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Special Right Triangles-Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008
11

Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

Dec 27, 2015

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Page 1: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

Special Right Triangles-Section 9.7Pages 405-412

Adam DecSection 830 May 2008

Page 2: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

Introduction

Two special types of right triangles. Certain formulas can be used to find the

angle measures and lengths of the sides of the triangles.

One triangle is the 30-60-90(the numbers stand for the measure of each angle).

The second is the 45-45-90 triangle.

Page 3: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

30- 60- 90

30 - 60 - 90 - Triangle Theorem: In a triangle whose angles have measures 30, 60, and 90, the lengths of the sides opposite these angles can be represented by x, x , and 2x, respectively.

To prove this theorem we will need to setup a proof.

3

Page 4: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Proof Given: Triangle ABC is equilateral, ray BD bisects angle ABC.

Prove: DC: DB: CB= x: x : 2x3

Since triangle ABC is equilateral, Angle DCB= 60, Angle DBC= 30 , Angle CDB= 90 , and DC= ½ (BC)

According to the Pythagorean Theorem, in triangle BDC:

x + (BD) = 2x

x + (BD) = 4x

(BD) = 3x

BD = x

Therefore, DC: DB: CB= x: x : 2x

30

6090

2x

x

2 2 2

2 2 2

2 2

3

3

Page 5: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

45- 45- 90

45 - 45 - 90 - Triangle Theorem: In a triangle whose angles have measures 45, 45, 90, the lengths of the sides opposite these angles can be represented by x, x, x , respectively.

A proof will be used to prove this theorem, also.

2

Page 6: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Proof

Given: Triangle ABC, with Angle A= 45 , Angle B= 45 .

Prove: AC: CB: AB= x: x: x

Both segment AC and segment BC are congruent, because If angles then sides( Both angle A and B are congruent, because they have the same measure).

And according to the Pythagorean theorem in triangle ABC:

x + x = (AB)

2x = (AB)

X = AB

Therefore, AC: CB: AB= x: x: x

2

x

x

2 2 2

2 2

2

2

Page 7: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Easy Problems

Page 8: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Moderate Problems

Page 9: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Difficult Problems

Page 10: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

The Answers

1a: 7, 7 ; 1b: 20, 10 ; 1c: 10, 5; 1d: 346, 173 ; 1e: 114, 114

5: 11 17a: 3 ; 17b: 9; 17c: 6 ; 17d: 1:2 21a: 48; 21b: 6 + 6 25a: 2 + 2 ; 25b: 2 27: [40(12 – 5 )] 23

3

3 3

3

2

3

3

2

3 6

3

Page 11: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.

Works CitedRhoad, Richard. Geometry for Enjoyment and Challenge. New. Evanston,

Illinois: Mc Dougal Littell, 1991.

"Triangle Flashcards." Lexington . Lexington Education. 29 May 2008 <http://www.lexington.k12.il.us/teachers/menata/MATH/geometry/trianglesflash.htm>.