# 60˚-90˚ triangle. ```8.2 Special Right Triangles
Geometry
Mr. Jacob P. Gray
Franklin County High School
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8.2 Special Right Triangles
Objectives
•
•
To use the properties of 45˚-45˚-90˚ triangles
To use the properties of 30˚-60˚-90˚ triangles
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Take notes on the following:
Essential Understanding
 Certain right triangles have properties that allow
you to use shortcuts to determine side lengths
without using the Pythagorean Theorem.
Right Isosceles Triangle
 The acute angles of an Isosceles Right Triangle are
both 45˚.  Another name for this is a 45˚-45˚-90˚
triangle.
x
x
45˚
45˚
y
𝒙𝟐 + 𝒙𝟐 = 𝒚𝟐 Use the Pythagorean Theorem
𝟐𝒙𝟐 = 𝒚𝟐 Simplify
𝒙√𝟐 = 𝒚
Take the positive square root
of each side
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Take notes on the following:
Theorem 8-5
 45˚-45˚-90˚ Triangle Theorem
 In a 45˚-45˚-90˚ triangle, both legs are congruent
and the length of the hypotenuse is √2 times the
length of a leg.
s√2
Hypotenuse = √2 &middot; leg
45˚
s
45˚
s
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Here is a problem to practice.
 Problem 1) Finding the Length of the Hypotenuse
 What is the value of each variable?
1
2
9
45˚
x
45˚
45˚
h
Hypotenuse = √2 &middot; leg
45˚
2√2
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Here is another problem to practice.
 Problem 2) Finding the Length of a Leg
 What is the value of x?
Hypotenuse = √2 &middot; leg
6
45˚
45˚
x
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Here is another problem to practice.
 Problem 3) Finding Distance
 Softball A softball diamond is a square. The
distance from base to base is 60 ft. To the nearest
foot, how far does the catcher throw the ball from
home to second base?
Hypotenuse = √2 &middot; leg
60 ft
d
𝒅 = 𝟔𝟎 &middot; √𝟐
Use a calculator.
𝒅 ≈ 𝟖𝟒. 𝟖𝟓𝟐𝟖…
The catcher throws the ball about 85 ft from
home plate to second base.
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Another type of special right triangle is a
30˚-60˚-90˚ triangle.
Theorem 8-6
 30˚-60˚-90˚ Triangle Theorem
 In a 30˚-60˚-90˚ triangle, the length of the
hypotenuse is twice the length of the shorter leg.
 The length of the longer leg is √3 times the length
of the shorter leg.
hypotenuse = 2 &middot; shorter leg
longer leg = √3 &middot; shorter leg
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30˚
2s
60˚
s
s√3
Here is a problem to practice.
 Problem 4) Using the Length of One Side
 Algebra What is the value of d in simplest radical
form?
5
d
60˚
30˚
f
𝒍𝒐𝒏𝒈𝒆𝒓 𝒍𝒆𝒈 = √𝟑 &middot; 𝒔𝒉𝒐𝒓𝒕𝒆𝒓 𝒍𝒆𝒈
𝟓 = 𝒅√𝟑
𝟓
𝟑
=
𝒅 =
𝒅 𝟑
𝟑
𝟓
𝟑
∙
√𝟑
𝟓√𝟑
=
𝟑
𝟑
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Here is the last problem to practice.
 Problem 5) Applying the 30˚-60˚-90˚ Triangle Theorem
 Jewelry Making An artisan makes pendants in the shape
of equilateral triangles. The height of each pendant is
18mm. What is the length s of each side of a pendant to the
nearest tenth of a millimeter?
𝑙𝑜𝑛𝑔𝑒𝑟 𝑙𝑒𝑔 = 3 &middot; 𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑙𝑒𝑔
18 = √3
1
𝑠
2
S
18mm
3
18 =
𝑠
2
2
3
∙ 18 = 𝑠
𝑠 ≈ 20.7846. .
Each side is about 20.8 mm long.
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Classwork
 Pages 503-505
 7-21 all
 23-29 all
 34, 35, 37, 38, 39
STOP
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