) = ( 2 ) = ( -2 ) = ( -3 ) = ( 0. Solutions Key Spatial Reasoning. 225 Holt McDougal Geometry ARE YOU READY? PAGE 651

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1 CHAPTER 10 Solutions Key Spatial Reasoning ARE YOU READY? PAGE D. C. A 4. E 5. b = AB = 5-0 = 5; h = - (-1 = 4 bh = (5(4 = 10 units A = 6. b = LM = 6 - (- = 8, h = KL = 7 - = 4 A = bh = (8(4 = units 7. r = PQ = - (-6 = 8 A = π r = π(8 = 64π units 8. C = π(8 = 16π cm A = π(8 = 64π cm 9. C = π(1 = 1π ft A = π(10.5 = 110.5π ft 10. C = π ( π = in. A = π ( 16 π = 56 π in 11. AB = (5 - (- + (6 - = 80 = units M = ( - + 5, + 6 = (, 8 = (1, 4 1. CD = ( - (-4 + (- - (-4 = units M = ( -4 +, -4 + (- = ( - 1. EF = ( (4-1 = 18 = 4. units M = ( 0 + (-, = ( -, 5, -7 = (-1, GH = (- - + (- - (-5 = 5 = 5 units M = ( + (-, -5 + (- 15. r = A π = 11π π 16. a = A P = (18 = 8 ft = ( 0 = 11 = 11 cm = (-1.5,.5, -7 = (0, b = c - a = (17 - (8 = 5 = 15 m 18. b = A h - b 1 = (60-8 = 0-8 = 1 in SOLID GEOMETRY, PAGES CHECK IT OUT! 1a. cone vertex: N edges: none base: circle M b. triangular prism vertices: T, U, V, W, X, Y edges: TU, TV, UV, TW, UX, VY, WX, WY, XY bases: TUV, WXY a. The net has a triangular central face and other triangular faces. So, it forms a triangular pyramid. b. The net has a rectangular face between circular faces. So, it forms a cylinder. a. The cross section is a hexagon. b. The cross section is a triangle. 4. Cut through mdpts. of three edges that meet at one vertex. THINK AND DISCUSS 1. Both prisms and cylinders have congruent parallel bases. The bases of a prism are polygons, and the bases of a cylinder are circles. The bases of a prism are connected by, and the bases of a cylinder are connected by a curved surface.. EXERCISES GUIDED PRACTICE 1. cylinder. cone vertex: A edges: none base: circle B. rect. prism vertices: C, D, E, F, G, H, J, K edges: GH, GK, HJ, JK, GF, HE, JD, KC, FC, CD, DE, EF bases: rect. CDEF, rect. GHJK 5 Holt McDougal Geometry

2 4. triangular pyramid vertices: L, M, N, P edges: LM, LN, LP, MN, MP, NP base: triangle LMP The net has, nonadjacent rect. faces, and remaining faces are. So, the net forms a rectangular prism. 6. The net has 1 circular face and 1 other curved face that is a sector of a circle. So, the net forms a cone The net has 6 square faces and folds up without overlapping. So, the net forms a cube. 8. cross section is a circle 9. cross section is a pentagon 10. cross section is a rectangle cut to bases 1. cut to bases PRACTICE AND PROBLEM SOLVING 1. cube vertices: S, T, U, V, W, X, Y, Z edges: ST, TU, UV, VS, SW, TX, UY, VZ, WX, XY, YZ, ZW bases: STUV, WXYZ 14. rect, pyramid vertices: A, B, C, D, E edges: AB, BC, CD, AD, AE, BE, CE, DE base: ABCD 15. cylinder vertices: none edges: none bases: circle Q, circle R 16. The net has pentagonal faces and 5 faces. So, it forms a pentagonal prism. 17. The net has a central triangular face, surrounded by other triangular faces. So, net forms a triangular pyramid. 18. The net has a rectangular face between circular faces. So, it forms a cylinder. 19. square 0. rectangle 1. rectangle. cut to ground. cut to ground 4 7. Possible answers: 4. cube 5. rectangular prism 6. cylinder 7. hexagonal prism 8. Possible answer: The figure is a hexagonal prism whose bases are regular hexagons with 7-in. sides. Height of the prism is 1 in. 9. Possible answer: The figure is a cylinder whose bases each have radius 1 ft. Height of the cylinder is 9 ft. 0. Possible answer: The figure is a square prism with 6 cm by 6 cm bases and a height of 108 cm. 7a. pentagonal prism c. one 8. B is incorrect; bases are reg. hexagons. So, opposite sides of cross section must be congruent Figure b; when figure is folded, shaded faces will overlap. TEST PREP 41. D 4. F 4. B 44. G CHALLENGE AND EXTEND Holt McDougal Geometry

3 a a. A and B, C and F, D and G, E and H b. b. One SPIRAL REVIEW 5. y = - x 5. y = x Possible answer: y = x 55. largest is opp. longest side: B smallest is opp. shortest side: C 56. largest: E smallest: D 57. largest: I smallest: H 58. no; : 4 8 : yes; 7 : 11.9 = 4 : 40.8 = 5 : 4.5 = 10 : REPRESENTATIONS OF THREE- DIMENSIONAL FIGURES, PAGES no THINK AND DISCUSS 1. All 6 views are squares.. No; vertical lines do not meet at a vanishing point.. CHECK IT OUT! 1.. EXERCISES GUIDED PRACTICE 1. perspective.. 7 Holt McDougal Geometry

4 one-point perspective: two-point perspective: 8. one-point perspective: 10. yes 11. no 1. no 1. no two-point perspective: PRACTICE AND PROBLEM SOLVING Holt McDougal Geometry

5 yes. yes 4. no 5. no one-point perspective: two-point perspective: 9a. b. c. at front + along sides + at rear = one-point perspective: 1. two-point perspective:.. Possible answer: a right cylinder and a right square prism in which the diameter of the cylinder is equal to the side length of the square base of the prism and the heights are the same. 4a. two-point perspective 9 Holt McDougal Geometry

6 b. Extend a pair of lines to meet at each vanishing point. 41a. TEST PREP 5. B 6. H 7. Possible answer: b. Some edges that are on the -dimensional object are not in perspective drawing. If they were extended, they would meet at the vanishing point of drawing. All the edges of prism are also in the isometric drawing. CHALLENGE AND EXTEND 8. c Check students drawings. 40. SPIRAL REVIEW 4. x + y = 0 (1 x - y = 0 ( x - y = 10 ( x + y = 0 + (1 x = 40 x = 0 y = 0 - x = 0-0 = x - y = 7 4x + y = 8 5x = 45 x = 9 (9 - y = 7 y = 46. slope of AB = = - _ 47. slope of AC = = - -1 = 45. y = x + 5 x + y = 5 x + y = x + 10 y = 10 y = 5 x = y - 5 = 5-5 = slope of AD = = - - = pentagons and squares 5 parallelograms triangles 0 Holt McDougal Geometry

7 10- FORMULAS IN THREE DIMENSIONS, PAGES CHECK IT OUT! 1a. V = 6, E = 1, F = =. d = = = 75 = cm. Graph center of base at (0, 0, 0. Since height is 7, graph vertex at (0, 0, 7. Radius is 5, so, base will cross x- axis at (0, 5, 0 and y-axis at (5, 0, 0. Connect vertex to base. b. V = 7, E = 1, F = = 4a. d = ( x - x 1 + ( y - y 1 + ( z - z 1 = (6-0 + (0-9 + (1-5 = = units M ( x 1 + x, y 1 + y, z 1 + z M ( 0 + 6, 9 + 0, M(, 4.5, 8.5 b. d = ( x - x 1 + ( y - y 1 + ( z - z 1 = (1-5 + ( (0-16 = = units M ( x 1 + x, y 1 + y, z 1 + z M ( 5 + 1, , M(8.5, 1, Locations of divers on surface can be represented by ordered triples (18, 9, 0 and (-15, -6, 0. d = ( x - x 1 + ( y - y 1 + ( z - z 1 = ( ( (0-0 = ft THINK AND DISCUSS 1. Find the difference of x-coordinates, the difference of y-coordinates, and the difference of z-coordinates. Square each result, and add. The distance is the square root of the sum.. EXERCISES GUIDED PRACTICE 1. because the bases are circles, which are not polygons. V = 6, E = 9, F = = 4. V = 10, E = 0, F = = 6. 1 = h h = h = = = 5.7 in. 8. Graph the center of base at (0, 0, 0. Since the height is 4, graph the vertex at (0, 0, 4. The radius is 8, so, the base will cross the x-axis at (0, 8, 0 and the y-axis at (8, 0, 0. Connect the vertex to base. 9. Graph the center of the bottom base at (0, 0, 0. Since the height is 4, graph the center of the top base at (0, 0, 4. The radius is, so the bottom base will cross the x- axis at (0,, 0 and the y-axis at (, 0, 0. Draw the top base parallel to the bottom base and connect the bases.. V = 6, E = 10, F = = 5. d = = = 4 = ft 7. d = = = 89 = 17 in. 1 Holt McDougal Geometry

8 10. cube has 8 vertices: (0, 0, 0, (7, 0, 0, (7, 7, 0, (0, 7, 0, (0, 0, 7, (7, 0, 7, (7, 7, 7, (0, 7, d = ( x - x 1 + ( y - y 1 + ( z - z 1 = (5-0 + (9-0 + (10-0 = = units M ( x 1 + x, y 1 + y, z 1 + z M ( 0 + 5, 0 + 9, M (.5, 4.5, 5 1. d = ( x - x 1 + ( y - y 1 + ( z - z 1 = (7-0 + (0 - + (14-8 = = units M ( x 1 + x, y 1 + y, z 1 + z M ( 0 + 7, + 0, M (.5, 1.5, d = ( x - x 1 + ( y - y 1 + ( z - z 1 = (9-4 + (1-6 + (15-10 = = units M ( x 1 + x, y 1 + y, z 1 + z M ( 4 + 9, 6 + 1, M (6.5, 9, Represent the locations of the starting point and the camp by ordered triples (0, 0, 0 and (, 7, 0.6. d = ( x - x 1 + ( y - y 1 + ( z - z 1 = ( (7-0 + (0.6-0 = km PRACTICE AND PROBLEM SOLVING 15. V = 8, E = 1, F = = 17. V = 11, E = 0, F = = = h h = h = = = 8 = 7 5. m 1. Graph the center of the bottom base at (0, 0, 0. Since the height is, graph the center of the top base at (0, 0,. The radius is 5, so, the bottom base will cross the x-axis at (0, 5, 0 and the y-axis at (5, 0, 0. Draw the top base to the bottom base and connect bases.. Graph the center of the base at (0, 0, 0. Since the height is 4, graph the vertex at (0, 0, 4. The radius is, so, the base will cross the x-axis at (0,, 0 and the y-axis at (, 0, 0. Connect the vertex to the base.. prism has 8 vertices: (0, 0, 0, (5, 0, 0, (5, 5, 0, (0, 5, 0, (0, 0,, (5, 0,, (5, 5,, (0, 5,, 16. V = 8, E = 18, F = = 18. d = = = 69 = yd 0. 8 = s + s + s 64 = s 8 = s s = 8 = cm 4. d = (4-0 + (4-0 + (4-0 = = 48 = units M ( 0 + 4, 0 + 4, = M(,, 5. d = (9 - + ( (10-7 = = units M ( + 9, + 10, = M(5.5, 6.5, 8.5 Holt McDougal Geometry

9 6. d = (8 - + (8-5 + (10 - = = units M ( + 8, 5 + 8, + 10 = M(5, 6.5, Let the cloud and the raindrop hitting ground have the coordinates (0, 0, 6500 and (-700, 500, 0. The distance traveled by the raindrop is d = ( ( ( = 4,990, ft. 8. original: d = = = 169 = 1 ft doubled: d = = = 676 = 6 ft The diagonal is doubled. 9. V - E + F = F = -4 + F = F = 6 0. V - E + F = V = V - 4 = V = 6 9. bottom base extends forward to (5,, 5 center of top base: (1,, 10 Draw the top base to bottom base. 40. base extends forward to (6,, 6 vertex: (,, vertices of cube: (4,,, (10,,, (10, 8,, (4, 8,, (4,, 9, (10,, 9, (10, 8, 9, (4, 8, 9 1. V - E + F = 7 - E + 7 = 14 - E = E = 1. V = n + n = n E = n + n + n = n F = 1 + n + 1 = n + V - E + F = n - n + (n + =. V = n + 1 E = n + n = n F = n + 1 V - E + F = (n n + (n + 1 = 4a. K(0,.5, 6, L(0, 5, 0, N(7,.5, 6, P(7, 5, base extends forward to (8, 7, 1 b. KP = (7-0 + ( (0-6 = ft 5. d = = 108 = 6 6. w = = 49 = 7 7. h = = 108 = 6 8. l = = Holt McDougal Geometry

10 44. bottom base extends forward to (7,, 7 Draw the top base parallel to bottom base d = (8-0 + ( (6-0 = units M ( 0 + 8, 0 +, = M (4, 1.5, d = ( ( - + (1 - = 8 =.8 units M ( 1 +, +, + 1 = M (,, d = ( ( (6-8 = units M ( 6 +, 1 +, = M (4, 1.5, 7 d = (7-4 + (4 - + (4 - = 11. units M ( 4 + 7, + 4, + 4 = M (5.5,.5, d = ( (1-7 + (5-8 = units M ( 4 +, 7 + 1, = M (.5, 4, 6.5 d = ( - + (6-8 + ( - 5 = 9 = units M ( +, 8 + 6, 5 + = M(.5, 7, = ( ( - (-1 + (z - (- 1 = ( ( - (-1 + (z - (- 169 = z + 6z = z + 6z = (z + 15(z - 9 z = 9 or Possible answer: 4 Holt McDougal Geometry

11 5. Possible answer: d in. 54. Possible answer: a segment that connects a vertex of one base to the opposite vertex of the other base AB = (1-0 + (0-0 + (0-0 = 1 unit AC = (1-0 + ( (0-0 = 5. units AG = (1-0 + ( ( - 0 = units 55. AB = (5 - + (8 - + (6 - (- = 11 = 11 AC = (- - + ( ( - (- = 11 = 11 BC = ( ( ( - 6 = 4 = 11 AB = AC and A B + AC = BC ABC is a right isoceles cm; the segment is the hypotenuse of a right in which one leg is a diameter of one base, and the opposite vertex is on the other base. Legs measure 6 cm and (4 = 8 cm, so, segment length is 10 cm. Segment is longest because a diameter is longest possible segment in a circle. TEST PREP 57. C 58. H d = = m 59. B d = (9-7 + ( (1-8 = 189 = 11.9 units CHALLENGE AND EXTEND 60. legs of rt. measure h and a + a + a = a d = (a + h = 4 a + h 61. AB = (1 - (-1 + (- - + (6-4 = 4 = 6 AC = ( - (-1 + ( (8-4 = 96 = 4 6 BC = ( (-6 - (- + (8-6 = 4 = 6 AB + BC = AC. So, points are collinear. 6. AB = ( x - x 1 + ( y - y 1 + ( z - z 1 AM = ( x 1 + x - x 1 + ( y 1 + y - y 1 + ( z 1 + z - z 1 = ( x - x 1 + ( y - y 1 + ( z - z 1, and MB = ( x - x 1 + x + ( y - y 1 + y + ( z - z 1 + z = ( x - x 1 + ( y - y 1 + ( z - z 1 So AM = MB. Also, AM + MB = ( x - x 1 + ( y - y 1 + ( z - z 1 _ 4 ( x - x 1 + _ 4 ( y - y 1 + _ = 4 ( z - z 1 = ( x - x 1 + ( y - y 1 + ( z - z 1 = AB So A, M, and B are collinear. Since M is on AB and AM = MB, M is the midpoint of AB by def. midpt. 6. AG = (a (b (c - 0 = a + b + c BH = (0 - a + (b (c - 0 = a + b + c AG = BH, so AG BH. By the definition of segs. Let M and N be mdpts. of AG and BH. M = ( 0 + a, 0 + b, 0 + c = ( a, b, c N = ( 0 + a, 0 + b, 0 + c = ( a, b, c M = N. These segs. have the same midpoint, so they bisect each other. SPIRAL REVIEW = yr old 66. A = b(h = bh 67. A = h ( 68. A = π(r = 9π r 69. cone 70. none 71. C 10A READY TO GO ON? PAGE 679 b 1 + b 1. hexagonal pyramid vertices: A, B, C, D, E, F, G edges: AB, AC, AD, AE, AF, AG, BC, CD, DE, EF, FG, BG base: hexagon BCDEFG. cone vertices: H edges: none base: circle J 5 Holt McDougal Geometry

12 . rectangular prism vertices: K, L, M, N, P, Q, R, S edges: KL, LM, MN, NK, PQ, QR, RS, PS, KP, LQ, MR, NS ; bases: KLMN, PQRS 4. cube 5. cone 6. pentagonal pyramid 7. square 8. rectangle 9. circle Let the coordinates of the nest and the bird be (0, 0, 0 and (6, - 7, 6. d = (6-0 + ( (6-0 = 11 = 11 ft 18. d = (4-0 + (6-0 + (1-0 = 196 = 14 units M ( 0 + 4, 0 + 6, = M(,, d = (5 - + ( (7 - (- = 11 = 11 units M ( + 5, 1 + (-5, = M(4, -,.5 0. d = (7 - + ( (0-9 = units M ( + 7, 5 +, = M(5,.5, SURFACE AREA OF PRISMS AND CYLINDERS, PAGES V = 8, E = 1, F = 6 V - E + F = = 15. V = 7, E = 1, F = 7 V - E + F = = 16. V = 4, E = 6, F = 4 V - E + F = = CHECK IT OUT! 1. L = Ph = (4s(s = 4(8(8 = 56 cm S = L + B = L + s = 56 + (8 = 84 cm. Step 1 Use the base area to find the radius. A = π r 49π = π r 49 = r r = 7 in. Step Use the radius to find the lateral area and the base area. height is times radius, or 14 in. L = πrh = π(7(14 = 196π in S = πrh + π r = 196π + (49π = 94π in. Surface area of right rectangular prism is S = Ph + B = 6(5 + (9(4 = 0 cm. A right cylinder is added to a rectangular prism. Lateral area of cylinder is L = πrh = π(( = 1π cm. The base area of the cylinder is not added, just raised through cm. The surface area of the composite figure is the sum of the areas of all surfaces on the exterior of figure. S = (prism surface area + (cylinder lateral area = 0 + 1π 9.7 cm 6 Holt McDougal Geometry

13 4. original dimensions: S = πrh + π r = π(11(14 + π(11 = 550π cm height, diameter multiplied by : S = πrh + π r = π(5.5(7 + π(5.5 = 17.5π cm Notice 17.5π = (550π. If the height and 4 diameter are multiplied by multiplied by 4., the surface area is 5. The 5 cm by 5 cm by 1 cm prism has a surface area of S = Ph + B = 0(1 + (5(5 = 70 cm, which is greater than the cm by cm by 4 cm prism and about the same as the half cylinder. It will melt faster than cm by cm by 4 cm prism and at about same rate as the half cylinder. THINK AND DISCUSS 1. Use the radius of the base to find the base area, then add twice the base area to the lateral area.. An oblique prism has at least one lateral face that is not a rectangle. All the lateral faces of a right prism are rectangles.. EXERCISES GUIDED PRACTICE 1. 5 lateral faces. L = Ph S = L + B = 4( = 7 ft = 7 + (5(7 = 14 ft. The base is a rt., because, 4, 5 is a Pythag. triple. L = Ph = 1( = 4 cm S = L + B = 4 + ( ((4 = 6 cm 4. L = Ph = (4s(s = 4(9(9 = 4 in S = L + B = L + s = 4 + (9 = 486 in 5. L = πrh = π((4 = 4π ft S = πrh + π r = 4π + π( = 4π ft 6. L = πrh = π(7.5(1 = 180π yd S = πrh + π r = 180π + π(7.5 = 9.5π yd 7. Step 1 Use the base area to find the radius. A = π r 64π = π r 64 = r r = 8 m Step Use the radius to find the lateral area and base area. The height is less than the radius, or 5 m. L = πrh = π(8(5 = 80π m S = πrh + π r = 80π + π(8 = 08π m 8. surface area of the right rectangular prism is S = Ph + B = 44(1 + (14(8 = 75 ft. A right cylinder is added to right rectangular prism. The lateral area of the cylinder is L = πrh = π(4(8 = 64π ft. The base area of the cylinder is not added, just raised through 8 ft. The surface area of the composite figure is the sum of the areas of all the surfaces on the exterior of the figure. S = (prism surface area + (cylinder lateral area = π 95.1 ft 9. surface area of cylinder is S = πrh + π r = π(14(14 + π(14 = 784π ft A right rectangular prism is removed from the cylinder. The lateral area is L = Ph = 40(14 = 560 ft. The base area is B = 14(6 = 84 ft. The surface area of the composite figure is the sum of the areas of all the surfaces on the exterior of the figure. S = (cylinder surface area + (prism rectangle area - (prism base area = 784π (84 = 784π ft 10. original dimensions: S = πrh + π r = π(8(4 + π(8 = 19π yd dimensions multiplied by : S = πrh + π r = π(4( + π(4 = 48π yd Notice 48π = 4 multiplied by 11. original dimensions: S = Ph + B = (6 + (8(8 = 0 yd (19π. If the dimensions are, the surface area is multiplied by dimensions multiplied by _ : S = Ph + B = 64 (4 + ( 16 ( 16 = 180 yd 9 Notice 180 = 4_ (0. If the dimensions are multiplied 9 9 by, the surface area is multiplied by in. bulb: L = πrh = π(1(16 = π in -in. bulb: L = πrh = π/4( = 4.5π in -in. bulb will produce more light PRACTICE AND PROBLEM SOLVING 1. L = Ph S = L + B = 0(10 = 00 cm = 00 + (5(5 = 50 cm 4. 7 Holt McDougal Geometry

14 14. L = Ph = 6(1(15 = 1080 m 15. L = Ph = (8(14 = 6 ft apothem of base is 6 m S = L + B (6 = m altitude of base is 4 ft S = L + B = ( = 6 + ( (8(4 = ft (7 16. L = πrh = π(5.5(7 = 77π in S = πrh + π r = 77π + π(5.5 = 17.5π in 17. L = πrh = π(4( = 184π cm S = πrh + π r = 184π + π(4 = 16π cm 18. Step 1 Use the base circumference to find the radius. C = πr 16π = πr r = 8 yd Step Use the radius to find the lateral area and the base area. The height is times the radius, or 4 yd. L = πrh = π(8(4 = 84π yd S = πrh + π r = 84π + π(8 = 51π yd 19. The base of the right triangular prism is a rt., because 6, 8, 10 is a Pythagorean triple. The surface area of the right triangular prism is S = Ph + B = 4(9 + ( (6(8 = 64 cm. A right cylinder is removed from the triangular prism. The lateral area of cylinder is L = πrh = π((9 = 6π cm. The base area of cylinder is B = π r = π( = 4π cm. The surface area of the composite figure is the sum of areas of all surfaces on the exterior of the figure. S = (prism surface area + (cylinder lateral area - (cylinder base area = π - (4π = π 5.0 cm 0. The surface area of right rectangular prism is S = Ph + B = 8(0.5 + (( = 1 ft. A right cylinder is added to the rectangular prism. The lateral area of cylinder is L = πrh = π(0.5( = π ft. The base area of cylinder is not added, just lowered through ft. The surface area of the composite figure is the sum of the areas of all surfaces on the exterior of figure. S = (prism surface area + (cylinder lateral area = 1 + π 18. ft 1. original: S = πrh + π r = π(4.5(11 + π (4.5 = 19.5π ft dimensions tripled: S = πrh + π r = π(1.5( + π (1.5 = 155.5π ft 155.5π = 9(19.5π. So, surface area is multiplied by 9.. original: S = Ph + B = 4( + (1(9 = 4 ft dimensions doubled: S = Ph + B = 84(6 + (4(18 = 168 ft 168 = 4(4. So, surface area is multiplied by 4.. left cell: S = Ph + B = 90(7 + (5(10 = 10 μm right cell: S = Ph + B = 5(15 + (15(11 = 1110 μm The cell that measures 5 μm by 7 μm by 10 μm should absorb at a greater rate. 4. S = πrh + π r 160π = π(5h + π(5 160π = 10πh + 50π 110π = 10πh h = 11 ft 5. S = Ph + B 86 = 6h + (10(8 86 = 6h = 6h h =.5 m 6. L = Ph 168 = 6(1h 168 = 7h h = 19 m 7. The bases are rt. with leg lengths and 5 units, and hypotenuse length + 5 = 9 units; height is 9 units. S = Ph + B = ( (9 + ( ((5 = = units 8. S = πrh + π r = π(9.55( π( mm 9. S = πrh + π r = π(10.605( π( mm 0. S = πrh + π r = π(8.955(1.5 + π( mm 1. S = πrh + π r = π(1.1( π( mm 8 Holt McDougal Geometry

15 . Possible answer: triple edge lengths Let l, w, and h represent the original dimensions. original: S = Ph + B = (l + wh + lw = lh + wh + lw tripled: S = Ph + B = ((l + (w(h + (l(w = 18lh + 18wh + 18lw = 9(lh + wh + lw. Possible answer: Multiply the radius and the height by. original: S = πrh + π r halved: S = π ( r ( h + π ( r = πrh + π r = ( πrh + π r 4 4. the triangular prism shaped frame S = (area of sides + (area of ends = (10(10 + ( (105 = ft half cylinder: S = (area of curved panel + (area of ends = (10 ( π(10 + ( π(5 = 50π + 5π = 75π 5.6 ft The triangular-prism-shaped frame requires more plastic. 5. area with given measurements: S = Ph + B = 1(6 + (( = 90 cm least possible area: S = 4lw + w = 4(5.95(.95 + (.95(.95 = cm greatest possible area: S = 4lw + w = 4(6.05(.05 + (.05(.05 = cm max. error below 90 cm is =.85 cm max. error above 90 cm is =.415 cm max. error <.415 cm 6. Find the area of each part of the net and add the areas. 7a. AB = 8-1 = 7 in. BC = = = 4 in. 5.7 in. b. AC + BC = AB AC + (4 = 7 AC + = 49 AC = 17 AC = 17 in. 4.1 in. c. S = Ph + B = 16 (4.1 + ( in TEST PREP 8. A S = (area of rectangle + (area of circles = (.0(8.1 + π( cm 9. F L = Ph = 4(5 = 10 in S = (.14rh + (.14 r = (.14(6(5 + (.14( in CHALLENGE AND EXTEND 41. 1st cylinder: S = πrh + π r = π(8( + π(8 = 176π cm nd cylinder: S = πrh + π r 176π = π(4h + π(4 176π = π(4h + π 144π = 8πh h = 18 cm 4. S = (area of sides + (area of ends = (1(1 + ( ( (1 + 18(9 = = 88 ft amount of paint = gal of paint are needed, costing 4($5 = $ L = Ph 144 = (l + wh 144 = ((w + w(w 144 = 16 w 9 = w w = cm l = w = ( = 9 cm, h = w = ( = 6 cm S = L + B = (9( = 198 cm SPIRAL REVIEW m s B C = AB + AC - (AB(ACcos A a = (7(8cos 45 a.804 a sin B AC = sin A BC sin B sin m B si n -1 8 sin 45 ( Holt McDougal Geometry

16 SURFACE AREA OF PYRAMIDS AND CONES, PAGES CHECK IT OUT! 1. Step 1 Find the base perimeter and the altitude. The base perimeter is (6 = 18 ft. The altitude is ft, so, the base area is bh = (6( = 9. ft Step Find the lateral area. L = Pl = (18(10 = 90 ft Step Find the surface area. S = Pl + B = ft. Step 1 Use the Pythag. Thm. to find l. l = = 10 cm Step Find the lateral area and the surface area. L = πrl = π(8(10 = 80π cm S = πrl + π r = 80π + π(8 = 144π cm. original: S = Pl + B = (60(1 + (15 = 585 ft base edge length and slant height multiplied by /: S = Pl + B = (40(8 + (10 = 60 ft Notice that 60 = 4_ (585. If the base edge length 9 and the slant height are multiplied by _ area is multiplied by 4_ 9., the surface 4. The base area of the cube is B = ( = 4 yd. The lateral area of the cube is L = Ph = (8( = 16 yd. By the Pythag. Thm., l = 1 + = 5 yd. The lateral area of the pyramid is L = Pl = (8 5 = 4 5 yd. A = (base area of cube + (lateral area of cube + (lateral area of pyramid = = yd 5. The radius of the large circle used to create the pattern is the slant height of the cone. The area of the pattern is the lateral area of the cone. The area of the pattern is also _ the large circle. So, πrl = _ 4 πl. πr(1 = _ 4 π(1 r = 9 in. 4 of the area of THINK AND DISCUSS 1. The lateral faces are all with an area of the base edge length times the slant height, the perimeter P is the sum of the base edge lengths. So, L = Pl.. A radius and the axis of a right cone form the legs of a right. Length of the hypotenuse is the slant height of the cone. The hypotenuse of a right is always the longest side, so slant height is greater than the height.. EXERCISES GUIDED PRACTICE 1. vertex and center of base. Step 1 Find the base perimeter and the apothem. The base perimeter is 6(8 = 48 cm. The apothem is 4 cm. So, base area is ap = (4 (48 = 96 cm. Step Find the lateral area. L = Pl = (48(1 = 88 cm Step Find the surface area. S = Pl + B = cm. Step 1 Find the base perimeter and the area. The base perimeter is 4(16 = 64 ft. The base area is (16 = 56 ft. Step Find the lateral area. The slant height is = 17 ft. L = Pl (64(17 = 544 ft Step Find the surface area. S = Pl + B = = 800 ft 4. Step 1 Find the base perimeter and the altitude. The base perimeter is (15 = 45 in. The altitude of base is 7.5 in. So, base area is bh = (15(7.5 = 56.5 in. Step Find the lateral area. L = Pl = = (45(0 = 450 in Step Find the surface area. S = Pl + B = in 40 Holt McDougal Geometry

17 5. Step 1 Use the Pythogorean Theorem to find the base radius r. r = 5-4 = 7 in. Step Find the lateral area and the surface area. L = πrl = π(7(5 = 175π in S = πrl + π r = 175π + π(7 = 4π in 6. L = πrl = π(14( = 08π m S = πrl + π r = 08π + π(14 = 504π m 7. Step 1 Use the base area to find the base radius r. A = 6π = π r 6 = r r = 6 ft Step Find the lateral area and the surface area. L = πrl = π(6(8 = 48π ft S = πrl + π r = 48π + π(6 = 84π ft 8. original: S = Pl + B = (4(10 + (6 = 156 in dimensions halved: S = Pl + B = (1(5 + ( = 9 in Notice that 9 = (156. If the dimensions are cut in 4 half, the surface area is multiplied by 9. original: S = πrl + π r = π(9(15 + π(9 = 16 cm dimensions tripled: S = πrl + π r = π(7(45 + π(7 = 1944 cm Notice that 1944 = 9(16. If the dimensions are tripled, the surface area is multiplied by The lateral area of the upper pyramid is L = Ph = ((15 = 40 ft. The lateral area of the upper pyramid is L = Ph = ((18 = 88 ft. S = (lateral area of upper pyramid + (lateral area of lower pyramid = = 58 ft 11. The lateral area of upper cone is L = πrl = π(1(6 = 1π m. The lateral area of cylinder is L = πrh = π(1(15 = 60π m. The lateral area of upper cone is L = πrl = π(1( = 84π m. S = (lateral area of upper cone + (lateral area of cylinder + (lateral area of lower cone = 1π + 60π + 84π = 1056π m The radius of the large circle used to create the hat is the slant height of the cone. The area of the hat is the lateral area of the cone. The area of the hat is also _ of the area of the large circle. So, πrl = _ 4 πl. πr(6 = _ 4 π(6 r = 4.5 d = (4.5 = 9 in. So, the hat will be too large. PRACTICE AND PROBLEM SOLVING 1. Step 1 Find the base perimeter and the area. The base perimeter is 4(6 = 4 ft. The base area is (6 = 6 ft. Step Find the lateral area. The slant height is + 4 = 5 ft. L = Pl (4(5 = 60 ft Step Find the surface area. S = Pl + B = = 96 ft 14. Step 1 Find the base perimeter and the altitude. The base perimeter is (40 = 10 cm. The altitude of base is 0 ft. So, the base area is bh = (40(0 = 400 cm. = 4 Step Find the lateral area. The slant height is 5-0 = 15 cm. L = Pl = (10(15 = 900 cm Step Find the surface area. S = Pl + B = cm 15. Step 1 Find the base perimeter and the apothem. The base perimeter is 6(7 = 4 ft. The apothem is.5 ft. So, the base area is ap = (.5 (4 = 7.5 ft. Step Find the lateral area. L = Pl = (4(15 = 15 ft Step Find the surface area. S = Pl + B = ft 16. L = πrl = π(11.5( = 64.5π cm S = πrl + π r = 64.5π + π(11.5 = 96.75π cm 17. l = = 7 in. L = πrl = π(1(7 = 444π in S = πrl + π r = 444π + π(1 = 588π in 18. h = (8-1 = 15 m; l = = 17 m L = πrl = π(8(17 = 16π m S = πrl + π r = 16π + π(8 = 00π m 41 Holt McDougal Geometry

18 19. original: S = Pl + ap = (4(1 + ( (4 = ( ft dimensions divided by : S = Pl + ap = (8(4 + ( (8 = ( ft ( = ( So, the surface area is divided by original: S = πrl + π r = π((5 + π( = 14π m dimensions doubled: S = πrl + π r = π(4(10 + π(4 = 56π m 56π = 4(14π. So, the surface area is multiplied by lateral area of left cone = π(7(4 = 168π in lateral area of right cone = π(7(17 = 119π in S = (lateral area of left cone + (lateral area of right cone = 168π + 119π = 87π in. lateral area of left pyramid = Pl = (6(15 = 70 cm lateral area of cube = Ph = (6(9 = 4 cm lateral area of right pyramid = Pl = (6(19 = 4 cm S = (lateral area of left pyramid + (lateral area of cube + (lateral area of right pyramid = = 96 cm. S = πrl = πl πr(6 = π(6 r = d = ( = 6 in. 4. B = s = 6. So, s = 6 cm and P = 4 cm S = Pl + B = (4(5 + 6 = 96 cm 5. B = as = ( s s = s 4 4 = s s = m P = ( = 6 m S = Pl + B = (6( + = 4 m 6. B = π r = 16π. So, r = 4 in. S = π(4(7 + π(4 = 44π in 7. B = π r = π, so r = 1 ft S = π(1( + π(1 = π ft 8a. l = = 116 = 9 cm S = Pl + s = (( 9 + (8 = cm b. S = πrl + π r = π(4.5l + π( π = 4.5πl π l = 1. cm 4.5π 9. S = πrl + π r π = πr(1 + π r = 1r + r 0 = r + 1r - 0 = (r - 8(r + 9 r = 8 m (since r is positive 0. P = 4s =. So, s = 8 ft S = Pl + s 56 = (l + (8 56 = 16l = 16l l = 1 ft 1. L = Pl P(10 = 5P P = 4 cm 10 =. r + h = l r + 7 = 5 r = 5-7 = 4 units S = πrl + π r = π(4(5 + π(4 = 1176π units. left pyramid: L left = Pl = (0(14 = 10 cm right pyramid: L right = Pl = (0(8 = 10 cm S = L left + L right = = 0 cm 4. left cone: l = = 10 m S = πrl + π r = π(6(10 + π(6 = 96π m right cone: l = = 6 m S = πrl + π r = π(10(6 + π(10 = 60π m S = 96π + 60π = 456π m 5. s = 00( ft = 600 ft and h = (10 = 0 ft. So, l = = 19, ft Pl (400( ,000 ft L = 6. A triangle is formed with vertices at the midpoints of opposite sides of the square base and the third vertex at the vertex of the pyramid. The side lengths of the triangle are l, l, and s, the edge length of the base. By the Triangle Inequality Theorem, l + l >s, so l>s. Therefore, l > s. 7. In an oblique cone, the distance from a point on edge of the base to the vertex is not the same for each point on the edge of the base. 4 Holt McDougal Geometry

19 TEST PREP 8. D A = ( t F tl = t 16 + tl (II = t 8 + l (III ( t L = Pl 16 = (4s(18 4 = 7s s = 6 cm S = L + B = 16 + (6 = 5 cm 40. B l = = 41 cm L = πrl = π(9(41 = 69π cm CHALLENGE AND EXTEND 41a. l 10 = l l = 10(l - 0 5l = 10l = 5l l = 40 cm S = πrl + π r = π(10(40 + π(10 = 500π cm b. l = 40-0 = 0 cm L = πrl = π(5(0 = 100π cm c. B = π(5 = 5π cm d. S = (surface area of cone + (area of top base - (lateral area of top of cone = 500π + 5π - 100π = 45π cm 4. L = 4 ( ( b 1 + b l = ( b 1 + b l 4a. c = πr b. C = πl c. c C = πr πl = r d. A = πl l L = c C (A = πl πrl SPIRAL REVIEW, PAGE 696 ( r l = 44. Since the area of a circle depends on the square of its radius, the surface area of a cone cannot be described by a linear function. 45. Since perimeter is 1-dimensional, the perimeter of a rectangle can be described by a linear function. 46. Since the area of a circle depends on the square of its radius, it cannot be described by a linear function. 47. area of ACEF = 4 = 16 cm area of BDG = (4( = 4 cm P = 4 16 = area of circle H = π( = 4π cm P = 4π 16 = π area of shaded region = (16-4π cm P = 16-4π = 4 - π S = L + B = ( (10 + ( (8(15 = = 50 in 51. S = L + B = ((8 + (10(15 + ((8(10 = = 700 cm 5. S = πrh + π r = π(( + π( = 0π 6.8 cm 10-6 VOLUME OF PRISMS AND CYLINDERS, PAGES CHECK IT OUT! 1. V = Bh = ( (5(7 (9 = yd. Step 1 Find the volume of the aquarium in cubic feet. V = lwh = (10(60(16 = 115,00 ft 1 gal Step Use the conversion factor to estimate the volume in gallons ft 115,00 ft 1 gal 859,70 gal 0.14 ft Step Use the conversion factor 8. lb to estimate the weight of water. 1 gal 859,70 gal 8. lb 7,161,18 lb 1 gal. V = π r h = π(8 (17 = 1088π in in 4. original dimensions: V = lwh = (4((1.5 = 18 ft length, width, and height doubled: V = lwh = (8(6( = 144 ft Notice that 144 = 8(18. If the length, width, and height are doubled, the volume is multiplied by The volume of cylinder is V = π r h = π( (5 = 45π cm. The volume of prism is V = Bh = ( (5 = 90 cm. The net volume of figure is difference of volumes. V = 45π cm THINK AND DISCUSS 1. In both formulas, the volume equals the base area times the height. The base area of a cylinder = π r, and the base area of a prism is given by the area formula for that polygon type.. An oblique prism has the same cross-sectional area at every level as a right prism with the same base area and height. By Cavalieri s Principle, the oblique prism and the right prism have the same volume. 4 Holt McDougal Geometry

20 . EXERCISES GUIDED PRACTICE 1. the same length as. V = lwh = (9(4(6 = 16 cm. V = Bh = 4. V = s ap h = = (8 = 51 ft ( (6(8 = m 5. Step 1 Find the volume of the ice cream cake in cubic feet. V = lwh = (19(9( = 4 ft 1 gal Step Use the conversion factor to estimate the volume in gallons ft 4 ft 1 gal 55 gal 0.14 ft Step Use the conversion factor 8. lb to estimate the weight of water. 1 gal 55 gal 4.7 lb 1,071 lb 1 gal 6. V = π r h = π(6 (10 = 60π ft ft 7. V = π r h = π( (5 = 45π m m 8. B = π r 5π = π r 5 = r r = 5 cm h = 5 + = 8 cm V = π r h = π(5 (8 = 00π cm 68. cm 9. original dimensions: V = lwh = (1(4(8 = 84 ft dimensions multiplied by 4 : V = lwh = ((1( = 6 ft Notice 6 = 1 (84. If the dimensions are multiplied 64 by, the volume is multiplied by original dimensions: V = π r h = π( (7 = 8π in dimensions tripled: V = π r h = π(6 (1 = 756π in Notice 756π = 7(8π. If the dimensions are tripled, the volume is multiplied by The volume of rect. prism is V = lwh = (1(14(6 = 1008 ft. The volume of cylinder is V = π r h = π(4 (4 = 64π ft. The total volume of the figure is the sum of its volumes. V = π ft 1. The volume of outside cylinder is V = π r h = π(10 (15 = 1500π in. The volume of inside cylinder is V = π r h = π(5 (15 = 75π in. The net volume of the figure is the difference of the volumes. V = 1500π - 75π = 115π 54. in PRACTICE AND PROBLEM SOLVING 1. V = Bh = ( (9(15 (1 = 810 yd 14. V = Bh = _ ap h = (50( m tan 6 ( B = s 49 = s s = 7 ft h = 7 - = 5 ft V = Bh = (49(5 = 45 ft 16. Step 1 Find the volume in cubic feet. V = lwh = (9(16 ( = 48 ft Step Use the conversion factor 1 yd to find the 7 ft volume in cubic yards. Then round up to nearest cubic yard. V = 48 1 yd 1.78 yd 7 ft Colin must buy yd of dirt. Step Use the cost per yard to find the cost of dirt. Cost = ($5 = $ V = π r h = π(14 (9 = 1764π cm cm 18. V = π r h = π(6 ( = 108π in 9. in 19. V = Bh = 4π(16 = 84π cm cm 0. original dimensions: V = π r h = π( ( = 1π yd dimensions multiplied by 5: V = π r h = π(10 (15 = 1500π yd 1500π = 15(1π. So, the volume is multiplied by Holt McDougal Geometry

21 1. original dimensions: V = Bh = (5 (10 = 50π m dimensions multiplied by _ : 5 V = π r h = π( (6 = 54π m 54π = 7 15 by (50π. So, the volume is multiplied. volume of lower cube: V = s = (8 = 51 cm volume of middle cube: V = s = (6 = 16 cm volume of upper cube: V = s = (4 = 64 cm total volume: V = = 79 cm. volume of square-based prism: V = Bh = (4 (1 = 19 ft volume of each half-cylinder: V = π r h = π( (4 = 8π ft total volume: V = 8π π = π 4. ft 4. radius r = in.: V = π r h = π( h = 4πh h = in. 4π radius r = 1.5 in.: V = π r h = π(1.5 h =.5πh h = in..5π 5a. V = π r h = π(5 ( = 75π 5.6 in b. V = lwh = ((1( = 9 in P = 9 75π a. ( h = h = 5 h = 1 h = in. b. V = π r h = π(1.5 ( 1.5 in 0.55 oz c..5 in 7. V = lwh 495 = (5(9h = 45h h = 11 ft. 1 in 17.9 oz 8. V = Bh 60 = B(9 B = 40 in 9. S = πrh + π r 10π = πr(8 + π r 10 = 16r + r 105 = 8r + r 0 = r + 8r = (r - 7(r + 15 r = 7 m V = π r h = π(7 (8 = 9π m 0. l = 7 units, w = units, h = 6 units V = lwh = (7((6 = 16 units 1. V = lwh = (1( ( = 6 ft or V = (1(4( = 576 in. V = Bh = (4 ( 4 = 4 in = 4 ( 1 in 4 servings. Step 1 Find the volume in cubic feet. V = π r h = π(45 (5 = 105,00π ft Step Convert the volume to gallons. 105,00π ft 1 gal,468,79 gal 0.14 ft 4. Step 1 Find the volume in cubic feet. V = lwh = (50(100 ( 4 = 150 ft Step Convert the volume to gallons. 150 ft 1 gal 98 gal 0.14 ft Step Convert to weight in pounds. 98 gal 8. lb 77,705 lb 1 gal 5. For a scale factor of k, the ratio of surface area to volume of the new prism is _ times the k ratio of the surface area to the volume of the old prism. A cube with edge length 1 has a surface area-to-volume ratio of 6:1. If each edge length is multiplied by, the surface area-to-volume ratio is 4:8 = :1. 6. V new = V old s new = ( s old s new = ( s old Multiply edge length by. TEST PREP 7. A V candle = π r h = π(.4 (6.0 = 69.6π cm V wax = lwh = (15(1(18 = 40 cm # candles π 14.9 # candles = F wire length = (# edges(edge length 96 = 1s s = 8 in. V = s = (8 = 51 in 9. B V prism = Bh = ( (9 = 81 in V cylinder = π r h = π(1.75 ( in 40. H upper rect. prism: V = (10-4(10(4 = 40 cm lower rect. prism: V = (10(10(10-4 = 600 cm total: V = = 800 cm 45 Holt McDougal Geometry

22 CHALLENGE AND EXTEND 41. V = lwh = (x + (x - 1(x = x + x - x 4. V = π r h = π(x + 1 (x = π x + π x + πx 4. B = _ ( (x x = x 4 V = Bh = ( x (x + 1 = x + x The volume is equal to the surface area, so π r h = π r + πrh. Solve for r to get r = h h -. If h <, then r < 0, so h must be greater than. Similarly, if you solve for h, you get h = r r -. If r <, then h < 0, so r must be greater than. SPIRAL REVIEW 45. m = r -100 (1 t = r + 40 ( r > m ( substitute (1 in (: r > r > r (4 substitute (4 in (: t = r + 40 < 140 t 19 So t = typing time is 5000 words = 15 min = ( min 40 wpm So, he takes 15-min breaks. Total time is 15 + (15 = 155 min or h 5 min. 47. ABC CDA m ABC = m CDA 0x - 10 = x + 4 7x = 14 x = m ABC = 0( - 10 = BC AD 49. AB DC BC = AD AB = DC z + 11 = _ z = y + 6 = 4y z 6 = y y = AB = ( + 6 = 8 BC = S = L + B z = 16 ( = 15 = Pl + s = ((10 + (8 = 4 in 51. S = L + B = _ Pl + _ ap = _ (0(8 + ( tan 6 (0 = cm tan 6 5. C = π = πr r = 0.5 ft S = πrh + π r = π(0.5( + π(0.5 = 1.5π ft.9 ft 10-7 VOLUME OF PYRAMIDS AND CONES, PAGES CHECK IT OUT! 1. Step 1 Find the area of the base. B = ap = ( (1 = 6 cm Step Use the base and the height to find the volume. The height is equal to the base. V = Bh = (6 (6 = 6 cm. First find the volume in cubic yards. V = Bh = (70 (66 = 107,800 yd Then, convert the answer to find the volume in cubic feet. Use the conversion factor 7 ft 1 yd in cubic feet. 107,800 y d 7 ft =,910,600 f t 1 yd. V = π r h = π(9 (8 = 16π m 4. original dimensions: to find the volume V = π r h = π(9 (18 = 486π cm radius and height doubled: V = π r h = π(18 (6 = 888π cm Notice that 888π = 8(486π. If the radius and height are doubled, the volume is multiplied by The volume of the rectangular prism is V = lwh = (5(1(15 = 4500 ft. The volume of the rectangular pyramid is V = Bh = ((5(1(15 = 1500 ft. The volume of the composite figure is the difference of the volumes. V = = 000 ft 46 Holt McDougal Geometry

23 THINK AND DISCUSS 1. The volume of a pyramid is the volume of a prism with the same base and height.. EXERCISES GUIDED PRACTICE 1. perpendicular. Step 1 Find the area of base. B = lw = (6(4 = 4 in. Step Use the base and the height to find the volume. V = Bh = (4(17 = 16 in. Step 1 Find the area of base. B = ap = ( (4 = 4 cm Step Use the base and the height to find the volume. V = Bh = 4. V = Bh = (4 (4 = 96 cm (5(9 = 75 ft 5. V = ( Bh = ( (5. 7 ( 65 mm 6. V = π r h = π(9 (14 = 78π cm cm 7. V = π r h = π(1 (0 = 1440π in 45.9 in 8. V = π r h = π(1 (0 = 960π m m 9. original dimensions: V = π r h = π(5 ( = 5π cm dimensions tripled: V = π r h = π(15 (9 = 675π cm Notice that 675π = 7(5π. If the dimensions are tripled, the volume is multiplied by original dimensions: V = Bh = ( 9 (15 = 405 cm dimensions multiplied by V = Bh : = ( 4.5 (7.5 = cm Notice that 405 = multiplied by ( If the dimensions are 8, the volume is multiplied by The volume of the cube is V = s = (1 = 178 cm. The volume of the rectangular pyramid is V = Bh = (1 (18 = 864 cm. The volume of the composite figure is the sum of the volumes. V = = 59 cm 1. The volume of the outer cone is V = π r h = π(8 (1 = 56π in. The volume of the inner cone is V = π r h = π(4 (6 = π in. The volume of the composite figure is the difference of the volumes. V = 56π - π = 4π in 70.7 in PRACTICE AND PROBLEM SOLVING 1. Step 1 Find the area of the base. B = lw = (8(6 = 48 ft Step Use the base and height to find the volume. V = Bh = (48(10 = 160 ft 14. Step 1 Find the area of the base = 1. So, the base is a right triangle with b = 1 m and h = 5 m. B = bh = (1(5 = 0 m Step Use the base and height to find the volume. V = Bh = (0(9 = 90 m 47 Holt McDougal Geometry

24 15. Step 1 Find the height. 6 + h = 10 h = 8 ft Step Use the height and base edge length to find the volume. V = Bh = (1 (8 = 84 ft 16. Step 1 Find the volume in cubic feet. The height is 5( = 15 ft. V = Bh = (45 (15 = 10,15 ft Step Convert the volume to cubic yards. Use the conversion factor 1 yd 7 ft. 10,15 ft 1 yd 75 yd 7 ft 17. V = π r h = π(9 (41 = 1107π m m 18. V = π r h = π( (4 = 16 π in 16.8 in 19. B = π r 6π = π r r = 6 ft h = r = (6 = 1 ft V = Bh (6π(1 = 144π ft 45.4 ft 0. original dimensions: = V = π r h = π(15 (1 = 1575π in dimensions multiplied by V = π r h = π(5 (7 = 175 π in Notice that 175 π = 1 (1575π. If the dimensions 7 are multiplied by, the volume is multiplied by 1 1. original dimensions: V = Bh = : ( 7 (4 = 196 dimensions multiplied by 6: V = Bh = ft 7. ( 4 (4 = 14,11 cm Notice that 14,11 = 16 ( 196. If the dimensions are multiplied by 6, the volume is multiplied by 16.. The volume of the cylinder is V = π r h = π(6 (10 = 60π ft. The volume of the cone is V = π r h = π(6 (10 = 10π ft. The volume of the composite figure is the difference of the volumes. V = 60π - 10π = 40π ft ft. The volume of the rectangular prism is V = lwh = (10(5( = 100 ft. The volume of each square-based pyramid is V = Bh = ( 5 ( = 5 ft. The volume of the composite figure is the sum of the volumes. V = = 150 ft 4. V = π r h = π( (7 = 1π in 5. r = d = 5_ m V = π r h = π ( 5 π m 6 6. r + h = l 8 + h = 5 V = 7. r = h = 45 ft π r h = d = r + h = l 1 + h = 1 V = V = ( = 5 π(8 (45 = 11,760π ft (6 = 1 cm h = 5 cm π r h = π(1 (5 = 40π ft 8. B = b h base = (10(5 = 5 ft Bh = (5 ( ft 9. V = Bh = (15 (18 = 150 m 0. Step 1 Find the apothem of the base. tan 6 = 4.5 a a = 4.5 tan 6 in. Step Find the area of the base. B = _ ap = 4.5 (45 = ( tan 6 tan 6 in Step Find the volume of the pyramid. V = _ Bh = _ ( ( in tan 6 1. B = ap = (4 (48 = 96 cm V = Bh = (96 ( 166. cm. V = Bh V = lwh 11 = ((8h 11 = 8h h = 14 m 48 Holt McDougal Geometry

25 . V = π r h 15π = π r (5 15π = 5_ π r 75 = r r = 5 cm C = πr = π (5 = 10π cm 4. r + h = l r + 8 = 10 r = 6 ft V = π r h = π(6 (8 = 96π ft 5. S = Pl + B 800 = (4s(17 + s 800 = 4s + s 0 = s + 4s = (s - 16(s + 50 s = 16 in. ( s + h = l 8 + h = 17 V = h = 15 in. Bh = (16 (15 = 180 in 6. V = π r h 1500π = π r (0 4500π = 0π r 5 = r r = 15 yd r + h = l = l l = 5 yd S = πrl + π r = π(15(5 + π(15 = 600π yd 7. The base is a right triangle with a base length of 5 units and a height of units. The height of the pyramid is 7 units. B = (5( = 7.5 units V = Bh = (7.5(7 = 17.5 units 8. A is incorrect because it uses the slant height of the cone instead of the height. 9. : ; The base areas are the same for both figures The volume of the prism is By, and the volume of the figure formed by pyramids is _ B(y. The ratio of the volumes is By : _ B(y, which is equivalent to :. 40. Possible answer: Substitute the given values for r and S into the surface area formula and solve for l. Then, use the Pythagorean Theorem and the values for r and l to solve for h. Substitute the values for r and h into the volume formula. 41a. V = π r h = b. V = π r h = π( (8 = π.5 in π(4 (8 = 18 π 14.0 in c. The large size holds 4 times as much. So, the price should be 4($1.5 = $5. TEST PREP 4. A 1 + h = H V = h = 9 cm π r h = L = Pl 50 = (4s(5 50 = 50s s = 7 m. 5 + h = 5 V = Bh π(1 (9 = 4π cm h = ( 7 ( m 44. B V = π r h = π( (6 = 18π in V = Bh 4 = ( h h 79 = h h = 9 cm CHALLENGE AND EXTEND 46. radius = apothem of regular triangle 1 s = 1. So, r = V = π r h = ( π = ft ( = _ 47. r = s = 1 ft V = π r h = π(1 ( = _ π ft 9 π ft 48. radius = apothem of regular hexagon s =. So, r = ft V = π r h = π ( ( = π ft 49. slant height is altitude of a face; s = 10. So, l = 5 cm. ( s + h = l 5 + h = (5 h = 50 = 5 cm V = ( Bh = (10 ( = cm 50. h = 9 in.; the volume of a cone with the same base and height as the cylinder is the volume of the cylinder. For a cone to have the same volume as the cylinder, the height of cone must be times the height of the cylinder. 49 Holt McDougal Geometry

26 SPIRAL REVIEW 51. x - y = 4 (1 x = y - 4 ( Substitute ( into (1: y y = 4 y = 8 y = 14 x = (14-4 = 8 The numbers are 8 and x + y = 197 (1 x = y + 0 ( 5. x + y = 88 (1 10x = 4y ( Solve ( for y and substitute into (1: y = 10x 4 = 5x x + 5x = 88 6x + 5x = x = 176 x = 16 y = 10(16 = 40 4 The numbers are 16 and 40. Substitute ( into (1: y y = 197 y y = 94 y = 54 y = 118 x = ( = 79 The numbers are 79 and DF AC = = ; EF BC = = ; C F triangles are by SAS AB 10 = AB = (10 = 0 AB = LM and PQ are both perpendicular to LN. So, LM parallel PQ. By Corr. Post., M NQP; by Reflex. Prop. of, N N. So are by AA. PQ LM = PN LN PQ 9 = PQ = 9(16 = 144 PQ = AB = (8-1 + (9-1 + (10 - = = M ( 1 + 8, 1 + 9, + 10 = M(4.5, 5, AB = (5 - (-4 + (1 - (-1 + (-4-0 = = M ( , , = M(0.5, 0, AB = (- - + ( - (- + (-4-4 = = M ( + (-, - +, 4 + (-4 = M(0, 0, AB = (-1 - (- + ( 5 - (-1 + ( 5 - = = 49 = (-1 M (, , + 5 = M(-,, SPHERES, PAGES CHECK IT OUT! 1. V = 4_ π r 04π = 4_ π r 178 = r r = 1 ft. hummingbird eyeball: V = 4_ π r = 4_ π( π c m human eyeball: V = 4_ π r = 4_ π(0. = 0.06π c m A human eyeball is about.604π 7. times as 0.06π great in volume as a hummingbird eyeball.. S = 4π r = 4π(5 = 500π cm 4. original dimensions: S = 4π r = 4π( = 6π m Notice that 4π = 9 by radius multiplied by : S = 4π r = 4π(1 = 4π m (6π. If the radius is multiplied, the surface area is divided by Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the surface area of the hemisphere, the lateral area of cylinder, and the area of the lower base of cylinder. S hemisphere = (4π r = π( = 18π ft L cylinder = πrh = π((5 = 0π ft B cylinder = π r = π( = 9π ft The surface area of composite figure is 18π + 0π + 9π = 57π ft. Step Find the volume of the composite figure. The volume of composite figure is difference of volume of cylinder and volume of hemisphere. V cylinder = π r h = π( (5 = 45π ft V hemisphere = ( 4_ π r = _ π( = 18π ft The volume of composite figure is 45π - 18π = 7π ft. THINK AND DISCUSS 1. The surface area is 4 times the area of the great circle.. Both the area of sphere and the area of the composite figure have a volume of 4_ π r. 50 Holt McDougal Geometry

27 . EXERCISES GUIDED PRACTICE 1. One endpoint is the center of the sphere, and the other is a point on the sphere.. V = ( 4_ π r = _. V = 4_ π r = 4_ π(1 = 4_ 4. V = 4_ π r 88π = 4_ π r 16 = r r = 6 cm 5. grapefruit: V = lime: V = ( 4_ π r = _ ( 4_ π r = _ π ( 5 π(11 = 66 π in π m π(5 = 50 π cm = 5 π cm 1 The grapefruit is 8 times as great in volume as the lime. 6. S = 4π r = 4π(8 = 56π yd 7. A = π r S = 4π r = 4A = 4(49π = 196π cm 8. S = 4π r 674π = 4π r 1681 = r r = 41 ft V = 4_ π r = 4_ π(41 = 75,684 π ft 9. original dimensions: S = 4π r = 4π(15 = 900π in dimensions doubled: S = 4π r = 4π(0 = 600π in Notice that 600π = 4(900π. If dimensions are doubled, the surface area is multiplied by original dimensions: dimensions multiplied by V = 4_ π r = 4_ V = 4_ π r = 4_ π(8 = π(8 048 π cm = π cm Notice that π = 1 64( 048 π. If the radius is multiplied by, the volume is multiplied by : 11. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the surface area of two hemispheres and the lateral area of cylinder. S hemisphere = (4π r = π( = 8π ft L cylinder = πrh = π((5 = 0π ft The surface area of the composite figure is (8π + 0π = 6π ft. Step Find the volume of the composite figure. The volume of the composite figure is the sum of the volume of the cylinder and the volume of the two hemispheres. V cylinder = π r h = π( (5 = 0π ft V hemisphere = ( 4_ π r = π( = 16 π ft The volume of the composite figure is 0π + ( 16 π = 9 π ft. 1. Step 1 Find the surface area of the composite figure. The surface area of the composite figure is the sum of the lateral area of the hemisphere and the surface area of the cylinder, less the area of the base of the hemisphere. S hemisphere = (4π r = π( = 8π in S cylinder = πrh + π r = π(8( + π(8 = 176π in B hemisphere = π r = π( = 4π in The surface area of the composite figure is 8π + 176π - 4π = 180π in. Step Find the volume of the composite figure. The volume of the composite figure is the difference of the volume of the cylinder and the volume of the hemisphere. V cylinder = π r h = π(8 ( = 19π ft V hemisphere = ( 4_ π r = π( = 16 π in The volume of the composite figure is 19π - 16 π = 560 π in. PRACTICE AND PROBLEM SOLVING 1. V = 4_ π r = 4_ π(9 = 97π cm 14. V = ( 4_ π r = _ 15. V = 4_ π r 7776π = 4_ π r π(7 = 686 π ft 58 = r r = 18 d = 6 in mm pearl: V = 4_ π r = 4_ π(4.5 = 11.5π mm 6-mm pearl: V = 4_ π r = 4_ π( = 108π mm The 9-mm pearl is.75 times as great in volume. 17. S = 4π r = 4π(1 = 1764π in 18. A = π r S = 4π r = 4A = 4(81π = 4π in 51 Holt McDougal Geometry