1 / 62

Space Figures and Cross Sections

Space Figures and Cross Sections GEOMETRY LESSON 11-1 (For help, go to Lesson 1-3.) For each exercise, make a copy of the cube at the right. Shade the plane that contains the indicated points. 1. A, B, and C 2. A, B, and G

johana
Download Presentation

Space Figures and Cross Sections

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. Space Figures and Cross Sections GEOMETRY LESSON 11-1 (For help, go to Lesson 1-3.) For each exercise, make a copy of the cube at the right. Shade the plane that contains the indicated points. 1. A, B, and C 2. A, B, and G 3. A, C, and G 4. A, D, and G 5. F, D, and G 6. B, D, and G 7. the midpoints of AD CD, EH, and GH Check Skills You’ll Need 11-1

  2. Space Figures and Cross Sections GEOMETRY LESSON 11-1 1.2. 3. 4. 5. 6. 7. Solutions 11-1

  3. AF, BG, CH, DI, EJ, AB, BC, CD, DE, EA, FG, GH, HI, IJ, and JF. Space Figures and Cross Sections GEOMETRY LESSON 11-1 How many vertices, edges, and faces of the polyhedron are there? List them. There are 10 vertices: A, B, C, D, E, F, G, H, I, and J. There are 15 edges: There are 7 faces: pentagons: ABCDE and FGHIJ, and quadrilaterals: ABGF, BCHG, CDIH, DEJI, and EAFJ Quick Check 11-1

  4. Space Figures and Cross Sections GEOMETRY LESSON 11-1 Use Euler’s Formula to find the number of edges on a solid with 6 faces and 8 vertices. F+V= E+ 2 Euler’s Formula 6 + 8 = E+ 2 Substitute the number of faces and vertices. 12 = ESimplify. A solid with 6 faces and 8 vertices has 12 edges. Quick Check 11-1

  5. Draw a net. Space Figures and Cross Sections GEOMETRY LESSON 11-1 Use the pentagonal prism from Example 1 to verify Euler’s Formula. Then draw a net for the figure and verify Euler’s Formula for the two-dimensional figure. Use the faces F = 7, vertices V = 10, and edges E = 15. F+V= E+ 2 Euler’s Formula 7 + 10 = 15 + 2 Substitute the number of faces and vertices. Count the regions: F = 7 Count the vertices: V = 18 Count the segments: E = 24 F + V = E + 1 Euler’s Formula in two dimensions 7 + 18 = 24 + 1 Substitute. Quick Check 11-1

  6. Space Figures and Cross Sections GEOMETRY LESSON 11-1 Describe this cross section. The plane is parallel to the triangular base of the figure, so the cross section is also a triangle. Quick Check 11-1

  7. Space Figures and Cross Sections GEOMETRY LESSON 11-1 Draw and describe a cross section formed by a vertical plane intersecting the top and bottom faces of a cube. If the vertical plane is parallel to opposite faces, the cross section is a square. Sample: If the vertical plane is not parallel to opposite faces, the cross section is a rectangle. Quick Check 11-1

  8. Space Figures and Cross Sections GEOMETRY LESSON 11-1 1. Draw a net for the figure. Sample: Use Euler’s Formula to solve. 2. A polyhedron with 12 vertices and 30 edges has how many faces? 20 • 3. A polyhedron with 2 octagonal faces and 8 rectangular faces has how many vertices? • 4. Describe the cross section. • 5. Draw and describe a cross section formed by a vertical plane cutting the left and back faces of a cube. 16 Circle Check students’ drawings; rectangle. 11-1

  9. Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 (For help, go to Lessons 1-9 and 10-3.) Find the area of each net. 1.2.3. Check Skills You’ll Need 11-2

  10. Solutions 1. Each square has an area of (4)(4) = 16 cm2. The total area is 6(16) = 96 cm2. 2. The area of each circle is r 2= (2)2 = 4 cm2. The area of the rectangle is bh = (4 )(8) = 32 cm2. The total area of the two circles and the rectangle is 2(4 ) + 32 = 8 + 32 = 40 , 125.7 cm2. 3. An altitude from any vertex of a triangle measures 3 3. The area of each triangle is bh = (6)(3 3) = 9 3 m2. The total area of the four triangles is 4(9 3) = 36 3 or about 62 m2. 1 2 1 2 Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 11-2

  11. Draw a net for the cube. Surface Area = sum of areas of lateral faces + area of bases = (121 + 121 + 121 + 121) + (121 + 121) Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 Quick Check Use a net to find the surface area of the cube. Find the area of one face. 112= 121 The area of each face is 121 in.2. = 6 • 121 = 726 Because there are six identical faces, the surface area is 726 in.2. 11-2

  12. Use the formula L.A. =ph to find the lateral area and the formula S.A. = L.A. + 2B to find the surface area of the prism. The area B of the base is ap, where a is the apothem and p is the perimeter. 1 2 Draw the base. Use 30°-60°-90° triangles to find the apothem. Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 Find the surface area of a 10-cm high right prism with triangular bases having 18-cm edges. Round to the nearest whole number. The triangle has sides of length 18 cm, so p = 3 • 18 cm, or 54 cm. 11-2

  13. 9 = 3 alonger leg  3  shorter leg 9 3 3 3 3 a== = 33Rationalize the denominator. 3354 = 81 3 1 2 1 2 B=ap = The area of each base of the prism is 81 3 cm2. S.A. =L.A. + 2BUse the formula for surface area. = 540 + 162 3 2(81 3 ) Substitute =ph + 2B 820.59223Use a calculator. = (54)(10) + 9 3 Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 Quick Check (continued) Rounded to the nearest whole number, the surface area is 821 cm2. 11-2

  14. 2( r2) =2rh + Substitute the formula for lateral area of a cylinder and area of a circle. = 2 (6)(9) + 2 (62) Substitute 6 for r and 9 for h. = 108 + Simplify. = 180 The surface area of the cylinder is 180 ft2. 72 Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 The radius of the base of a cylinder is 6 ft, and its height is 9 ft. Find its surface area in terms of . S.A. =L.A. + 2BUse the formula for surface area of a cylinder. Quick Check 11-2

  15. Cornmeal Container Barley Container S.A. =L.A. + 2B Use the formula for surface area of a cylinder. S.A. =L.A. + 2B Substitute the formulas for lateral area of a cylinder and area of a circle. =2rh + 2 r 2 =2rh + 2 r 2 d 2 Find the surface area of each container. Remember that r= . Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 A company sells cornmeal and barley in cylindrical containers. The diameter of the base of the 6-in. high cornmeal container is 4 in. The diameter of the base of the 4-in. high barley container is 6 in. Which container has the greater surface area? 11-2

  16. Substitute the formulas for lateral area of a cylinder and area of a circle. =rh + r 2 =rh + r 2 2 2 2 2 Substitute for r and h. = (2)(6) + (22) = (3)(4) + (32) 2 2 2 2 Simplify. = 24 + = 24 + 8 = 32 = 42 Because 42 in.2 32 in.2, the barley container has the greater surface area. 18 Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 (continued) Cornmeal Container Barley Container S.A. =L.A. + 2B Use the formula for surface area of a cylinder. S.A. =L.A. + 2B Quick Check 11-2

  17. Use the prism below for Exercises 1 and 2. 1. Use a net to find the surface area. 2. Use a formula to find the surface area. 3. The height of a prism is 5 cm. Its rectangular bases have 3-cm and 9-cm sides. Find its surface area. 4. The radius of the base of a cylinder is 16 in., and its height is 4 in. Find its surface area in terms of . 5. A contractor paints all but the bases of a 28-ft high cylindrical water tank. The diameter of the base is 22 ft. How many square feet are painted? Round to the nearest hundred. S.A. = 216 ft2 640 in.2 Surface Areas of Prisms and Cylinders GEOMETRY LESSON 11-2 S.A. = L.A. + 2B = 168 + 48 = 216; 216 ft2 174 cm2 1900 ft2 11-2

  18. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 (For help, go to Lesson 8-1.) Find the length of the hypotenuse in simplest radical form. 1.2.3. Check Skills You’ll Need 11-3

  19. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 Solutions 1. Use the Pythagorean Theorem: a2 + b2 = c2 (8)2 + (13)2 = x2 64 + 169 = x2x2 = 233 x = 233 in. 2. Use the Pythagorean Theorem: a2 + b2 = c2 (7)2 + (9)2 = x2 49 + 81 = x2x2 = 130 x = 130 m 3. Use the Pythagorean Theorem: a2 + b2 = c2 (12)2 + (13)2 = x2 144 + 169 = x2x2 = 313 x = 313 cm 11-3

  20. You are given = 12 ft and you found that p= 30, so you can find the lateral area. 1 2 L.A. = p Use the formula for lateral area of a pyramid. 1 2 = (30)(12)Substitute. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 Find the surface area of a square pyramid with base edges 7.5 ft and slant height 12 ft. The perimeter p of the square base is 4 X 7.5 ft, or 30 ft. = 180Simplify. 11-3

  21. Find the area of the square base. S.A. = L.A. BUse the formula for surface area of a pyramid. = 180  56.25Substitute. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 (continued) Because the base is a square with side length 7.5 ft, Bs2 7.52 56.25. = 236.25Simplify. The surface area of the square pyramid is 236.25 ft 2. Quick Check 11-3

  22. 1 2 Use the formula L.A. =p to find the lateral area of the pyramid. The altitude of the pyramid, apothem of the base, and altitude of a lateral face form a right triangle, so you can use the Pythagorean Theorem to find the slant height . = 202 + (4 3)2 = 400 + 48 = 448 Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 Find the lateral area of the hexagonal pyramid below. Round your answer to the nearest whole number. 11-3

  23. 1 2 L.A. =pUse the formula for lateral area. 1 2 = (48)( 448) Substitute. 507.98425 Use a calculator. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 (continued) The lateral area of the hexagonal pyramid is about 508 m2. Quick Check 11-3

  24. = r+r 2Substitute the formulas for L.A. and B. = (5)(13) + (5)2Substitute. = 65 + 25Simplify. = 90 The surface area of the cone is 90 in.2. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 Find the surface area of the cone in terms of . S.A. = L.A.+ BUse the formula for surface area of a cone. Quick Check 11-3

  25. Use the formula L.A. =r to find the lateral area of the cone. The altitude of the cone, radius of the base, and slant height form a right triangle. Use the Pythagorean Theorem to find the slant height . = 0.52  1.52 = 0.25  2.25 = 2.5 Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 Leandre uses paper cones to cover her plants in the early spring. The diameter of each cone is 1 ft, and its height is 1.5 ft. How much paper is in the cone? Round your answer to the nearest tenth. The cone’s diameter is 1 ft, so its radius r is 0.5 ft. 11-3

  26. L.A. =r Use the formula for lateral area of a cone. = (0.5) 2.5 Substitute 0.5 for r and 2.5 for . 2.4836471 Use a calculator. Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 (continued) Find the lateral area. The lateral area of the cone is about 2.5 ft.2 Quick Check 11-3

  27. 1. Find the slant height of a square pyramid with base edges 12 cm and altitude 8 cm. 2. Find the lateral area of the regular square pyramid to the right. 3. Find the surface area of the pyramid to the right whose base is a regular hexagon. Round to the nearest whole number. 4. Find the surface area of a cone with radius 8 cm and slant height 17 cm in terms of . 5. The roof of a building is shaped like a cone with diameter 40 ft and height 20 ft. Find the area of the roof. Round to the nearest whole number. 200 cm2 Surface Areas of Pyramids and Cones GEOMETRY LESSON 11-3 10 cm 56 in.2 1517 ft2 1777 ft2 11-3

  28. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 (For help, go to Lessons 1-9 and 10-1.) Find the area of each figure. Round to the nearest tenth if necessary. 1. square - side length 7 cm 3. circle - radius 10 mm 5. rectangle - 14 in. by 11 in. 7. equilateral triangle - side length 8 in. 2. circle - diameter 15 in. 4. rectangle - 3 ft by 1 ft 6. triangle - base 11 cm, height 5 cm Check Skills You’ll Need 11-4

  29. 1.A = s2 = 72 = 49 cm2 2.A = r2 = (7.5)2 = 56.25 176.7 in.2 3.A = r2 = (10)2 = 100 314.2 mm2 4.A = w = (3)(1) = 3 ft2 5.A = w = (14)(11) = 154 in.2 6.A = bh = (11)(5) = 27.5 cm2 7. A = bh = (8)(4 3) = 16 3 27.7 in.2 1 2 1 2 1 2 1 2 Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Solutions 11-4

  30. The area of the base B=w= 3  5 = 15. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Find the volume of the prism below. V= Bh Use the formula for volume. = 15 • 5Substitute 15 for B and 5 for h. = 75Simplify. The volume of the rectangular prism is 75 in.3. Quick Check 11-4

  31. Use the Pythagorean Theorem to calculate the length of the other leg. 292 – 202 = 841  400 = 441  21 Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Find the volume of the prism below. The prism is a right triangular prism with triangular bases. The base of the triangular prism is a right triangle where one leg is the base and the other leg is the altitude. 11-4

  32. 1 2 1 2 The area B of the base is bh= (20)(21) = 210. Use the area of the base to find the volume of the prism. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 (continued) V= Bh Use the formula for the volume of a prism. = 210 •40Substitute. = 8400Simplify. The volume of the triangular prism is 8400 m3. Quick Check 11-4

  33. The formula for the volume of a cylinder is V= r 2h. The diagram shows h and d, but you must find r. 1 2 r= d = 8 V= r 2h Use the formula for the volume of a cylinder. = • 82•9Substitute. = 576Simplify. The volume of the cylinder is 576 ft3. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Find the volume of the cylinder below. Leave your answer in terms of . Quick Check 11-4

  34. You can use three rectangular prisms to find the volume. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Find the volume of the composite space figure. Each prism’s volume can be found using the formula V = Bh. 11-4

  35. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 (continued) Volume of prism I = Bh = (14 • 4) • 25 = 1400 Volume of prism II = Bh = (6 • 4) • 25 = 600 Volume of prism III = Bh = (6 • 4) • 25 = 600 Sum of the volumes = 1400 + 600 + 600 = 2600 The volume of the composite space figure is 2600 cm3. Quick Check 11-4

  36. Volumes of Prisms and Cylinders GEOMETRY LESSON 11-4 Find the volume of each figure to the nearest whole number. 1. 2. 3. 4.5. 62 m3 1800 ft3 45 in.3 63 m3 1800 mm3 11-4

  37. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 (For help, go to Lesson 8-1.) Use the Pythagorean Theorem to find the value of the variable. 1. 2. 3. Check Skills You’ll Need 11-5

  38. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 Solutions 1. a2 + b2 = c2h2 + (9)2 = (15)2 h2 + 81 = 225h2 = 225 – 81 = 144 h = 144 = 12 cm 2. One leg is half the length of the side of the base, 6 in. a2 + b2 = c2h2 + (6)2 = (10)2h2 + 36 = 100 h = 64 = 8 in. 3. a2 + b2 = c2 (2)2 + (1.5)2 = 2 4 + 2.25 = 2 2 = 6.252 = 6.25 = 2.5 m 11-5

  39. 1 3 1 3 V=BhUse the formula for volume of a pyramid. = (225)(22) Substitute 225 for B and 22 for h. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 Find the volume of a square pyramid with base edges 15 cm and height 22 cm. Because the base is a square, B= 15 • 15 = 225. = 1650 Simplify. The volume of the square pyramid is 1650 cm3. Quick Check 11-5

  40. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 Find the volume of a square pyramid with base edges 16 m and slant height 17 m. The altitude of a right square pyramid intersects the base at the center of the square. 11-5

  41. Because each side of the square base is 16 m, the leg of the right triangle along the base is 8 m, as shown below. 172= 82h2Use the Pythagorean Theorem. 289 = 64 h2Simplify. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 (continued) Step 1: Find the height of the pyramid. 225 =h2Subtract 64 from each side. h= 15 Find the square root of each side. 11-5

  42. 1 3 1 3 V=BhUse the formula for the volume of a pyramid. = (16  16)15Substitute. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 (continued) Step 2: Find the volume of the pyramid. = 1280 Simplify. The volume of the square pyramid is 1280 m3. Quick Check 11-5

  43. Find the volume of the cone below in terms of . 1 2 r= d = 3 in. 1 3 V=r 2hUse the formula for volume of a cone. 1 3 = (32)(11) Substitute 3 for r and 11 for h. = 33 Simplify. The volume of the cone is 33 in.3. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 Quick Check 11-5

  44. 1 3 d 2 V= r 2h Use the formula for the volume of a cone. r= = 2 1 3 V= (22)(7)Substitute 2 for r and 7 for h. 29.321531Use a calculator. Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 An ice cream cone is 7 cm tall and 4 cm in diameter. About how much ice cream can fit entirely inside the cone? Find the volume to the nearest whole number. About 29 cm3 of ice cream can fit entirely inside the cone. Quick Check 11-5

  45. 1470 mm3 3 m3 Volumes of Pyramids and Cones GEOMETRY LESSON 11-5 Find the volume of each figure. When appropriate, leave your answer in terms of . 1.2. 3. square pyramid with base edges 24 in. long and slant height 15 in. 4. cone with diameter 3 m and height 4 m 5. 60 ft3 1728 in.3 150 ft3 11-5

  46. Surface Areas and Volumes of Spheres GEOMETRY LESSON 11-6 (For help, go to Lesson 1-9.) Find the area and circumference of a circle with the given radius. Round your answers to the nearest tenth. 1. 6 in. 2. 5 cm 3. 2.5 ft 4. 1.2 m 5. 15 yd 6. 12 mm Check Skills You’ll Need 11-6

  47. 2 r 2 r 2 r 2 r 2 r 2 r Surface Areas and Volumes of Spheres GEOMETRY LESSON 11-6 Solutions 1. A = r 2 = (6)2 = 36 113.1 in.2, C = = 2 (6) = 12 37.7 in. 2. A = r 2 = (5)2 = 25 78.5 cm2, C = = 2 (5) = 10 31.4 cm 3. A = r 2 = (2.5)2 = 6.25 19.6 ft2, C = = 2 (2.5) = 5 15.7 ft 4. A = r 2 = (1.2)2 = 1.44 4.5 m2, C = = 2 (1.2) = 2.4 7.5 m 5. A = r 2 = (15)2 = 225 706.9 yd2, C = = 2 (15) = 30 94.2 yd 6. A = r 2 = (12)2 = 144 452.4 mm2, C = = 2 (12) = 24 75.4 mm 11-6

  48. S.A. = 4 r 2Use the formula for the surface area of a sphere. 18 2 = 4 • 92Substitute r== 9. = 324 Simplify. The surface area of the sphere is 324 ft2. Surface Areas and Volumes of Spheres GEOMETRY LESSON 11-6 Find the surface area of the sphere below. Leave your answer in terms of . Quick Check 11-6

  49. Step 1: Use the circumference to find the radius in terms of . C=2 r Use the formula for circumference. 13=2 r Substitute 13 for C. 13 2 13 2 = r Solve for r. S.A. = 4r 2Use the formula for the surface area of a sphere. 13 2 = 4• ( )2Substitute for r. 169 =Simplify. 53.794371Use a calculator. Surface Areas and Volumes of Spheres GEOMETRY LESSON 11-6 Quick Check The circumference of a rubber ball is 13 cm. Calculate its surface area to the nearest whole number. Step 2: Use the radius to find the surface area. To the nearest whole number, the surface area of the rubber ball is 54 cm2. 11-6

  50. 4 3 V= r 3Use the formula for the volume of a sphere. 30 2 4 3 = • 153Substitute r = = 15. = 4500Simplify. The volume of the sphere is 4500 cm3. Surface Areas and Volumes of Spheres GEOMETRY LESSON 11-6 Find the volume of the sphere. Leave your answer in terms of . Quick Check 11-6

More Related