Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.
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Transcript of Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.
![Page 1: Special Right Triangles- Section 9.7 Pages 405-412 Adam Dec Section 8 30 May 2008.](https://reader033.fdocuments.us/reader033/viewer/2022042822/56649e385503460f94b28f48/html5/thumbnails/1.jpg)
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.](https://reader033.fdocuments.us/reader033/viewer/2022042822/56649e385503460f94b28f48/html5/thumbnails/2.jpg)
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.
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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.
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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
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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.
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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
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The Easy Problems
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The Moderate Problems
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The Difficult Problems
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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
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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>.