1 / 14

Geometry

Geometry. 8.2 The Pythagorean Theorem (This section, along with 8.4, are very important as they are utilized throughout the second semester). Radical Review. Simplify each expression. = 8/3. = 5. You try!. = 28. = 9/5. Do you know these?. 2. 2. 14 =. 196. 7 =. 49. 2. 8 =. 64.

menefer
Download Presentation

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. Geometry 8.2 The Pythagorean Theorem (This section, along with 8.4, are very important as they are utilized throughout the second semester)

  2. Radical Review • Simplify each expression. = 8/3 = 5 You try! = 28 = 9/5

  3. Do you know these? 2 2 14 = 196 7 = 49 2 8 = 64 2 13 = 169 2 2 15 = 225 9 = 81 2 12 = 2 144 10 = 100 2 16 = 256 2 11 = 2 121 17 = 289

  4. Pythagorean Theorem • In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. A c b C a B

  5. Pythagorean Theorem Proof This is in the state standards and may be on the STAR test! It is important to get this down! c Area of large square = Area of large square c ½ ab a a large square = 4 triangles + small square b ½ ab (b – a) 2 2 2 4(½ ab) + (b – 2ab + a ) c = c (b – a) b c c 2 2 b – 2ab + a 2 2 b 2 c = 2ab + b – 2ab + a c (b – a) ½ ab (b – a) 2 2 2 c = b + a a 2 2 b 2 c a + b = ½ ab a c c (b – a)(b – a) 2 2 b – 2ab + a

  6. Find the value of x together. # BC AC AB x 1) 8 6 x = 10 B 15 2) x 9 x = 12 7) 12 x x = A C

  7. Find the value of x on your own. Try #3, 4 and 8. Who can solve these on the board? # BC AC AB x = x B 3) 5 5 4) x 3 x = 6 5) x 1 x = 1 2 6) 1 x x = A C 12 8) x 8 x =

  8. 6 x 3 Solve for x. 9) 2 2 2 3 + 6 = x 2 9 + 36 = x 6 2 45 = x 9 5 3 3

  9. 17 x 15 6 Solve for x. 16) y 2 2 2 15 + y = 17 2 225 + y = 289 2 y = 64 y = 8 2 2 2 6 + 8 = x You may recognize this one, x = 10.

  10. 18) The diagonals of a rhombus have lengths 18 and 24. Find the perimeter of the rhombus. A rhombus has perpendicular diagonals. 2 2 2 9 + 12 = x x 15 15 2 12 81 + 144 = x 2 225 = x 9 x = 15 15 15 Thus, the perimeter is 60. A rhombus is a parallelogram. Diagonals of a parallelogram bisect each other.

  11. Reminders • The diagonals of a rhombus are perpendicular to each other. • The altitude drawn to the base of an isosceles triangle is perpendicular to the base at its midpoint.

  12. HW • Complete #1-20 from the packet on a separate sheet. For any ones we have done already, write “Did in Class” • P. 292 #19-22 • P. 289 (32-38 Even)

  13. Part of the HW, Answers to exercises #10-20 Skip 14, 16, 18, 20. 10) x = 17 11) x = 4 12) x = 3 13) x = 8 14) 15) 17) 19) 20) x = 7

  14. Let’s review the triangle proportion formulas from 8.1 and do # on page

More Related