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Area: Parallelograms

Area: Parallelograms. Lesson 10-1. Objectives: 1. to find areas of rectangles 2. to find areas of parallelograms. Area: Parallelograms. Lesson 10-1. New Terms: 1. area – the area of a figure is the number of square units it encloses

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Area: Parallelograms

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  1. Area: Parallelograms Lesson 10-1 Objectives: 1. to find areas of rectangles 2. to find areas of parallelograms

  2. Area: Parallelograms Lesson 10-1 New Terms: 1. area – the area of a figure is the number of square units it encloses 2. altitude – an altitude is a line segment perpendicular to the line containing a base of the figure and drawn from the side opposite that base Tips: look for two segments that form a right angle when determining a base and height

  3. Area: Parallelograms Lesson 10-1 Find the area of the rectangle. Step 1: Change the units so that they are the same. 150 cm = 1.5 m Change 150 centimeters to meters. Step 2: Find the area. A = bhUse the formula for area of a rectangle. = (4)(1.5) Replace b and h with the dimensions 4 and 1.5. = 6 Simplify. The area of the rectangle is 6 m2.

  4. Area: Parallelograms A = bharea formula A = bh = (8)(2) Substitute. = (2.5)(6) = 16 Simplify. = 15 Lesson 10-1 Find the area of each parallelogram. a. b. The area is 16 m2. The area is 15 in.2.

  5. Area: Triangles and Trapezoids Lesson 10-2 Objectives: 1. to find areas of triangles 2. to find areas of trapezoids

  6. Area: Triangles and Trapezoids Lesson 10-2 New Terms: 1. altitude of a triangle – is the perpendicular segment from a vertex of a triangle to the line containing the opposite side. The height is the length of the altitude. Tips: bases of a trapezoid are the two sides parallel to each other, even if the figure is turned to find the area of an irregular figure, you may need to use more than one area formula. (make sure to know all the area formulas)

  7. Area: Triangles and Trapezoids A = bhUse the formula for area of a triangle. = • 13 • 6 Replace b with 13 and h with 6. 1 2 1 2 Lesson 10-2 Find the area of the triangle. = 39 Simplify. The area is 39 in.2.

  8. Area: Triangles and Trapezoids Area of triangle Area of rectangle A = bh A = bh = • 45 • 20 = 45 • 30 = 450 = 1,350 1 2 1 2 Lesson 10-2 Find the area of the figure. Add to find the total: 450 + 1,350 = 1,800. The area of the figure is 1,800 cm2.

  9. Area: Triangles and Trapezoids A = h(b1 + b2) A = • 3(31 + 40) Use the formula for the area of a trapezoid. Replace h with 3, b1 with 31, and b2 with 40. = • 3(71) Simplify. = • 213 1 2 1 2 1 2 1 2 Lesson 10-2 Suppose that, through the years, a layer of silt and mud settled in the bottom of the Erie Canal. Below is the resulting cross section of the canal. Find the area of the trapezoidal cross section. = 106.5 The area of the cross section is 106.5 ft2.

  10. Area: Circles Lesson 10-3 Objectives: 1. to find areas of circles 2. to find areas of irregular figures that include parts of a circle

  11. Area: Circles Lesson 10-3 New Terms: 1. circumference – is the measurement around the outside of the circle. 2 times the radius times pi ~ 2πr or πd. 2. radius – is half the diameter. 3. diameter – segment that goes from one point of the circle to another point, and goes through, contains, the center. Tips: remember to follow the order of operations when finding the area of a circle or semi-circle. Square the radius first, and then multiply by π. when asked to find the “exact area” of a circle (or circular shape) do not substitute 3.14 in for π, leave π as π. make sure not to confuse the radius as the diameter…know the difference.

  12. A = r 2 1 2 = (10)2r = d; r = 10 = 100 Simplify. The area is 100 in.2. Area: Circles Lesson 10-3 Find the exact area of a circle with diameter 20 in.

  13. A = r 2 = (50)2r = d; r = 50 = 2,500 exact area 1 2 (2,500)(3.14) Use 3.14 for . Area: Circles Lesson 10-3 A TV station’s weather radar can detect precipitation in a circular region having a diameter of 100 mi. Find the area of the region. = 7,850 approximate area The area of the region is about 7,850 mi2.

  14. Area of region that is one fourth of a circle: area of circle = r 2 area of quarter circle = r 2 A (3.14)(15)2 Replace with 3.14 and r with 15. = 176.625 ft2 1 4 1 4 Area: Circles Lesson 10-3 A pound of grass seed covers approximately 675 ft2. Find the area of the lawn below. Then find the number of bags of grass seed you need to buy to cover the lawn. Grass seed comes in 3-lb bags.

  15. 1,302 ÷ 675 1.93 Divide to find the number of pounds of seed. Area: Circles Lesson 10-3 (continued) Area of region that is a rectangle: area of rectangle = bh A = 45 • 25 Replace b with 45 and h with 25. = 1,125 The area of the lawn is about 177 ft2 + 1,125 ft2 = 1,302 ft2. You need to buy one 3-lb bag of grass seed.

  16. Space Figures Lesson 10-4 Objectives: 1. to identify common space figures 2. to identify space figures from nets New Terms: 1. space figure – are three-dimensional figures, solids net– is a pattern (2-D) you can form into a space figure Tips: a “cube” is a rectangular prism with six congruent square faces Lateral means “on the side”. The lateral faces of a prism or pyramid are the surfaces that connect with a base.

  17. Space Figures Lesson 10-4 A prism has two parallel bases that are congruent polygons, and lateral faces that are parallelograms A pyramidhas a base that is a polygon. The lateral faces are triangles.

  18. Space Figures Lesson 10-4 A cylinder has two parallel bases that are congruent circles. A cone has one circular base and one vertex.

  19. Space Figures Lesson 10-4 A sphere is the set of all points in space that are a given distance from a given point called the center.

  20. Space Figures Lesson 10-4 Describe the bases and name the figure. The bases are circles. The figure is a cylinder.

  21. Space Figures Lesson 10-4 Name the space figure you can form from the net. With two hexagonal bases and rectangular sides, you can form a hexagonal prism.

  22. Surface Area: Prisms and Cylinders Lesson 10-5 Objectives: 1. to find surface areas of prisms 2. to find surface areas of cylinders

  23. Surface Area: Prisms and Cylinders Lesson 10-5 New Terms: 1. Surface Area (SA) – is the sum of the areas of the base(s) and the lateral faces of a space figure. Lateral Area (LA)– is the sum of the areas of the lateral faces. Tips: SA = surface Area, LA = lateral area, B = area of the base, p= perimeter of the base One way to find the surface area of a space figure is to find the area of its net. Surface Area – is measured in square units. SA of a cylinder can also be written as SA= 2πrh + 2πr2

  24. Find the area of each rectangle in the net. Surface Area: Prisms and Cylinders Lesson 10-5 Find the surface area of the rectangular prism using a net. Draw and label a net. 60 + 60 + 150 + 90 + 150 + 90 = 600 Add the areas. The surface area is 600 cm2.

  25. Step 1: Find the lateral area. L.A. = phUse the formula for lateral area. = (5 + 6 + 5 + 6)20 p = 5 + 6 + 5 + 6 and h = 20 = 440 Surface Area: Prisms and Cylinders Lesson 10-5 Find the surface area of the rectangular prism. Step 2: Find the surface area. S.A. = L.A. + 2B Use the formula for surface area. = 440 + 2(5 • 6) L.A. = 440 and B = 5 • 6 = 440 + 60 = 500 The surface area of the rectangular prism is 500 in.2.

  26. Step 1: Find the lateral area. L.A. = 2 rhUse the formula for lateral area. 2(3.14)(8)(15) 754 Step 2: Find the surface area. S.A. = L.A. + 2B Use the formula for surface area. = L.A. + 2( r 2) 754 + 2(3.14)(8)2 1,156 Surface Area: Prisms and Cylinders Lesson 10-5 Find the surface area of the cylindrical water tank. The surface area of the water tank is about 1,156 ft2.

  27. Surface Area: Pyramids, Cones, and Spheres Lesson 10-6 Objectives: 1. to find surface areas of pyramids 2. To find surface areas of cones and spheres

  28. Surface Area: Pyramids, Cones, and Spheres Lesson 10-6 New Terms: 1. slant height (l)– is the height of a face, used to find the area of the lateral faces. Tips: the slant height is not perpendicular to the base of a pyramid or cone. In a pyramid, it is perpendicular to the base of a triangular face. In a cone, it is the shortest segment that joins the vertex to a point on the circle base there are a number of analogies that one can use to remember not only the lateral-area formulas, but also the surface-area formulas, each of which involves adding lateral area and base area(s).

  29. Step 1:L.A. = p Use the formula for lateral area. = • 20 • 8 = 80p = 4(5) and = 8. 1 2 1 2 Surface Area: Pyramids, Cones, and Spheres Lesson 10-6 Find the surface area of the square pyramid. Step 2: S.A. = L.A. + BUse the formula for surface area. = 80 + 52 Lateral area = 80 and B = 52. = 80 + 25 = 105 The surface area of the pyramid is 105 m2.

  30. Step 1:L.A. = r Use the formula for lateral area. 3.14(3)(7)r = 3 and = 7. = 65.94 Step 2: S.A. = L.A. + B Use the formula for surface area. 65.94 + 3.14(3)2L.A. 65.94 and B = (3)2. = 65.94 + 28.26 = 94.2 Surface Area: Pyramids, Cones, and Spheres Lesson 10-6 Find the surface area of the cone. The surface area of the cone is about 94 m2.

  31. S.A. = 4 r 2Use the formula for surface area. 4(3.14)(3,963)2r 3,963 197,259,000 Round to nearest 1,000. Surface Area: Pyramids, Cones, and Spheres Lesson 10-6 Earth has an average radius of 3,963 mi. What is Earth’s approximate surface area to the nearest 1,000 mi2? Assume that Earth is a sphere. = 197,259,434.64 Multiply. The surface area of Earth is about 197,259,000 mi2.

  32. Volume: Prisms and Cylinders Lesson 10-7 Objectives: 1. to find volumes of prisms 2. to find volumes of cylinders

  33. Volume: Prisms and Cylinders Lesson 10-7 New Terms: 1. Volume (V) – is the number of cubic units needed to fill a 3- dimensional figure. cubic unit– is the space occupied by a cube with edges 1 unit long. (ie. in3) Tips: read the question carefully, you can calculate both the volume and surface area of a 3-D figure, make sure to understand the concepts of both and use the correct formula.

  34. 1 2 = 63 • 20B = • 9 • 14 = 63 cm2 Volume: Prisms and Cylinders Lesson 10-7 Find the volume of the triangular prism. V = Bh Use the formula for volume. = 1,260Simplify. The volume is 1,260 cm3.

  35. V = r 2h B = r 2 3.14 • 3.42 • 16Replace r with 3.4, and h with 16. Volume: Prisms and Cylinders Lesson 10-7 Find the volume of the juice can, to the nearest cubic centimeter. V = Bh Use the formula for volume. = 580.7744 Simplify. The volume is about 581 cm3.

  36. Volume: Pyramids, Cones, and Spheres Lesson 10-9 Objectives: 1. to find volumes of pyramids and cones 2. to find volumes of spheres

  37. Volume: Pyramids, Cones, and Spheres Lesson 10-9 Tips: the formula for the volumes of cones and pyramids uses each figure’s height, not slant height. Remember, the height of a cone or pyramid is the length of the segment from the vertex perpendicular to the base. the volume of a sphere is the only one (studied so far) that involves a third power.

  38. V = Bh Use the formula for volume. V = r 2h B = r 2 (3.14)(2)2(12)Replace r with 2 and h with 12. 1 3 1 3 1 3 Volume: Pyramids, Cones, and Spheres Lesson 10-9 Find the volume of the cone. = 50.24Simplify. The volume of the cone is about 50 in.3.

  39. V = Bh Use the formula for volume. V = s 2h B = s2 = (8)2(12) Replace s with 8 and h with 12. 1 3 1 3 1 3 Volume: Pyramids, Cones, and Spheres Lesson 10-9 Find the volume of the square pyramid. = 256 Simplify. The volume of the pyramid is 256 in.3.

  40. V = r 3Use the volume formula. (3.14)(3,963)3Replace r with 3,963. 260,579,713,159Simplify. 4 3 4 3 Volume: Pyramids, Cones, and Spheres Lesson 10-9 Additional Examples Earth has an average radius of 3,963 mi. What is Earth’s approximate volume to the nearest 1,000,000 mi3? Assume that Earth is a sphere. The volume of the Earth is about 260,580,000,000 mi3.

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