank you.
```

```
Date: 01/20/99 at 12:35:48
From: Doctor Peterson
Subject: Re: Integrated math 2b

Hi, Noel. Thanks for a carefully explained question!

The two special kinds of triangles are special because two sides are
related in a simple way. For the 45-45-90 triangle,

+
| \
|45 \
|     \
A|       \ C
|         \
|           \
|_            \
| |           45\
+-----------------+
A

the two sides opposite the 45-degree angles are equal. You can get the
other side from the Pythagorean Theorem:

C = sqrt(A^2 + A^2) = sqrt(2A^2) = A * sqrt(2)

That is, the hypotenuse is the square root of 2 times the other sides.

For the 30-60-90 triangle, the important thing is that it's exactly
half of an equilateral triangle:

+                        +
|\                      /|\
| \                    /6|0\
|30\                  /  |  \
|   \A              A/   |   \A
B|    \              /    |    \
|     \            /     |     \
|_     \          /      |      \
| |   60\        /60     |     60\
+--------+      +--------+--------+
A/2                  A

That means that the side opposite the 30-degree angle is half the
length of the hypotenuse. Again, you can get the length of the third
side by the Pythagorean Theorem:

B = sqrt(A^2 - (A/2)^2) = sqrt(A^2 - (A^2)/4)
= sqrt((3/4)A^2) = A * sqrt(3)/2

So in your problem, a is half of 14, or 7 (just imagine completing the
equilateral triangle by reflecting the triangle in side b if this
isn't clear.) You can either use the Pythagorean Theorem directly on
these numbers, or multiply 7 by the square root of 3 from the formula

If you prefer, just try memorizing these pictures:

+
| \
|45 \
|     \
1|       \ sqrt(2)
|         \
|           \
|_            \
| |           45\
+-----------------+
1

+                        +
|\                      /|\
| \                    /6|0\
|30\                  /  |  \
|   \ 2            2 /   |   \ 2
sqrt(3)|    \              /    |    \
|     \            /     |     \
|_     \          /      |      \
| |   60\        /60     |     60\
+--------+      +--------+--------+
1                    2

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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