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Math in the Real World Mrs. Cecil Period 1-2 LT1: Parallel - PowerPoint PPT Presentation

Math in the Real World Mrs. Cecil Period 1-2 LT1: Parallel Lines and Transversals Andrew is making a toothpick bridge for his science project. He is modeling his bridge after the one shown in the picture. Find m 2 if m 1 = 40


  1. Math in the Real World Mrs. Cecil Period 1-2

  2. LT1: Parallel Lines and Transversals Andrew is making a toothpick bridge for his science project. He is modeling his bridge after the one shown in the picture. Find m ∠ 2 if m ∠ 1 = 40 ⁰ . ∠ 1 and ∠ 2 are Alternate Interior Angles, this means their angles measures are equal. 1 m ∠ 1 = 40 ⁰ 2 m ∠ 1 = m ∠ 2 m ∠ 2 = 40 ⁰

  3. LT1: Parallel Lines and Transversals A diagonal brace strengthens the wire fence. The brace makes a 40° angle with the wire as shown. Find x . The angles shown are corresponding angles, which means their angle measures are equal. 2 x + 8 = 120 120° -8 -8 2 x + 8 2 x = 112 2 2 x = 56

  4. LT2: Interior and Exterior Angles of Triangles Romeo looks up at Juliet from below her balcony. His angle of elevation is 65°. What is Juliet’s angle of depression? The sum of the angles of a triangle equal 180°. x x + 65 + 90 = 180 x + 155 = 180 -155 -155 x = 25 Juliet’s angle of depression is 25°. 65°

  5. LT2: Interior and Exterior Angles of Triangles Sam used wood to build an A-frame house in Lake Tahoe. The base of the house forms a 102° angle with the ground. 3 Find the measure of ∠ 3. The sum of the two remote interior angles of a triangle equal the measure of the exterior angle of the triangle. 102 = ∠ 2 + ∠ 3 102° 1 2 102° 102 = (180-102) + ∠ 3 102 = 78 + ∠ 3 -78 -78 24 = ∠ 3 m ∠ 3 = 24°

  6. LT3&4: Pythagorean Theorem A receiver and defensive back both start on the offensive team’s 30 yard line, 15 yards away from each other. How far would the defensive back have to run to tackle the receiver at the defensive team’s 30 yard line? a² + b² = c² 15² + (20 + 20)² = x ² [ 20 yds] [ 20 yds] 15² + 40² = x ² 225 + 1600 = x ² 1,825 = x ² 40 yds √ 1,825 = √ x² 42.7 = x 15 yds x The defensive back would have to run 42.7 yards.

  7. LT3&4: Pythagorean Theorem a² + b² = c² Eviley wants to buy a new TV for her room. The 30² + 50² = x² wall she wants to put the TV on has a length of 900 + 2500 = x² 3,400 = x² 50” and a height of 30”. The size of the TV is √3,400 = √x² determined by the length of the TV’s diagonal. 58.3 = x Given the TV sizing chart, determine the biggest A 55” TV is the biggest size TV that will fit Eviley’s wall. TV that will fit Eviley’s wall. 30” 50”

  8. LT5&6: Transformations Jose is playing Tetris and doesn’t know where to place his next piece. Joe should rotate the green piece 90° counterclockwise and translate the piece all the way to the right. This will fill in the empty space in the bottom right corner.

  9. LT7: Congruency by Transformations Mrs. Cecil is putting together a toy for her son and is trying figure out which part she needs to match the instructions (Part A). Part A Part 1 Part 2 Part 3 Part 3 is the part Mrs. Cecil needs because if you rotate Part 3 90° counterclockwise and translate it to the left, Part 3 will be the same size, shape, and position as Part A.

  10. LT8: Similarity by Transformations Monica wants to buy a frame for her 5x7 inch family photo, but she doesn’t know which frame to buy. She wants to get a frame with a mat border inside. Which frame will her picture fit in perfectly? 5 = 7 11.2 12 8 11.2 11.5(5) = 8(7) 56 = 56 Frame 1 is similar to 7.5 Frame 2 8 Frame 1 Monica’s photo and will fit her picture perfectly. 5 = 7 7.5 12 12(5) = 7.5(7) 60 = 52.5 Frame 2 is NOT similar to Monica’s photo.

  11. LT9: Volume of Cylinders, Cones, and Spheres How much lip balm is in a tube of Chapstick with a radius of .75 cm and a height of 7.2 cm? Volume of a Cylinder: V = π r ² h V = π (.75 ²) (7.2) V = π (.5625)(7.2) V = π (4.05) V = (3.14)(4.05) V = 12.72cm ³ The tube of Chapstick can hold 12.72cm³ of lip balm.

  12. LT9: Volume of Cylinders, Cones, and Spheres Spaceship Earth at Disney World’s Epcot theme park has an approximate radius of 82.5 feet. What is the volume of Spaceship Earth? Volume of a Sphere: V = 4 π r ³ 3 V = 4 π (82.5 ³) 3 V = 4 π (561,515.625) 3 V = (2,246,062.5) π 3 V = (748,687.5) π V = 2,350,878.75 ft ³ The volume of Spaceship Earth is approx. 2,350,878.75 ft³.

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