Chapter 9. b: 196:1 c: 9:1

Transcription

1 9.1.1: Chapter a: Solutions vary. Here is a possibility: b: Answers vary. Assuming there are no hidden cubes, V = 11!un a: 4 25 b: 196:1 c: 9: Since the perimeter is 100, each side is 10. The central angle is = 36. The right triangle has acute angles 18 and 72. Area = 769.4!un (36! 9" ) + 2 # 3.86!un a: (4,!2) b: (!6,!4) c: ( 1 2,12) A 9.1.2: a:! square mm b: 400% is 4 times as large. Therefore its area increases by a factor of 4 2 ; 987!16 " 15,778.6 square mm a: b: 42!un a: 151 b: Yes; a regular 36-gon has interior angles of 170. c: 27!180 = 4, a: 2x + 4x! 3 + 7x! 6 + 3x x + 10 = 540,!x = 31 b: 4x x! 2 = 180,!x = a: 2 2 units b: (!1,6) c: (8,5) d: (2,!3) D

2 9.1.3: square units SA = 574!cm 2 ;V = 882!cm Yes, by AAS! a: 6 or!6 b: no solution because absolute value cannot be negative c: x = 3or! !(6) 2 (14.5) = 522!!in 3 522!!in3 ;! " 1!gallon 1 231!in # 7.1!gal B 9.1.4: If she needs the balloon to double in width, then the volume will increase by a factor of 8. That means the balloon requires 24 breaths to blow it up. Since she has already used 3 breaths, she needs 21 more to fill the balloon a: SA = 180! " 565.5!in 2 ;!V = 324! " !in 3 b: 324! " 27 = 27, !in circumference of each circle = 10! ; total distance 20! " 62.8 feet a: x! 10.3 b: no solution because the hypotenuse must be the longest side of a right triangle c: the length of the base of the composite triangle must be 6 3. The smaller right triangle has a base length 6 3! 5 " 5.39;!x " The line should be solid and the shading should be below the line D

3 9.1.5: a: 2 b: 24 and 96; ratio = 4 ; It is the square of the linear scale factor. c: 6 and 48; ratio = 8 = 2 3. It is the cube of the linear scale factor height of the tank = 6 3! 10.4 in, so V = 7!13! 6 3 = " 945.7!in a: False (isosceles trapezoid) b: True c: True d: False (parallelogram) (12x + 7) = 30x! 4, so x = a:!2 b: 5, 1 2 c: 2 d:!2, D 9.2.1: a: a rhombus b: They are perpendicular and intersect each other at their midpoints. c: 32 3! 55.4!un a: b: 120!cm 3 c: 158!cm x! 5 = 2x + 9 ;!x = a: Not congruent since r! 1 b:!abc "!DBC (HL! ) c:!abc "!MLK (SSS! ) y = 6x! B

4 9.2.2: b: It got longer. c: Although the change is minute, the line segment got a little longer. d: The line segment will continue to lengthen but will never be exactly Area of whole circle= 16!!un 2 ; Area for dog to roam = ! (16! ) = " 33.51!un height 9 3, so A = (20+14)(9 3) 2 = 153 3! 265!un DA! CB and m!dab! m!cba (given), AB! BA (reflexive property), so!dab "!CBA (SAS ). Then!C!!D (! "s #! parts ) equilateral triangle: 360 3!sides = 120 ; regular heptagon: (7!2)"180 7 # 128.6, so the measure of the interior angle of a regular heptagon is greater C 9.2.3: a: b: 24 cubic units c: The volume of the new solid must be 1 the original, so the reduced volume 8 must be 3 cubic units !ABC "!EDF (given), so!b!!d (! "s #! parts ), so DG! BG (if base angles of a triangle are congruent, then the two sides opposite those angles are also congruent) and!dbg is isosceles (definition of an isosceles triangle) a:! 74.2!m 2 b:! 36.2 meters Using equations: n + q = 14 and 0.05n q = 2.90, so n = 3and q = It has 8 sides B

5 9.2.4: a: (25! )(21) = 525! " 1,649.3!un 3 b: BA! 19.3, sov! (19.3)(7)! 135.2!un She should construct an arc centered at P with radius k so that it intersects n and m each once (call these intersections R and S). She should then construct two more circles with radius k, centered at R and S. The fourth vertex lies where these two circles intersect ,000!180 = 180, ! square miles a: b: C

6 10.1.1: Chapter V = (6)(5)(8)! "(2 2 )(8) # 139.5!in 3 SA = ! 2(2 2 )" + 4"(8) # 311.4!in a: 70 b: 50 c: 2x She is constructing an angle bisector height = 6 3, area= (15)(6 3) (144! ) = ! " 231.3!un2, perimeter = (24! ) = ! " 66.6!un They intersect only once at (3,5) A : a: 64 b: 128 c: 64 d: 180 e: 128 f: central angle = 3.6, A! square units a: 5m + 1 = 3m + 9,!m = 4 b: 2(x + 4 ) = 3x! 9,!x = 17 c: (p! 2) = p 2,! p = 10 d: 18t = 360,!t = a: D(0, 4) and E(4, 7) b: DE = 5 units, so AC should be 10 units long. c: = 10 units b: 108 b: 72 c: D

7 10.1.3: a: 3 b: 6 c: 2 d: 1 e: 4 f: a: They have the same measure. b: CD!, it is a fraction of a circle with a larger diameter. (28! ) " 14.7 c: OY! KY! EY! PY (all radii are congruent),!pyo!!eyk (arc measures are equal), so!poy "!EKY (SAS ). Therefore, PO! EK because! "s #! parts It must be a rhombus a: yes, (AA~),!ABC ~!LKH b: not enough information c: yes, (AA~),!ABC ~!EDC C : a: 50 b: 50 c: 67 d: 126 e: 54 f: a: x! 31.9,!y! 10.5 b: x! 3.7 c: x! 34.7,!y = 250! a: 4 times b: 360 c: = area of the regular pentagon! 61.9!ft 2 ; total SA! 2(61.9) + (5)(6)(12)! ft 2 volume! (61.9)(12)! 743.3!ft a: It is a square. Students should demonstrate that each side is the same length and that two adjacent sides are perpendicular (slopes are opposite reciprocals). b: C! is at (!5,!8) and D!! is at (!7, 4) C

8 10.1.5: MA = = 39, MB = 6, so AB = ER and ER = 1485! 38.5 feet ! 1 6 = (9)(12) = 1 (20)h,!h = 5.4" The ratio of the volumes must be 1 5 ( ) 3 1, so the volume must be ( ) 3 (500! ) = 4!!cm a: A! sq. cm b: A! 74.21sq. cm c: A! sq. cm E : a: 6 ;! 1 2 b: 5 (60) = 50 times 6 c: Answers vary, but two squares should have an even number, no numbers can be 3, and all numbers must be less than a: 13 b: 6.5 c:! 67.4 d:! a: Have Ken s analyzed for $30, and use the ratios of similarity to calculate the data for Erica s nugget themselves. b: Since r = 5, the ratio of the areas is 25. Thus, 110! 25 = 2750!cm 2. c: Since r = 5, the ratio of the volumes is 5 3 = 125. Thus, Erica s nugget weighs 125! 56 = 7000!g, which is about 14 pounds! x + 3x + 4x + 5x = 360,!x! If a point makes the inequality true, it must be in the shaded region of the graph. a: true b: false c: false d: true B

9 10.2.2: Region A is 1 4 of the circle, so it should result 1 (80) = 20 times. Regions B and 4 C have equal weight (which can be confirmed with arc measures), so they should each result (80! 20) 2 = 30 times a: = 40 b: mad! = 2(97 ) = 194, m!c = 0.5(194 ) = 97 c: mab! = 125 and the length of AB! = 125 (16! ) " 17.5"; area = 125 (64! ) " 69.8!in Methods vary, but a variety of relationships could be used, such as Parallel Line Angle Conjectures, the Exterior Angle Theorem, or the Triangle Sum Theorem; x = 109,!y = 71,!z = a: 34 b: 4 3 c:!5 d: She is incorrect, which can be tested by substituting both answers into the equation; w =!6 or A

10 10.2.3: a: 19 4 b: 12 c:! Note: Equivalent equations can vary, depending on what constant both sides of the equation are multiplied by. a: one possible equivalent equation: 8x! 3 = 6x,!x = 1.5 b: one possible equivalent equation: 7x + 4 = 110,!x = 106 7! a: The top & bottom views are the same. Same is true for the right & left views and the front & back views. b: V = 7!un 3, SA = 28!un 2 ; Methods vary. c: Since this solid has no holes, the surface area can be calculated by adding the areas of each of the views b: AB AC c: BC AB BC d: AC e: AB AC BC f: AC a: If x represents the length of chord AC, then x 2 = ! 2(10)(10)cos80 ; x! 12.9 b: C : The expected value is 1 4 ($3) + 3 ($1) = $1.50 per spin, so each player should pay 4 $1.50 so that there is no net gain or loss over many games a: x = y b: y = 2x or x = 1 y c: 3y = 5x d: x + y = a: yes, because of the Triangle Angle Sum Theorem, 180! 64! 26 = 90 b: yes, because = V = (16)(16)(16) = 4096!un 3 ; SA = 6(16)(16) = 1536!un a: not similar because there are not three pairs of corresponding angles that are congruent b: similar (AA ~) c: similar (SSS ~) d: similar (AA ~) B

11 Chapter : a: Yes; the sum of the interior angles only depends on the number of sides of the polygon. b: No; only regular hexagons have a guarantee of having an angle with the measure 120. c: Yes; that is the definition of hexagon lateral surface area = circumference of the base times the height = 8! "15 = 120! # 377!cm Equations may vary. a: 2x = 180! 106,!x = 37 b: x + 67 = 180,!x = 113 c: sin 73 = sin57 = sin50,!x! 7.9,!y! x y d: 4x! 2 + 2(8x! 9) = 180,!x = a: A = 144 square units, P = 84 units b: A = 16 square units, P = 28 units a: a sphere b: a cylinder with a cone on top and bottom c: a double-cone: two cones attached at the vertices B a: 4 b: 6 c: (10)(12)h = 840, so h = = 7 mm (x! 4) 2 + (y! 2) 2 = a: x 2 + y 2 = r 2 b: sin! = y r ;!y c: cos! = x r ;!x If the circle s center is C and if the midpoint of AB is D, then!adc is a triangle. Then the radius, AC is 10 units long and the area of the circle is 100! " units C

12 11.1.2: The expected value per throw is 1 4 (2) (3) (5) = 15 = 3.75, so her expected 4 winnings over 3 games are 3(3.75) = ; yes, she should win enough tickets to get the panda bear The areas are all equal because the triangles have the same base and height PA = PA (Reflexive Property), m!pba = m!pca = 90 (tangents are to radii drawn to the point of tangency), PB = PC (radii of a circle must be equal), so!pab "!PAC (HL). Therefore AB = AC (! "s #! parts ) a: x = 12.9 b: x = 2 c: x = 0 d: x = 23 9! a: 10 b: tan! = 8,!! " D

13 11.1.3: a: Yes. One way is to cut off a corner so the cross-section is a triangle. b: a tetrahedron (also called a triangle-based pyramid) V = 1 3 (62 )(4) = 48 cubic units; slant height is = 5 ; SA = 4( 1 2! 6! 5) + 62 = 96 square units a: 8 faces, 12 edges, and 6 vertices b: a square a: x = 12,!y = 7.5 b: 4 3 c: 48 square units m!ecb = m!ead (given), AE! CE (definition of midpoint),!dea!!bec (vertical angles are congruent),!aed "!CEB (AAS ), so AD! CB (! "s #! parts ) D See graph at right. x-intercepts: (4, 0) and ( 4, 0), y-intercepts: (0, 2) and (0, 8) a:!(1.5) 2 (4.5) " 31.8!in 3 b: Volume of the pot is!(7) 2 (10) " !in 3. Therefore, Aimee would need = 48.4 or 49 cans of soup to fill the pot. c: 2!(1.5)(4.5) " 42.4 in A = 16! square units, C = 8! units Yes, the game is fair because the expected value is 2 6 (10) (!5) = a: 18x = 540,!x = b: x2 + (x + 17) 2 = 25 2,!x = E c: 2(x + 4 ) = 134,!x = 63 d: x x + 15 = 180,!x = 39

14 11.1.4: a: BA = 1 2 (7)(24) = 84, V = (84)(12) = 1008!un3, SA = (2)(84) + (12)( ) = 840 un b: BA = 25!, V = 1 3 (25! )(12) = 100!!un3, lateral SA =!(5)(13) = 65!!un 2, SA = 25! + 65! = 90! " 282.7!un a: vertically through the vertex of the cone b: a circle c: a sphere a:!abc ~!FED (AA~) b:!abc ~!MKL (SSS~) c: not similar because the zoom factors for corresponding sides are not equal The graph should include a circle with radius 5, center (0, 0), and a line with slope 1 and y-intercept (0, 1); (3, 4) and ( 4, 3) a: corresponding angles,! b: alternate interior angles,! c: straight angle, supplementary d: alternate exterior angles, neither because lines intersect A : r = 4 cm, SA = 4!(4 2 ) = 64! " 201.1!cm 2, V = 4 3!(4)3 = 256! " cm a: An icosahedron has 20 faces, so the surface area is (20)(45) = 900!mm 2. b: Since it has 12 faces, = 9!cm 2. c: The area of each face is 1 2 (6)(3 3) = 9 3! 15.6!in2 so total ! 29 SA = 4(15.6)! 62.4!in ! square units a: ( 4,10) b: (!1,9) B

15 11.2.1: a: Earth s circumference: 8000! " 25,132.7 miles. Therefore the distance to the moon is 238, ,132.7! 9.5 times greater. b: cos89.85 = 238,900,!x! 91,253,182.4 miles x C! 102.6, so mab!! 32 (102.6)! 9.1 mm a: 1 3 (92 )(12) = 324!cm 3 b: 12!cm a: 4 b: 1 c:! Answers vary. Sample responses: the rings in a tree, the ripples created when a stone is tossed in a pond, the rings of a dartboard, etc D : radius of slice = 3! " in., x! 2.52 " ! 2.3 inches The surface area of the moon! 4"(1080) 2! 14,657, which makes it larger than Africa and smaller than Asia Central angle = 36, apothem = un., A = (307.77)(10)! sq. un V = 324! 12 = 312!cm !ART ~!PIT by AA ~ D

16 11.2.3: a: x = 270 b: x = 132,!y = 15.7 c: 3(x + 2) = 6x,!x = a: a = 44,!b = 28,!c = 56 b: c is smaller. But some students may notice that the vertical angles (72 ) are each the average of the two arcs they intercept a:! 436,000 b: The sun s radius is almost double the distance between the Earth and the moon. That means that if the sun were placed next to the Earth, its center would be farther away than the moon! c: 1,295, x 2 + y 2 = base area= 36 un 2, slant height = 109! units, lateral SA = ! 125.3!un 2, total SA! ! 161.3!un C a: x = 117,!y = 88! b: r =! 11.8,!z = 310 sin25 c: 9(9 + a) = 8(21),!a = V(prism)(34)(84)(99) = 282,744!un 3, V(cylinder)!(38) 2 (71) " 322,088.6!un 3, so the cylinder has more volume x = 2(62 ),!x = (3) (!1) (10) = 11 6 = $ The solution is shown with dashed lines in the diagram D

17 12.1.1: Chapter Since 2!r = 40 feet, then r = 20 " 6.4 feet; SA! 4"(20! " )2! 509.3square feet; V = 4 3!(20! )3 " !ft a: 6 b: 4 or!4 c: 4 d: 30 e: 3 or!3 f: 3 or! Answers vary. Typical cross-sections: regular hexagon, circle, rectangle, etc a: Yes; the graph includes the circle and all of the points inside the circle. b: No; the graph only includes the points outside the circle. The circle itself would be dashed B : V = 820( 1 2 )3 = 102.5!cm a: x = 33,!y = 120 b: a! 36.9,!b = 4 c: z = 12,!w = 5 d: x = FG = 3.5!cm, BC = 14!cm base radius = 14 in; V = 1 (196! )(18) = 1176! " !in B

18 12.1.3: a: V = 1 3!(32 )(10) = 30! " 94.2!un 3 b: One possible method: BA = (21)(18)! (12)(12) = 234!un 2, V = (234)(10) = 2340!un a: x 2 + y 2 = 9 b: a: x = (x + 2) 2 ;!x = 15 b: tan!1 ( 8 ) " 28.1,!180! 90! 28.1 " 61.9 ; Sample tools: trigonometry and 15 the sum of the angles of a triangle a: He should lose $1.50 on average per flip b: He is expected to lose $ a: 124 b: 25! un 2 c:! C, by SAS! : a: BA! 77.25!un 2, LA = (8)(4)(16) = 512!un 2, total SA! 2(77.25) + 512! !un 2 b: slant height = 12 units, LA = 4( 1 2 )(10)(12) = 240!un2, BA = (10)(10) = 100!un 2, total SA = = 340!un a: u =!1 b: x = 5,!! 8 3 c: k = 1 or 9 d: p =!4.5 or a: Since the hypotenuse is 1, sin! = y 1, and y = sin!. Also cos! = x, so x = cos!. 1 b: It must be 1 because of the Pythagorean Theorem. c: Yes, this appears to be true for all angles a: b is larger, even though we are not told that b is a central angle b: the missing angle = 180! 62! 70 = 48, and since the angle opposite side a is bigger, a must be larger than b c: a = 9 3! 15.6!un 2 and b = 16 un 2, so b is larger than a a: 4; 6; 5.25 b: when y =!4, x = 0 or 3; when y = 0, x =!1or A

19 12.1.5: a: The slant height of the cone is! 9.22 m, LA(cone)! 6"(9.22)! !m 2, and LA(cylinder) = 12!(11) = 132! " !m 2, so total surface are is! = !m 2 b: V(cylinder) 36!(11) = 396! " !m 3 and V(cone) = 1 3 (36! )(7) = 84! " !m3, so total volume is! = !m a: E(1, 3) and F(7, 3) ; AB = 9, DC = 3, EF = 6 ; EF seems to be the average of AB and CD. b: Yes; EF = 4, while AB = 6 and CD = 2 c: Sample response: The midsegment of a trapezoid is parallel to the bases and has a length that is the average of the lengths of the bases a: sin27 = x sin102,!x! 8.17 b: = sin62,!x! x c: tan x = 6,!x! a: Vertical angles are equal, 2x + 9 = 4x! 2,!x = 5.5 b: The sum of the angles of a quadrilateral is 360, (3x + 8) (2x! 1) = 360, x = 41 c: When lines are parallel, same-side exterior angles are supplementary, so 7x! 3 + 4x + 12 = 180 and x = B

20 12.2.1: ; One method: The base angles of!prs must add up to 40 so that the sum of all three angles is 180. Then add the 40 and 35 of!qps and!qrs, respectively, and the sum of the base angles of!prs must be 115. Thus, m!q must be 180! 115 = A! 1,459,379.5 square feet a: No; the triangle is equilateral, so all angles must be 60 b: yes; = region B, since area! 17.46!un 2 and perimeter! 24.3 units a: This has one solution, because 4( 23! 3) = 11 4 b: This has no real solution because x 2 must be positive or zero c: This has two solutions because x = ±6 d: This has no solution because the absolute value must be positive or zero C : a: a = 120, b = 108, so a is greater b: not enough information is given since we do not know that the lines are parallel c: third side is approximately 8.9 units, so b is opposite the greater side and must be greater than a d: a is three more than b, so a must be greater e: a = 7 tan 23! 2.97 and b = 2 cos 49! 3.05, so b is greater than a x = 8 and y = a: y-intercept: (0,6), x-intercepts: (3,0) and (!1,0) b: (1,8) c: f (100) =!19,594 and f (!15) =! ! square units SA = 76!un 2 ; V = 40!un B

21 12.2.3: Side length = 4 so height of triangle is 2 3. Thus the y-coordinate of point C could be 2 ± 2 3 ; (5, 7.46)or(5,!1.46) length of diagonal = 4! " 6.47, height of shaded triangle = (4!) 2 " 2 2 # units, Area of triangle! 4(6.16) 2! 12.31units Jamila s product does not equal zero. She cannot assume that if the product of two quantities is three, then one of the quantities must be 3; Correct solution: x =!6 or C : a: 60 sq. un. b: side length! un, so area! (7.282)(7)! sq. un a: 5 b: 6! 5 6 = " 69.4% c: 5 (49! ) " square cm. 6 d: 1! 1 4! 2 3 = 1 12, 1 (360 ) = a: It was 80 feet above ground because y = 80 when x = 0 b:!16(3) (3) + 80 = 128 feet,!16( 1 2 )2 + 64( 1 ) + 80 = 108 feet 2 c:!16x x + 80 = 0 ; x = 5 seconds They all are triangles, so they are all similar to each other. In addition, the triangles in (a) and (c) are congruent because corresponding sides have equal length.

22 a: b: 11 cubic units c: The volume of the new solid must be 5 3 = 125 times the original, so the increased volume must be 11(5 3 ) = 1375 cubic units Students can add a height from C to AB, draw a segment from C that bisects!c, or draw a median from C to AB. Either way, they can prove the small triangles are congruent (HL! if the height is drawn, SAS! if the angle bisector is used, and SSS! if the median is used). Then!A!!B (! "s #! parts ) base length! units, perimeter! units, area! square units a: m!pcq = 46, tan 46 = 5 5, CP = CQ! 4.83, sin 46 =, CR! 6.95, so CP CR x! 6.95 " 4.85! 2.12 units b: k 2 = (8)(18) = 144,!k = 12 c: x 2 = ! 2(7)(7)cos102,!x = a: It is a prism with dimensions 2x3x4 units b: SA = 52 un 2, V = 24!un 3 c: SA = 52(3 2 ) = 468!un 2, V = 24(3 3 ) = 648!un The angles, from smallest to largest, measure 64, 90, 116, 130, and 140, so the probability is Answers vary. Possible cross-sections: square, rectangle, triangle, line, point B