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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

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Problem Presentation set 2 problem 12

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  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

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