The document explains the properties and relationships of 45-45-90 and 30-60-90 special right triangles. It provides formulas for calculating the lengths of sides based on leg and hypotenuse values, utilizing the Pythagorean theorem and specific triangle theorems. Examples illustrate how to apply these principles to solve for unknown sides in various triangle configurations.

1. Obj. 24 Special Right Triangles
The student is able to (I can):
• Identify when a triangle is a 45-45-90 or 30-60-90
triangle
• Use special right triangle relationships to solve problems

2. Consider the following triangle:
To find x, we would use a2 + b2 = c2, which
gives us:
What would x be if each leg was 2?
1
1 x
2 2 2
2
1 1 x
x 1 1 2
x 2
+ =
= + =
=

3. Again, we will use the Pythagorean Theorem
Simplifying the radical, we can factor
to give us
Do you notice a pattern?
2
2 x
2 2 2
2
2 2 x
x 4 4 8
x 8
+ =
= + =
=
8
2 2.

4. Thm 5-8-1 45º-45º-90º Triangle Theorem
In a 45º-45º-90º triangle, both legs are
congruent, and the length of the
hypotenuse is times the length of
the leg.
2
x
x x 245º
45º

5. Example Find the value of x. Give your answer in
simplest radical form.
1.
2.
3.
45º
x
8
8 2
x
7
7 2
9 2x
9 2
9
2
=

6. If we know the hypotenuse and need to find
the leg of a 45-45-90 triangle, we have to
divide by . This means we will have to
rationalize the denominator, which means
to multiply the top and bottom by the
radical.
The shortcut for this is to divide the
hypotenuse by 2 and then multiply by
2
16 x
16 16 2
x
2 2 2
  
= =   
  
16 2
8 2
2
= =
2.
16
x 2 8 2
2
= =

7. Examples Find the value of x.
1.
2.
x
45º 20
20
x 2 10 2
2
= =
x
5
5
x 2
2
=

8. Thm 5-8-2 30º-60º-90º Triangle Theorem
In a 30º-60º-90º triangle, the length of
the hypotenuse is 2 times the length of
the shorter leg, and the length of the
longer leg is times the length of the
shorter leg.
Note: the shorter leg is always opposite
the 30º angle; the longer leg is always
opposite the 60º angle.
3
x
2xx 3
60º
30º

9. Examples Find the value of x. Simplify radicals.
1. 2.
3. 4.
7
x
60º
30º
11x
9
x
60º
1616
60º
x
9 3
16
3 8 3
2
=
14
11
5.5
2
=

10. Examples
To find the shorter leg from the longer leg:
Find the value of x
1.
2.
9
x
60º
10
x
30º
longer leg 3 longer leg
3
3 3 3
  
=  
  
9
x 3 3 3
3
= =
10
x 3
3
=