1 / 11

BELLRINGER

W. 1. BELLRINGER. P. Y. 45°. PR = ?, ∠ RPS = ?. 2. 95°. S. V. ?. 4k-2. 4. 132°. ?. k+10. WX = ? ∠ VZY = ?. Q. R. Z. T. X. ALTITUDE. This is the newest type of segment we will learn that exists in a triangle.

derron
Download Presentation

BELLRINGER

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. W 1. BELLRINGER P Y 45° PR = ?, ∠RPS = ? 2. 95° S V ? 4k-2 4 132° ? k+10 WX = ? ∠VZY = ? Q R Z T X

  2. ALTITUDE • This is the newest type of segment we will learn that exists in a triangle. • This segment has one endpoint at a vertex of the triangle, and its other endpoint perpendicular to the opposite side. X XY is an altitude. Y

  3. For Example, Every Triangle has THREE altitudes. The altitude is called the HEIGHT of a triangle.

  4. Example 1 37° 122 + 52 = AC2 144 + 25 = AC2 169 = AC2 AC = √169 = 13 A CA D 53° AF is an altitude of length 12 cm. CF = 5cm. EC = ? ∠ADE = ? B 53° Triangle Angle Sum E 90° But, since DE is a midsegment, EC is half the length of AC. Thus, EC = 13/2 = 6.5 cm. F C

  5. Example 2 What is the height of the triangle? HYPOTENUSE LEG Use the pythagorean theorem: leg2 + leg2 = hypotenuse2 Therefore, 72 + height2 = 252. 49 + h2 = 625.  h2 = 625 - 49  h2 = √576,  h = 24 feet. 25 ft 7 ft LEG

  6. How many altitudes are in an obtuse triangle?? • See sketchpad example . . . • 3!

  7. Where are the Altitudes in a Right Triangle? The LEGS of a right triangle are ALTITUDES. 2 1 3

  8. Altitudes help us find the area of a triangle. • AREA OF A TRIANGLE = (base)(height) / 2. • Altitude = Height and allows us to find the area. Observe Skethpad example.

  9. Example 3 • Find the area of the triangle. First, find the base YZ. Use P.T. twice— WY2 + 102 = 152 WY2 +100 = 225 WY2 = 125  WY = 11.2 A = (b ∙ h) / 2 A = (29.7)(10) / 2 A = 297 / 2 A = 148.5 cm2 X 21 cm 15 cm Then, using PT again, WZ = 18.5 cm 10 cm Thus, YZ = 11.2 + 18.5 = 29.7 cm Y Z W

  10. Example 4 X The altitude is 7 feet. The area of Triangle XYZ is 70 ft2 A = (b ∙ h) / 2 70 = 7b / 2 70 ∙ 2 = 7b 140 = 7b b = 140/7 b = 20 feet. Z Y What is the length of YZ? 20 FEET!

  11. EXERCISES! Draw 3 altitudes! A 2. 1. D 4(x + 2) A AB is an altitude. Find the area of triangle AXZ 3. X +20 C 8 in 10 in 17 in Z E B AB = ? X B 3. DO page 418, #1, 5, 6. DO page 465, #1, 4, 6, 10.

More Related