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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.

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BELLRINGER

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  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.

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