Chapter : 5.1.2:

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1 5.1.1: Chapter x! 7.50 and y! 8.04 units; Sine or cosine could be used to get the first leg, then any one of the trig ratios or the Pythagorean Theorem to get the other a: False (a rhombus and square are counterexamples) b: True c: False (it does not mention that the lines must be parallel, so a counterexample with two non-parallel lines cut by a transversal can be drawn.) 5-9. a: (0.8)(0.8) = 0.64 = 64% b: (0.8)(0.2) = 0.16 = 16% area = 74 sq. un, perimeter = units a: x =!3 b: m = 10 c: p =!4 or 2 3 d: x = : a: sin22 = x 17,!!x! 6.37 b: tan 49 = 7,!!x! 6.09 x c: cos60 = x 6,!!x = ! feet a: a possible area model: b: 1 4 c: = 25 36! 69% 5-19: a: False (a triangle is a counterexample) b: False (this is only true for rectangles and parallelograms) c: True 5-20.!ABC ~!EFD by SAS ~

2 5.1.3: a: x = ±5 b: all numbers c: x = 2 d: no solution 5-27.! 11.5 seconds a: 3 8 b: 1 8 c: 3 8 d: 1 8 The sum must be equal to one area! square units, perimeter! units minutes 5.1.4: All of the triangles are similar. They are all equilateral triangles Since tan(33.7 )! 2 3,!y = 2 3 x a: sin! = b a b: tan! = a b c: cos! = a b a: 20 = b:, Since the sum of the probabilities of finding the ring and not finding the 5 ring is 1, you can subtract 1! 1 5 = 4 5. c: No, his probability is still 4 20 = 1 because the ratio of the shaded region to the 5 whole sandbox is unchanged a: cos23 = 18 x or = 18 x b: Since 67 is complementary to 23, then sin67 = cos23. So sin 67!

3 5.2.1: a: A = 1 square units, P = units b: A = ! square units, P = ! units a: y = 111,!x = 53 b: y = 79,!x = 47 c: y = 83,!x = 53 d: y = 3,!x = 3 2 units a: 4 2 units; students can use the Pythagorean Theorem or can use the fact that it is a triangle. b: It is a trapezoid; 24 square units a: Answers vary. Sample responses: x < 3, x is even, etc. b: the length of each leg is 6 units a; not similar b: SAS ~ c: SSS ~ d: SSS ~ or SAS ~ 5.2.2: a: 16 units b: 4 units and 4 2 units c: 24 units d: 10 and 10 3 units a: m!a = 35,!m!B = 35,!m!ACB = 110, m!d = 35, m!e = 35, m!dce = 110 b: Answers vary. Once all the angles are found, State which pairs of corresponding angles have equal measure, such as m!d = m!a, to reach the conclusion that!abc ~!DEC by AA ~ or SAS ~. c: They are both correct. Since both triangles are isosceles, we cannot tell if one is the reflection or the rotation of the other (after dilation) cos52 = b c,!tan52 = a b,!cos 38 = a c = x,!x! inches a: 1 2 b: 0 c: 3 4 d: 1

4 5.3.1: 5-67.! a: impossible because a leg is longer than the hypotenuse b: impossible because the sum of the angles is more than William is correct a: A!("3,"6),! B!("5,"4),! C!(0,"4) b: A!(3,3),! B!(1,1),! C!(1,6) a: x = 16 5 b: no solution c: x =!11 or 3 d: x = b is correct; if two sides of a triangle are congruent, the angles opposite them must be equal : They must have equal length. Since a side opposite a larger angle must be longer than a side opposite a smaller angle, sides opposite equal angles must be the same length ! 10.6 units square units % (!2, 4)

5 5.3.3: a: 29% b: cos29 = y,!y! a: (!1,!2) b: (4,!4) c: (3, 4) a: It uses circular logic. b: The arrow between Marcy likes chocolate and Marcy likes Whizzbangs should be reversed. The arrow connecting Marcy likes chocolate and Whizzbangs are 100% chocolate should be removed tan!1 ( 3 4 ) " a: 1 12 b: C 5.3.4: a: The diagram should be a triangle with sides marked 116 ft. and 224 ft. and the angle between them marked 58. b:! 190 feet a: Corresponding angles have equal measure. b: The ratio of corresponding sides is constant, so corresponding sides are proportional y = (tan25 )x + 4 or y! 0.466x It must be longer than 5 and shorter than 23 units years B

6 5.3.5: The third side is 12.2 units long. The angle opposite the side of length 10 is approximately 35.45, while the angle opposite the side of length 17 is approximately x! 11.3 units; Methods include using the Pythagorean Theorem to set up the equation x 2 + x 2 = 16 2, using the triangle shortcut to divide 16 by 2, or to use sine or cosine to solve using a trigonometric ratio No, because to be a rectangle, the parallelogram needs to have 4 right angles. Possible counterexample: a parallelogram without 4 right angles a: P! units, A = 72 sq. units b: P = 30 units, A = 36 sq. units A(2, 4),!B(6,2),!C(4,5) y = 3 4 x a: m!abe = 80,!m!EBC = 60,!m!BCE = 40,!m!ECD = 80!, m!dec = 40,!m!CEB = 80,!m!BEA = 60 b: a:! 8.64 cm b: PS = SR = 5.27 cm, so the perimeter is! 25.5 cm area! sq. units, perimeter! units a: (0.7)(0.7) = 0.49 = 49% b: (0.3)(0.7) = 0.21 = 21% a: ! units b:! c: (!2,0) a: x = 45 = b: x =!10 or x = 10 c: x = 1.3 d: no solution

7 6.1.1: Chapter a: alternate interior angles b: vertical angles c:!u!&!!z,!s!&!!x,!v!&!!w, and!t!&!!y 6-5. a: They are similar by SAS~. b: Yes, because they are similar and the corresponding sides have a ratio of x = 180,!x = = a: 8 b:! c:! a and 1b: statements ii and iv 2: The cupcakes are burned 3: The fans will not buy the cupcakes because they are burned 4: The team will not have enough money for uniforms 6-9. A 6.1.2: a = 97,!b = 15,!c = 68,!d = a:! 3.75, tangent b: 7 2! 9.9, Pythagorean Theorem or 45! 45! 90! ratios c:! 9.54, Law of Cosines a: 25 units b: 56 square units and 350 square units a: A!("2,"7),! B!("5,"8),! C!("3,"1) b: A!!(2,7),! B!!(5,8),! C!!(3,1) c: reflecting across the y-axis Let B represent the measure of angle B. Then (3B + 5 ) + B + (B! 20 ) = 180, so m!a = 122, m!b = 39, and m!c = A

8 6.1.3: Reasoning can vary. Sample responses: a: a = 123, when lines are //, corr.! s are =, b = 123, when lines are //, alt. int.! s are =, c = 57, suppl.! s b: all = 98, suppl.! s, then, when lines are //, alt. int.! s = and corres. or vert.! s = c: g = h = 75,, when lines are //, alt. int. or corres.! s =, then vert.! s = a: 40% b: To make only one freethrow, Vicki needs to make the first and miss the second. This probability is = 0.24 or 24%. The probability of her scoring two free-throws is = 0.36 or 36%. Therefore, she has a greater chance of scoring two points than one. This is a counter-intuitive result that could benefit from some discussion in class at a later point if time allows a: x =!4 and y = 0 b: No solution; the lines are parallel = 5,!x = 7.5 x b: The perimeter of both triangles is! units c:! 32.9 and! C

9 6.1.4: Justifications and order may vary: a = 53, given; b = 55, straight angle (with!g ); c = 72, triangle angle sum; d = 53, when lines are parallel, alternate interior angles are equal; e = 55, when lines are parallel, alternate interior angles are equal; f = 127, straight angle (with!a ), so they are supplementary x-intercept (4,0), y-intercept (0,6) a: For left-hand triangle: c 2 = ! 2 " 3" 6 cos60, c = 3 3! units; For right-hand triangle: c 2 = ! 2 " 6 " 3 3 cos 30, c = 3 units; They are congruent. b: Yes; by SSS! or SAS! a: converse: If the ground is wet, then it is raining. Not always true. b: converse: If a polygon is a rectangle, then it is a square. Not always true. c: converse: If a polygon has four 90 angles, then it is a rectangle. Not always true. d: converse: If a polygon is a triangle, then it has three angles. Always true. e: converse: If vertical angles are congruent, then two lines intersect. Always true b: 22 = 10,!x! x sin 40 = h,!h! feet : a: x! b: x! c: x! d: x = square units; One way: Find AC = 5 and then calculate 1 (5)(3.6), or can use 2 BC as the base and calculate 1 (2 + 4)(3) a: m = 33,!n = 36 b: area (small) = 378 square units, perimeter (small) = 80 units, area (big) = square units, and perimeter (big) = 120 units a: similar because of AA~ b: neither because angles not equal c: congruent because of ASA! or AAS! a:! b: y = x + 3 c: (1,4) D

10 6.2.2: a: lines l and m are parallel because alternate interior angles are equal b: line n is perpendicular to line m because w + k = 180 and if w = k, then each is 90 c: no special statements can be made because vertical lines are always equal d: lines l and m cannot be parallel because otherwise z + k = a:!abc ~!DEF (AA~) b:!mon "!PQR (AAS! or ASA! ) c: neither congruent nor similar because m!j! 62. If m!j = 62, then m!l = 180! 2 " 62 = 56. Since sin56! sin 72, this triangle cannot exist. 5 8!##"!##" a: It is a trapezoid. The slope of WZ equals the slope of XY. b:! 18.3 units c: (!9,1) d: Because alternate interior angles are congruent, the angle of depression equals the angle formed by the line of sight and the ground. Then tan! = 52,!! " c 2 and a 2 + b B 6.2.3: a: Yes, because parallel lines assure us that the alternate interior angles are congruent. Since corresponding angles in the triangles have equal measure, the triangles are similar by AA ~. x b:,!x = = x a: x = 4 b: x = 55 c: x = 23 and y = 43 d: x = 5.5 and y = area! square units, perimeter! units a: 288 feet by 256 feet b: area of shape = 59.5 square units, area of island = 60,928 square units a: Yes because of AAS! or ASA! ;!DEF "!LJK b: reflection and rotation c: x! C

11 6.2.4: a: congruent (HL! or SAS! ) b: congruent (AAS! ) c: not necessarily congruent d: congruent (SAS! ) a: x + 4x! 2 = 90,!x = 18.4 (complementary angles) b: 2m m! 1 + m + 9 = 180,!m = (Triangle Angle Sum Theorem) c: 7k! 6 = 3k + 18,!k = 6 (vertical angles are equal) x d: 16 = 8,!x! 9.8 (corresponding parts of similar figures have equivalent ratios) x = 11;!m!ABC = a: 4 or 75% b: 3 or 15% c: 1 or 100% c = 10 because of substitution D 6.2.5: a: 5x + 3 = 4x + 9 because if lines are parallel, then alternate interior angles are equal, x = 6 b: q = t because if lines are parallel, then corresponding angles are equal; c + t = 180 because if lines are parallel, then same-side interior angles are supplementary; 66 c: 180! 88 = 92 ;!g + q = 180 because when lines are parallel, same-side interior angles are supplementary Methods vary. a: x! b: x! 7.86 c: x! a: x = 4 b: x! 8.1 c: not enough information d: x! a: x = 15 b: k = 5 c: t = 9 and w = 131 d: x! B

12 7.1.1: Chapter a. They are congruent by ASA! or AAS! b: AC! 9.4 units and DF = 20 units 7-7. Relationships used will vary, but may include alternate interior angles, Triangle Angle Sum Theorem, etc.; a = 26,!b = 65,!c = 26,!d = width = 60 mm, area = 660mm a quadrilateral a: (6,!13) b: not possible, these curves do not intersect 7.1.2: Using the Pythagorean Theorem, AB = 8 and JH = 5. Then, since 3 6 = 4 8 = 5 10,!ABC ~!HGJ because of SSS ~ units a: 3m = 5m! 28,!m = 14 b: 3x x! 8 = 180,!x = 15 c: 2(n + 4) = 3n! 1,!n = 9 un d: 2(3x + 12) = 11x! 1,!x = 5 un Rotating about the midpoint of a base forms a hexagon (one convex and one non-convex). Rotating the trapezoid about the midpoint of either of the non-base sides forms a parallelogram a: 10 units b: (!1,4) c: 5 units, it must be half of AB because C is the midpoint of AB.

13 7.1.3: a: The 90 angle is reflected so m!xz Y! = 90. Then m!yz Y! = 180. b: They must be congruent because rigid transformations (such as reflection) do not alter shape or size of an object. c: XY! X Y ", XZ! XZ, YZ! Y " Z,!Y!! Y ",!YXZ!! Y " XZ and!yzx!! Y " ZX M (0,7) ; A variety of methods are possible a: The triangles are similar because corresponding sides are proportional (SSS ~). b: The triangles are similar because parallel lines assure that corresponding angles have equal measure (AA ~). c: Not enough information is provided. d: The triangles are congruent by AAS or ASA a: It is a parallelogram; opposite sides are parallel. b: 63.4 ; They are equal. c: AC :!y = 1 2 x + 1,!BD :!y =!x + 5 : No 2 d: (3,2) a: No solution; lines are parallel. b: (0,3) and (4,11) Side length = 50 units, diagonal is 50! 2 = 100 = 10 units a: It is a rhombus. It has four sides of length 5 units. b: HJ :!y =!2x + 8 and GI :! 1 2 x + 3 c: They are perpendicular. d: (6,!1) e: 20 square units a: 6n! 3 = n + 17,!n = 4 b: 7x! x + 14 = 180 so x = Then 5y! 2 = 7(18.5)! 19, so y = c: 5w w = 180,!w = 18 d: k 2 = ! 2(15)(25)cos120,!k = ! units a: 1 8 b: 5 6

14 7.1.4: = 10 sides b: regular decagon If the diagonals intersect at E, then BE = 12 mm, since the diagonals are perpendicular bisectors. Then!ABE is a right triangle and AE = 15 2! 12 2 = 9 mm. Thus, AC = 18 mm Yes, she is correct. One way: Show that the lengths on both sides of the midpoint are equal and that (2, 4) lies on the line that connects ( 3, 5) and (7, 3) (a) and (c) are correct because if the triangles are congruent, then corresponding parts are congruent. Since alternate interior angles are congruent, then AB! DE AB = 40! 6.32,!BC = 34! 5.83, therefore C is closer to B : a: x = 8.5 b: x = 11 c: x = a: = 5 sides b : = ! 36.4 feet from the point on the street closest to the Art Museum a: x + x + 82 = 180,!x = 49 b: 2(71 ) + x = 180,!x = a: similar (SSS ~) b: congruent (ASA! or AAS! ) c: congruent, because if the Pythagorean Theorem is used to solve for each unknown side, then 3 pairs of corresponding sides are congruent; thus, the triangles are congruent by SSS! d: similar (AA ~) but not congruent since the two sides of length 12 are not corresponding

15 7.2.2: x! 1 = x + 8,!x = 3;!!5y + 2 = 22,!y = a: 83 b: a: It is a parallelogram, because MN! PQ and NP! MQ b: (1,!5)

16 7.2.3: a: congruent (SSS! ) b: not enough information c: congruent (ASA! ) d: congruent (HL! ) a: It is possible. b: Same-side interior angles should add up to 180. c: One pair of alternate interior angles are equal, but the other is not for the same pair of lines cut by a transversal; or, the vertical angles are not equal a: Yes, HL! b: 18, 4 c: tan18 = 4,!AD! 12.3 units AD d:! 49.2 square units a: Parallelogram because the opposite sides are parallel. b:! AC ##" :!y = 3 4 x;!bd!##" :!y =! 3 2 x a: 2x + 52 = 180,!64 b: 4x! 3 + 3x + 1 = 180,!26 sin 77 sin 72 c: =,!x! 8.2 x 8 d: 5x + 6 = 2x + 21,!x = : ! 62.4 square units No; using the Pythagorean Theorem and the Law of Cosines, the perimeter of the triangle is! 26.3 feet a: congruent (SAS! ) and x = 2 b: congruent (HL! ) and x = A = 24 square units 7-76.! meters

17 7.2.5: b: Since corresponding parts of congruent triangles are congruent, 2y + 7 = 21 and y = m!a = 132,!m!b = 108,!m!c = 120,!m!a + m!b + m!c = AB! CD and AB! CD (given), so!bac!!dca (alt. int. angles). AC! CA (Reflexive Property) so!abc "!CDA (SAS! ).!BCA!!DAC (! " s!" parts). Thus, BC! AD BC // AD (if alt. int. angles are congruent, then the lines cut by the transversal are congruent) A = 42 square units, P! 30.5 units a:!adc ; AAS! or ASA! b:!sqr ; HL! c: no solution, only angles are congruent d:!tzy ; SAS! and vertical angles e:!gfe ; alternate interior angles equal and ASA! f:!def, SSS!

18 7.2.6: a: The triangles should be! by SSS! but 80! 50. b: The triangles should be! by SAS! but 80! 90 and 40! 50. c: The triangles should be! by SAS! but 10! 12. d: Triangle is isosceles but the base angles are not equal. e: The large triangle is isosceles but base angles are not equal. f: The triangles should be! by SAS! but sides 13! a: 12 b: 15 c: This problem is similar to the Interior Design problem (7-19). Her sink should be located 3 2 feet from the right front edge of the counter. This will make the 3 perimeter! 25.6 feet, which will meet industry standards : a: (4.5, 3) b: (!3,1.5) c: (1.5,!2) a:!shr ~!SAK by AA~ b: 2HR = AK,!2SH = SA,!SH = HA c: 6 units a:!ced ; vertical angles are equal, ASA! b:!efg ; SAS! c:!hjk;!hi + IJ = LK + KJ,!!J "!J ; SAS! d: not!, all corresponding pairs of angles equal is not sufficient No, her conclusion in Statement #3 depends on Statement #4, and thus must follow it a: must be a quadrilateral with all four sides of equal length b: must be a quadrilateral with two pairs of opposite sides that are parallel

19 7.3.2: Multiple answers are possible. Any order is valid as long as Statement #1 is first, Statement #6 is last, and Statement #4 follows both Statements #2 and #3. Statements #2, #3, and #5 are independent of each other and can be in any order as long as #2 and #3 follow Statement # a: 6 b: 3 c:! a: yes, by SAS~ b:!fgh!!fij,!!fhg! FJI c: Yes, because corresponding angles are congruent and because of the Triangle Midsegment Theorem. d: 2(4x! 3) = 3x + 14, so x = 4 and GH = 4(4)! 3 = 13 units a: a right triangle. Some students may also call it a slope triangle. b: B! is at (2,7). ABC B! is a kite a: Must be: trapezoid. Could be: isosceles trapezoid, parallelogram, rhombus, rectangle, and square. b: Must be: parallelogram. Could be: rhombus, rectangle, and square : a: (8,8) b: (6.5,6) c: (1,8.5) d: (2,4) a: X and Y b: Y and Z a: Must be: none. Could be: right trapezoid, rectangle, square. b: Must be: none. Could be: Kite, rhombus, square a: = 20 sides b: It can measure 90 (which forms a square). It cannot be 180 (because this polygon would only have 2 sides) or 13 (because 13 does not divide evenly into 360 ) It must be a triangle because it is a right triangle and the hypotenuse is twice the length of a leg.

20 8.1.1: Chapter a: 110 b: 70 c: 48 d: a: The measure of an exterior angle of a triangle equals the sum of the measures of its remote interior angles. b: a + b + c = 180 (the sum of the interior angles of a triangle is 180 ), x + c = 180 (straight angle); therefore, a + b + c = x + c (substitution) and a + b = x a + b = x (subtracting c from both sides) x = 72 and y = = a:! (SAS! ), x = 79 b: cannot be determined c:! (AAS! ), x! 5.9 units d:! (SAS! ), x! a: True b: False (counterexample is a quadrilateral without parallel sides) c: True d: True e: False (counterexample is a parallelogram that is not a rhombus) 8.1.2: a: isosceles right triangle, because AC = BC and AC! BC b: 45, methods vary a = 87,!b = 83,!c = 96,!d = 94 ;! A = 40.5!un 2, P! 27.7 units (!5,1), (!3,7), and (!6,2) B

21 8.1.3: a: A = 36 sq. ft, P = 28 ft b: A = 600 sq. cm, P! 108.3cm QP = RS and PR = SQ (given), QR = QR (Reflexive Property), so!pqr "!SRQ (SSS! ) and!p!!s (! "s #! parts ) a: isosceles triangle b: The central vertex must be = 36. The other two angles must be equal since the triangle is isosceles. Therefore, (180! 36 ) 2 = 72. c: 10!14.5 = 145 square inches (6.5,5) a: The region can be rearranged into a rectangle with dimensions 14 and 7 units. b: 14(7) = 98 square units B 8.1.4: The reflections are all congruent triangles with equal area. Therefore, the total area is (6)(11.42) = square inches a: non-convex b: convex c: convex d: non-convex a: 64un 2 b:! 27.0un 2 c: 8 3! 13.9un a: 3 b: 15 c: 4 d: a: A = 192cm 2, P = 70cm b: The length of each side is 9 times the corresponding side in the floor plan. A = 15,552!cm 2 and P = 630 cm D c: The ratio is 9 = 9 ; the ratio of the perimeters equals the zoom factor 1 d: The ratio of the areas is 81 = 81. The ratio of the areas equals the square of 1 the zoom factor (9 2 ).

22 8.1.5: a: The interior and exterior angles must be supplementary. Therefore, 180! 20 = 160. b: Possible ways: Use = 18 sides or solve the equation 180(n!2) n = 160 to find n = a: x = 18,!y = 9 3 b: x = 24 2,!y = 24 c: x = 8 3 = 8 3 3,!y = 16 3 = Since the diagonals of a parallelogram bisect each other, they must intersect at the midpoint of BD. Thus, they intersect at (6, 21) A = 100 3! 173.2!mm a: ± 17! ±1.84 b: w! 2.17 and! c: no solution possible E a: 60 b: 82 c: 14 d: a: equilateral triangle b: rectangle c: nonagon d: rhombus or kite The x-coordinate must be 6, but the y-coordinate could be 6 3 or! a: Yes; since BC = BC (Reflexive Property), AB! DC (given), and!abc! DCB (given), then!abc "!DCB (SAS! ). Therefore, AC = DB (! "s #! parts ). b: No; the relationships in the figure are true even if points B and C were hinged, as long as the two angles remain congruent. See the diagram for problem 8-28 for a similar diagram a: (!2.5,0) and (3,0) b: The graph of y =!(2x 2! x! 15) would be the reflection of y = 2x 2! x! 15 across the x-axis because each y-value would have its sign changed D

23 8.2.1: a: A = 34un 2, P! 25.7 units b: A = 306 un 2, P! 77 units c: ratio of the perimeters = 3; ratio of the areas = inches or! 6.67 feet The area of the hexagon! 23.4 ft 2. Adding the rectangles makes the total area! 41.4 ft a: Reasoning will vary, but it is most likely that you will earn more extra credit if the class spins the spinner with the options of 5 and 10 points. b: Reasoning will vary, but now the first spinner is definitely more attractive x 2 = 2x x! 30,!x = 2.5 or 6: yes, there are two possible answers B 8.2.2: a: 3 4 b: rp c: ar a:! 403.1!cm 2 b:! cm a: = 6, so two sides will collapse on the third side. b: Answers vary. One solution is 2, 5, and a: AAS!,!"ABC! "DCB b: ASA!,!"ABC! "EDC D

24 8.3.1: Area of the entire pentagon! !un 2, so the shaded area! 3 5 (172.05)! !un a: x = 14 3, pattern b: x! 5.78, Law of Sines c: No solution because the hypotenuse must be the longest side d: 24 units, triangle angle formula BC! DC and!a!!e (given) and!bca!!dce (vertical angles are! ). So!ABC "!EDC (AAS! ) and AB! ED (! "s #! parts ) a: (1.5,5) b: y = 4 x + 3 c: 15 units B 8.3.2: (100! 25" ) 4 # 5.37 square units a: 8 b: a: x = 26 ; if lines are parallel and cut by a transversal, then alternate interior angles are equal. b: x = 33,!n = 59 ; if lines are parallel and cut by a transversal, then same-side exterior angles are supplementary a: 20 b:! 126.3!un The area of the hexagon is 24 3 units, so the side length of the square is D 24 3! 6.45 units.

25 8.3.3: ! 40 = 320, so = 8 9! 89% a: C = 28!!un,!A = 196!!un 2 b: C = 10!!un,!A = 25!!un 2 c: diameter = 100 un, radius = 50 un a! 3, b! 1, c! 4, d! a:!3 b:!4 c: 3 and!3 d: 2 and! D a: (55)(60) + 900! " square feet b: ! " feet, 298.5! 8 = $ or approximately $2,388 c: Area is four times as big! 24,509.6 square feet; perimeter is twice as big! 597 units a: x + x = 540,!x = 100 b: 6x + 18 = 2x + 12,!x = a: CD = 22, BC = 7, and ED = 6 ; the perimeter is = 48 units b: 54(4) = 216!cm a: 16 3! square units b: 36 square units, more c: 24 3! square units; its area is greater than both the square and the equilateral triangle. d: a circle a: 2!r = 24;!r = 12 ;!A = 144 " square units!! b: 2!r = 18!;!r = 9;!A = 81! " square units E

Lesson a: They are congruent by ASA or AAS. b: AC 9.4 units and DF = 20 units

Lesson a: They are congruent by ASA or AAS. b: AC 9.4 units and DF = 20 units Lesson 7.1.1 7-6. a: They are congruent by ASA or AAS. b: AC 9.4 units and DF = 20 units 7-7. Relationships used will vary, but may include alternate interior angles, Triangle Angle Sum Theorem, etc.;

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