1 / 12

GBK Geometry

GBK Geometry. Jordan Johnson. Today’s plan. Greeting Review HW Proofs, Midsegment Proof Area Proof Work Homework / Questions Clean-up. Puzzle: Cube Dissection Progress?. 3” x 3” wooden cube  twenty-seven 1” x 1” cubes Possible with fewer than six cuts?.

wenda
Download Presentation

GBK Geometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. GBK Geometry Jordan Johnson

  2. Today’s plan • Greeting • Review HW Proofs, Midsegment Proof • Area Proof Work • Homework / Questions • Clean-up

  3. Puzzle: Cube DissectionProgress? 3” x 3” wooden cube  twenty-seven 1” x 1” cubes Possible with fewer than six cuts?

  4. The Midsegment Theorem:Proof Strategy • (Theorem 37): A midsegment of a triangle is parallel to the third side and half its length. • Given:Triangle ABC with midsegment BC. • Prove:BC║BC and BC = BC⁄2

  5. The Midsegment Theorem – Proof Part I: Create parallelogram BB'PC. Extend B'C', choose point P so that B'C' = C'P.(OK by the Ruler Postulate.) Draw CP. (Two points determine a line.) AC'B' = PC'C (vertical angles) AC'B'  PC'C (by SAS), so CP = AB (by congruence). CP = B'B by transitivity. ACP = A (again, by congruence). AB║PC (ACP and A are equal alternate interior angles on AB and PC). BB'PCis a parallelogram because CP║B'B and CP = B'B.

  6. The Midsegment Theorem – Proof Part II: Prove BC║B'C' and B'C' = BC⁄2 BC║BP by definition of parallelogram, and BP is the same segment as B'C', so BC║B'C'. BC = B'P = 2B'C' since opposite sides of a parallelogram are equal. By division, B'C' = BC⁄2, completing the proof.

  7. About Area 1 • We have: • the Area Postulate • the rectangle area postulate • the square area theorem • We want: • Area theorems for: • Maybe also theorems for: 2  A1 = A2 = = + 3 4 5 4 5

  8. Random Groups (1) • Ruth / Trevor • Atticus / Jade • Ally / Bridget • Keelan / Izzy • Callan / Celeste • Colton / Nathan / Jared

  9. Random Groups (2) • Graden/ Shelton • Colby / Eva • Kai / Emily • Lucas / Kaitlyn • Sydney / Chester • Collin / Claire / Grace

  10. Random Groups (7) • Erin / Avery • Kesh / Nima • Megan / Sam / Olivia

  11. Homework • Take 20 minutes tonight to do this: • Tell how to find the area of a triangle. • State that formula as a theorem, and add it to your conjectures log. • Set up the proof for it, and think about how you might prove it. • Log any other conjectures you come up with. • Bring in what you’ve written and drawn, for me to check off in class.

  12. Clean-up / Reminders • Pick up all trash / items. • Push in chairs (at front and back tables). • See you tomorrow!

More Related