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Welcome to MM150 Unit 6

Welcome to MM150 Unit 6. Seminar. Line AB AB Ray AB AB Line segment AB AB. Plane. Any three points that do not lie on the same line determine a plane. (Since 2 points determine a line, a line and a point not on the line determine a unique plane).

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Welcome to MM150 Unit 6

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  1. Welcome to MM150 Unit 6 • Seminar

  2. Line ABAB • Ray ABAB • Line segment ABAB

  3. Plane • Any three points that do not lie on the same line determine a plane. (Since 2 points determine a line, a line and a point not on the line determine a unique plane). • 2. A line in a plane divides the plane into 3 parts: the line and 2 half-planes. • 3. The intersection of 2 planes is a line.

  4. 3 Definitions • Parallel planes – 2 planes that do not intersect • Parallel lines – 2 lines IN THE SAME PLANE that do not intersect • Skew lines – 2 lines NOT IN THE SAME PLANE that do not intersect.

  5. Angle D Side Vertex Side A F

  6. Angle Measures Acute Angle 0 degrees < acute < 90 degrees Right Angle 90 degrees Obtuse Angle 90 degrees < obtuse < 180 degrees Straight Angle 180 degrees

  7. More Angle Definitions 2 angles in the same plane are adjacent angles if they have a common vertex and a common side, but no common interior points. Example: [ang]BDL and [ang]LDM Non-Example: [ang]LDH and [ang]LDM 2 angles are complementary angles if the sum of their measures is 90 degrees. Example: [ang]BDL and [ang]LDM 2 angles are supplementary angles if the sum of their measures is 180 degrees. Example: [ang]BDL and [ang]LDH L M H B D

  8. If the measure of [ang]LDM is 33 degrees, find the measures of the other 2 angles. Given information: [ang]BDH is a straight angle [ang]BDM is a right angle L M H B D

  9. If [ang]ABC and [ang]CBD are complementary and [ang]ABC is 10 degrees less than [ang]CBD, find the measure of both angles. [ang]ABC + [ang]CBD = 90 Let x = [ang]CBD Then x – 10 = [ang]ABC X + (x – 10) = 90 2x – 10 = 90 2x = 100 X = 50 [ang]CBD = 50 degrees X – 10 = 40 [ang]ABC = 40 degrees D C B A

  10. Polygons

  11. Sum of Interior Angles 2 * 180 = 360 degrees 4 - 2 = 2 3 * 180 = 540 degrees 5 - 2 = 3 6 - 2 = 4 4 * 180 = 720 degrees

  12. The sum of the measures of the interior angles of a n-sided polygon is • (n - 2)*180 degrees What is the sum of the measures of the interior angles of a nonagon? n = 9 (9-2) * 180 = 7 * 180 = 1260 degrees

  13. EVERYONE: How many sides does a polygon have if thesum of the interior angles is 900 degrees? • (n - 2) * 180 = 900 • Divide both sides by 180 • n - 2 = 5 • Add 2 to both sides • n = 7 The polygon has 7 sides.

  14. Similar Figures Y B 80[deg] 80[deg] 4 4 2 2 A X Z 1 2 C 50[deg] 50[deg] 50[deg] 50[deg] [ang]A has the same measure as [ang]X [ang]B has the same measure as [ang]Y [ang]C has the same measure as [ang]Z XY = 4 = 2 AB 2 YZ = 4 = 2 BC 2 XZ = 2 = 2 AC 1

  15. Page 238 # 73 • Steve is buying a farm and needs to determine the height of a silo. Steve, who is 6 feet tall, notices that when his shadow is 9 feet long, the shadow of the silo is 105 feet long. How tall is the silo? 9 = 6 105 ? 9 * ? = 105 * 6 9 * ? = 630 ? = 70 feet The silo is 70 feet tall. ? 6 ft 9 ft 105 feet

  16. Area of a Trapezoid 3 m 2 m 4 m A = (1/2)h(b1 + b2) A = (1/2)(2)(3 + 4) A = (1/2)(2)(7) A = 1(7) A = 7 square meters

  17. Circle radius is in green diameter is in blue 2r = d Twice the radius is the diameter Circumference C = 2∏r or 2r∏ Since 2r = d C = ∏d Area A = ∏r2

  18. Prisms Pyramids

  19. Examples Page 263 #8 V = Bh V = (6 sq yd)*(6 yard) V = 36 cubic yards Page 263 #14 V = (1/3)Bh V = (1/3)(78.5 sq ft)(24 ft) V = 628 cubic feet

  20. Surface Area • Remember surface area is the sum of the areas of the surfaces of a three-dimensional figure. • Take your time and calculate the area of each side. • Look for sides that have the same area to lessen the number of calculations you have to perform.

  21. Examples Page 263 #8 Area of the 2 Bases 3 yd * 2 yd = 6 sq yd Area of 2 sides 2 yd * 6 yd = 12 sq yd Area of other 2 sides 3 yd * 6 yd = 18 sq yd Surface area 6 + 6 + 12 + 12 + 18 + 18 = 72 sq yd Page 263 #14 Surface area of a cone SA = [pi]r2 + [pi]r*sqrt[r2 + h2] SA = 3.14 * (5)2 + 3.14 * 5 * sqrt[52 + 242] SA = 3.14 * 25 + 3.14 * 5 * sqrt[25 + 576] SA = 78.5 + 15.7 sqrt[601] SA = 78.5 + SA = sq ft

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