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Section 1.3

Section 1.3. By BayLynn Ellsworth Becky Fallk Morgan Palmiter. Explanation of Content. Section 1.3: Collinearity , Betweenness , and Assumptions. Collinearity. Collinear points are points that lie on the same line. Noncollinear points are points that do not lie on the same line. A. B.

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Section 1.3

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  1. Section 1.3 By BayLynn Ellsworth Becky Fallk Morgan Palmiter

  2. Explanation of Content Section 1.3: Collinearity, Betweenness, and Assumptions

  3. Collinearity • Collinear points are points that lie on the same line. • Noncollinear points are points that do not lie on the same line. A B M S Q

  4. NONCOLLINEAR COLLINEAR the points don’t lie on the same line the points lie on the same line

  5. Betweenness of Points • All points must be collinear in order to say that a point is between two other points. B L X

  6. Triangle Inequality There are only two possibilities for any three points: • The points are collinear, which means that one point is between the other two and two of the distances add up to the third. • They are noncollinear, which means that the three points determine a triangle.

  7. *Something to Remember* • In a triangle, the sum of the lengths of any two sides is always greater than the length of the third side. 20 15 30

  8. Assumptions from Diagrams

  9. Sample Problems Difference < x < Sum x Identify range of values for x in each triangle. 6 11 5 17 _____ < x < ______

  10. Sample Problems (cont.) For each diagram, tell whether M is between B and R. 1.) M No R B 2.) B No M X R 3.) YES B M R

  11. Sample Problems (cont.) ↔ YES Should we assume that GF is a straight line? O straight lines CAN be assumed Should we assume that <O is congruent to <ORG? G F R No congruent angles can NOT be assumed

  12. Practice Problem #1 F O O FO is twice as long as FR. The perimeter of FOUR is 156. Find: OU ‾‾ ‾‾ > < ‾‾ O R U

  13. Practice Problem #1 Answer Let FR = x Let FO = 2x 2x + 2x + x + x = 156 6x = 156 x = 26 • ‾‾‾‾ Since FR is congruent to OU, and FR = x, OU = x. Since x = 26, OU is 26. • ‾‾‾‾ • ‾‾‾‾ • ‾‾‾‾ • ‾‾‾‾ • ‾‾‾‾ • ‾‾‾‾ OU = 26 • ‾‾‾‾

  14. Practice Problem #2 W <WOR = 81˚ 19’ 10” <WOD = 119˚ 31’ 43” <ROD = ? R O D

  15. Practice Problem #2 Answer 119˚ 31’ 43” - 81˚ 19’ 10” 38˚ 12’ 33” <ROD = 38˚ 12’ 33” Since <WOD = 119˚ 31’ 43” , and <WOR = 81˚ 19’ 10”, we need to subtract <WOR from <WOD to find <ROD. - = 119˚ 31’ 43” - HOT 81 ˚ 19’ 10” - warm 38 ˚ 12’ 33” - cool

  16. Practice Problem #3 2 3 1 If <1, <2, and <3 are in a ratio of 2 : 4 : 6 Find m <1, m <2, m <3

  17. Practice Problem #3 Answer 2x + 4x + 6x = 180 12x = 180 x = 15 2x = 30 = <1 4x = 60 = <2 6x = 90 = <3 Since we can assume straight angles, the sum of <1, <2, and <3 = 180. When solving for x, <1 = 30, <2 = 60, and <3 = 90.

  18. Practice Problem #4 Given: <ROD is a right angle <AOS is a right angle <ROA is (3x + 2y)° <AOD is (3x + 3y)° <DOS is (6x) Find: x and y A R O D S

  19. Practice Problem #4 Answer 3x + 2y +3x + 3y = 90 6x + 5y = 90 3x + 2y + 6x =90 9x + 3y = 90 3x + y = 30 y = -3x + 30 6x + 5(-3x + 30) = 90 6x – 15x + 150 =90 -9x + 150 = 90 -9x = -60 x = 6 2/3 6x + 5y =90 6 (6 2/3) + 5y = 90 40 + 5y = 90 5y = 50 y = 10 Since <ROD and <AOS are right angles, the two angles that make up the right angles are complementary and have a sum of 90°. Therefore, < ROA (3x + 2y) and <AOD (3x + 3y) = 90. The equation formed from that is 6x + 5y = 90. Then the sum of <AOD (3x + 3y) and <DOS (60) = 90. That equation simplifies to y = -3x + 30. When substituting y into the first equation, x = 6 2/3. After putting x back into the original equation, y = 10. A R O D S

  20. Works Cited Page Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry for Enjoyment and Challenge. Boston: McDougal Little/Houghton Mifflin, 1991.

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