1 / 9

Problem Presentation set 2 problem 12

Problem Presentation set 2 problem 12. Ellen Dickerson. Before we start on the problem lets review a little bit…. Line segment Distance of a line segment Angle Pythagorean theorem. In the figure to the right, Δ ABC is a right triangle, with a right angle at B, BD is

nelle-pickett
Télécharger la présentation

Problem Presentation set 2 problem 12

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. Problem Presentation set 2 problem 12 Ellen Dickerson

  2. Before we start on the problem lets review a little bit…. • Line segment • Distance of a line segment • Angle • Pythagorean theorem

  3. In the figure to the right, ΔABC is a right triangle, with a right angle at B, BD is perpendicular to AC. If CD=9, and AD=3, then find the exact value BC+AB+BD. AD=3 CD=9

  4. Lets start finding the three right triangles in the picture • ΔABC (the right angle is at ∠ ABC) • ΔADB (the right angle is at ∠ ADB) • ΔBDC (the right angle is at ∠ BDC) AD=3 CD=9

  5. Now we will use the Pythagorean Theorem (a2+b2=c2)to find a formula for each triangle • ΔABC: (AB)2 + (BC)2 = 122 • ΔADB: 32 + (DB)2 = (AB)2 • ΔBDC: 92 +(DB)2 = (BC)2 AD=3 CD=9

  6. ΔABC: (AB)2 + (BC)2 = 122 • ΔADB: 32 + (DB)2 = (AB)2 • ΔBDC: 92 + (DB)2 = (BC)2 • (AB)2 +(BC)2 = 122 • (AB)2 + [ 92 +(DB)2] = 122 • (AB)2 + 81 + (BD)2 = 144 • [32 + (BD)2] + 81 +(BD)2 = 144 • 90 + 2(BD)2 = 144 • 2(BD)2 = 54 • (BD)2 = 27 • BD = √27 AD=3 CD=9

  7. ΔABC: (AB)2 + (BC)2 = 122 • ΔADB: 32 + (DB)2 = (AB)2 • ΔBDC: 92 + (DB)2 = (BC)2 • 32 + (DB)2 = (AB)2 • 9 +27 = (AB)2 • 36 = (AB)2 • 6 = AB AD=3 BD=√27 CD=9

  8. ΔABC: (AB)2 + (BC)2 = 122 • ΔADB: 32 + (DB)2 = (AB)2 • ΔBDC: 92 + (DB)2 = (BC)2 • 92 + (DB)2 = (BC)2 • 81 + 27 = (BC)2 • 108 = (BC)2 • √108 = BC AD=3 BD=√27 CD=9 AB=6

  9. The problem asked us to find BC+BD +AB • √108 + √27 + 6 • √(36 *3) + √(9*3) + 6 • 6√3 + 3√3 + 6 • 9√3 + 6 • . AD=3 BD=√27 CD=9 AB=6 BC=√108

More Related