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Transcript
Geometry Construction Project Rubric Ms. Chaney ~ East Paulding High School For an “A” (90-100%) grade: 20 to 24 constructions all are defined/described by a conjecture or a compass is used theorem no freehand lines or freehand arcs a straightedge is used white paper is used the drawing is clean all constructions are in a three-ring binder compass “marks” and arcs are evident For a “B” (80-89%) grade: 15 to 19 constructions all are defined/described by a conjecture a compass is used or theorem no freehand lines or freehand arcs a straightedge is used white paper is used the drawing is clean all constructions are in a three-ring binder compass “marks” and arcs are evident For a “C” (70-79%) grade: 10 to 14 constructions all are defined/described by a conjecture a compass is used or theorem no freehand lines or freehand arcs a straightedge is used white paper is used the drawing is clean all constructions are in a three-ring binder compass “marks” and arcs are evident Anything less than the above performances will result in a failing “F” grade and CANNOT be submitted for the optional assignment! All constructions must be in a THREE-RING BINDER. There are no exceptions! The three ring binder will have a title page, this rubric and the constructions list. Highlight the names of the constructions you have drawn. Please number each page of your project. Place the constructions in the order they are on the sheet in this packet. Each construction should be followed immediately with its construction write-up. This project is due on: TUESDAY, OCTOBER 6, 2015 DURING YOUR ASSIGNED CLASS TIME. (It will be received ONLY on or before the due date!! NO EXCEPTIONS) REMEMBER!!! This is your project!! YOU MAY NOT RECEIVE ANY OUTSIDE HELP (this includes a tutor). As always, I am available to assist you, but this must be your work. Geometry Constructions 1. Subtracting Line Segments 2. Adding Angles 3. Constructing Congruent Angles 4. Square 5. Rectangle 6. Rhombus 7. Pentagon 8. Octagon 9. Isosceles Acute Triangle 10. Isosceles Right Triangle 11. Isosceles Trapezoid 12. Constructing a Line Perpendicular through a Point Not On the Line 13. Constructing a Line Parallel to Another Line 14. Two Parallel Lines with Alternate Exterior Angles 15. Two Parallel Lines with Alternate Interior Angles 16. Two Parallel Lines with Corresponding Angles 17. Two Parallel Lines with Same Side Exterior Angles 18. Two Parallel Lines with Same Side Interior Angles 19. Bisecting an Angle (Angle Bisector Conjecture) 20. Perpendicular Bisector of a Segment 21. Constructing a Triangle from Three Line Segments 22. Constructing a 30-60-90 Triangle 23. Constructing a 45-45-90 Triangle 24. Constructing the Perpendicular Bisector of a Triangle 25. Constructing the Altitude of a Triangle 26. Constructing the Angle Bisector of a Triangle Geometry Construction Write-Up (EXAMPLE) Use a separate plain (unlined) white sheet of 8.5 * 11 paper for your construction. This sheet is for the following: Name of Construction: Constructing a 45-45-90 Triangle Method: (Explain how you made the construction) I drew two points on the paper and connected them with a ruler to make AB . (By postulate 2.1) Next, I constructed a perpendicular bisector of AB by placing my compass at A and adjusting my compass to a 1 AB and drew arcs below and above AB (definition on a line segment). Using the 2 same compass setting, I placed the compass at B and drew arcs above and below AB so that they intersected the previously drawn arcs. I labeled these points C and D. Using a straightedge to draw CD . I labeled the point where AB intersected CD as point E. E is the midpoint of AB , and CD is a perpendicular bisector of AB by theorem 2.1 and the definition of a perpendicular line. After drawing CD with my straightedge (def. of a line) I labeled AEC as 90 , by the definition of a perpendicular line. CD is perpendicular to AB and by the definition of perpendicular lines we know that they form a right angle. Thus, AEC is a right angle and the measure of AEC is 90 . width that was greater than Next, I bisected AE by placing my compass at A and adjusting my compass to a width that was greater than 1 AE and drew arcs below and above AE (definition on a line segment). Using the same compass setting, I 2 placed the compass at E and drew arcs above and below AE so that they intersected the previously drawn arcs. I labeled these points G and H. Using a straightedge to draw GH . I labeled the point where AE intersected GH as point F. F is the midpoint of AE , and GH is a bisector of AE by theorem 2.1 and the definition of a perpendicular line. After drawing GH with my straightedge (def. of a line) I labeled EFG as 90 . GH is perpendicular to AE and by the definition of perpendicular lines we know that they form a right angle. Thus, EFG is a right angle and the measure of EFG is 90 . After that I constructed an angle bisector for AEC angle. To do this I placed the sharp point of my compass on the vertex of AEC and drew an arc intersecting both sides of AEC , I labeled these intersection points I and J. With the compass on point I, I drew an arc in the interior of AEC . Keeping the same compass setting, I placed the compass on point J and dew an arc the intersected the arc from point I, I then labeled the point K. Next, I drew a ray connecting point E thru point K by the definition of a ray. I then extended EK until it intersected GH . I labeled that point L. Therefore, by the definition of an angle bisector I know that FEL is a 45 degree angle. Now, by Theorem 4.1 I know LFE is 180 degrees. Thus, FLE must be a 45 degree angle as well, by the triangular sum theorem. Hence, creating a 45-45-90 degree triangle. GEOMETRY CONSTRUCTION WRITE-UP Name of Construction (on page _____): Method: (Explain how you made the construction) Page _______