1 / 22

Warm Up for Section 1.1 (Tuesday, August 7)

45 o. 45 o. 7. Warm Up for Section 1.1 (Tuesday, August 7) Simplify: (1). (2). Find the two missing edge lengths in each triangle: (3). (4). (5). . 45 o. 45 o. 7. Warm Up for Section 1.1 (Tuesday, August 7)

naasir
Télécharger la présentation

Warm Up for Section 1.1 (Tuesday, August 7)

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. 45o 45o 7 Warm Up for Section 1.1 (Tuesday, August 7) Simplify: (1). (2). Find the two missing edge lengths in each triangle: (3). (4). (5).

  2. 45o 45o 7 Warm Up for Section 1.1 (Tuesday, August 7) Simplify: (1). (2). Find the two missing edge lengths in each triangle: (3). (4). (5).

  3. Work for Answers to WU, Section 1.1 (1). (2).

  4. Special Right Triangles Standard: MM2G1a, b Section 1.1 Day 2 Essential Question:What is the relationship between the lengths of the edges in a 30°–60°–90°triangle?

  5. Investigation 2: With your partner, complete the following regarding equilateralABC where AB =10: Step 1: Label the length of each edge. Step 2: Label the measure of B and C. Step 3: Using a straightedge, draw and label altitude . Step 4: Label the length of and . Step 5: Label the measure of BAD and CAD. Step 6: Label the measure of ADC. Step 7: Using the Pythagorean Theorem, find AD.

  6. a2 + b2 = c2 A 52 + x2 = 102 25 + x2 = 100 75 = x2 30° 30° 10 10 x 60° 60° B C 5 5 D 10

  7. Investigation 2: Note: the two legs of a 30o-60o-90o triangle are NOT equal in measure. The longer leg will always be opposite the ___o angle. The shorter leg will always be opposite the ___o angle. 60 30

  8. R 30° RT ST 12 60° S T 6 Consider the 30o-60o-90o right triangle created from an equilateral triangle pictured at right. (2). The long leg is segment ______ and the short leg is segment _______. (3). Use the Pythagorean Theorem to find RT.

  9. a2 + b2 = c2 62 + x2 = 122 R 36 + x2 = 144 108 = x2 30° 12 60° S T 6

  10. 30° 2x 60° x Summary: In a 30o-60o-90o triangle: Length of hypotenuse = length of short leg times 2 Length of long leg: length of short leg times Length of short leg: half the length of hypotenuse or the length of the long leg divided by

  11. Check for Understanding: Find the missing edge lengths for each triangle: Example 4:

  12. 60o 30o Check for Understanding: Find the missing edge lengths for each triangle: Example 5:

  13. Check for Understanding: Find the missing edge lengths for each triangle: Example 6:

  14. Check for Understanding: Find the missing edge lengths for each triangle: Example 7: 30o 60o

  15. Check for Understanding: Find the missing edge lengths for each triangle: Example 8: 60o 30o

  16. Check for Understanding: Find the missing edge lengths for each triangle: Example 9: 60o 30o

  17. Application problems: (7). Find the exact area of an equilateral triangle whose edge length is 12 cm. Round your answer to the nearest tenth. Recall: A = ½bh. A = ½bh A = ½(12) A = A ≈ 62.4 cm2 60o 12 12 h 60o 60o 6 6 12

  18. Application problems: (8). Find the exact perimeter of square ABDC if FB = 22 meters A B P = 4s P = 4 P = 22 45o F 22 45o D C

  19. 30° 60° Formula Sheet: Length of long leg = length short leg ∙ _____ Length of hypotenuse = length short leg ∙ _____ Length of short leg = length long leg ÷ ______ Length of short leg = length hypotenuse ÷ ______

  20. 10 x 7 Pythagorean Theorem: a2 + b2 = c2 72 + x2 = 102 49 + x2 = 100 x2 = 51 x =

  21. Triangle Sum Property: Sum of interior s = _____ x° 30° 25° x = 180o – 25o – 30o = 125o

  22. 120° Linear Pair: x° x = 180o – 120o = 60o

More Related