Introduction to Trigonometric Identities page 594. Find sin θ if cos θ = 0.5. sin 2 θ + cos 2 θ = 1 sin 2 θ + (0.5) 2 = 1 sin 2 θ + 0.

Transcription

1 LESSON 91 Name Introduction to Trigonometric Identities page 594 New Concepts Math Language Trigonometric identities are expressions that relate any two trigonometric functions. Trigonometric Identities The most commonly used trigonometric identities are below. tan θ = sin θ cos θ sin x + cos x = 1 Example Find sin θ if cos θ = 0.5. sin θ + cos θ = 1 sin θ + (0.5) = 1 sin θ = 1 Relating Sine and Cosine sin θ 0.87 Trigonometric Identity Substitute. Simplify. Solve. Example Building More Identities Express tan θ using only sin θ. Rearrange the second identity to solve for cos θ. sin θ + cos θ = 1 cos θ = 1 - sin θ cos θ = 1 - sin θ cos θ = 1 - sin θ Substitute this identity into the tangent identity. tan θ = sin θ cos θ sin θ tan θ = 1 - sin θ Relationships between the trigonometric functions: Function In terms of sine In terms of cosine In terms of tangent sin θ = sin θ 1 - cos θ cos θ = 1 - sin θ cos θ tan θ = _ sin θ 1 - sin θ 1 - cos θ cos θ tan θ 1 + tan θ tan θ tan θ Adaptations Lesson 91 Saxon Geometry 573

2 Lesson Practice page 597 a. If sin θ = 0.67, what is the value of cos θ to the nearest hundredth? + cos θ = 1 cos θ b. Express cos θ in terms of tan θ and show each step. tan θ = sin θ cos θ tan θ = 1 - cos θ (tanθ) = ( 1 - cos θ ) tan θ = 1 - cos θ tan θ θ = 1 - cos θ tan θ θ + = Factor out cos θ. cos θ ( ) = cos θ = 1 c. Let θ = angle Becky, so sin θ = 1. sin θ + cos θ = 1 ( ) + cos θ = 1 cos θ = cos θ = cos θ = ratio of Ivan s distance from Becky to Ruby s distance from Becky 574 Saxon Geometry Adaptations Lesson 91

3 Practice page Sketch a tessellation using this kite.. Find the value by graphing functions and looking for their. θ =, 3. y < x + 4. vertical: 4 y sin 7 = x -4 - O 4 x x = sin horizontal: cos 7 = y y = cos 7 5. S T V U 7. S = 4π r P U' T' V' S' 6. After a rotation of 15, the figure coincides with itself. number of sides = number of sides = 8. Draw a sketch on a coordinate grid. By its sides, it is classified as a triangle. S = Adaptations Lesson 91 Saxon Geometry 575

4 Practice, continued page See page 3 of the Student Reference Guide. 10. See page 34 of the Student Reference Guide. other leg length: hypotenuse length: A = Draw a sketch on a coordinate grid. The student is. u w The components will both be because 196 from A the positive x-axis is in Quadrant. C B 13. For solids, a cross section of a point is made when the plane cuts through the outermost tip of the solid. On a sphere this is any. On a cube, this is at. On a cylinder, this is a tipped tangent to the edge of one of the. 15. See page 8 of the Student Reference Guide. The ratio of the measure of an arc to 14. If two chords intersect in a circle, then the products of the chord segments are equal. = (x - )(x - ) = 0 The length of a segment cannot be 0. x = 16. See page 3 of the Student Reference Guide. r = the measure of its inscribed angle is :. S = Each patch is about in. 576 Saxon Geometry Adaptations Lesson 91

5 Practice, continued page Use the Hinge Theorem. SR QR 18. S = 1 Pl + B l = S = 19. Complete. 0. See page 30 of the Student Reference Guide. B A C A' C' B' Answer: 1. Find the volume of the cylinder. r =. area of = area of 31.4 = 3. lw = A ( x - )( x + ) = x + 8 x - x - = x + 8 x - x - = 0 r Adaptations Lesson 91 Saxon Geometry 577

6 Practice, continued page tan θ = 1 - cos θ tan θ = 5. Draw a sketch. Use the properties of right triangles. 6. Draw a sketch on a coordinate grid. = s width of the square beam: distance = 7. cos θ = 3 sin θ 1 = 3 sin θ 8. Corresponding sides of similar triangles are proportional. Let x = height of flagpole 1 = sin θ 1 = θ = θ 9. a + b = a + b tension load of cable: tan θ = x = 30. Desi made an error in step. He needed to use a and incorrectly with a binomial. angle of cable and horizontal: 578 Saxon Geometry Adaptations Lesson 91

7 LESSON 9 Name Quadrilaterals on the Coordinate Plane page 600 New Concepts The coordinates of a quadrilateral s vertices can be used to determine properties of the quadrilateral. The length and slope of the sides can be found from the coordinates to classify any quadrilateral in the coordinate plane. Example Proving Quadrilaterals are Parallelograms Is ABCD a parallelogram? Use the vertices given. Find the slopes of each side. Slope = y - y 1 x - x 1. slope of AB = = _ 3 4 slope of BC = = - y O A(1, 4) D(, ) B(5, 7) C(6, 5) x slope of CD = = _ 3 4 Lesson Practice page 60 a. slope of CD = slope of EF = CD EF and DE CDEF a parallelogram. slope of AD = = - The slopes of AB and CD are equal, so AB CD. The slopes of BC and AD are equal, so BC AD. Therefore, ABCD is a parallelogram. slope of DE = slope of FC = y 4 D(-1, 4) - O - C(1, 0) -4 E(3, 1) x F(5,-3) Adaptations Lesson 9 Saxon Geometry 579

8 Lesson Practice, continued page 603 b. slope of PQ = slope of RS = PQ but QR PQRS is a. slope of QR = slope of PS = P S -4 y R Q x c. Determine if the quadrilaterals are congruent. Use the distance formula to show that the corresponding sides and a corresponding diagonal are congruent. JK = ST = KL = TU = LM = UV = JM = SV = JL = SU = JKLM d. slope of GH = slope of JK = GH but HJ Practice congruent to STUV. GHJK is a. 1. Plot the coordinates of E. Circle the answer. D -4 - y 8 6 O slope of HJ = slope of GK = E 4 F x page 603. Draw a sketch. v x cos 1 = O y J K H x v y sin 1 = G A (0, -1) B (-1, 0) C (1, 0) D (1, 1) v y, v x 580 Saxon Geometry Adaptations Lesson 9

9 Practice, continued page Find the corresponding side lengths and slopes. AB = = EF = = BC = = FG = = CD = = GH = = AD = = EH = = m AB =, m BC =, m CD =, m AD =, m EF =, m FG =, m GH =, m EH =. Therefore, ABCD EFGH. 4. Find the side opposite the largest angle of any of the exterior sides. A B 94 C 110 The longest side in the figure is. E D 5. parallel line: y = 7. perpendicular line: y = 33 = 1 _ ( - m BF ) m BF = - = 33 = 1 _ ( - m CE ) 6. Placing n points on a line forms 8. Find the slopes. m WX = m YZ = WX WXYZ segments., m XY =,, m WZ =, and XY a parallelogram. m CE = - = Adaptations Lesson 9 Saxon Geometry 581

10 Practice, continued page Line l a tangent because this would Theorem P(-3, 6) 8 y S(3, 6) It does intersect C at two points, one of which 4 is P, the other is Q. Since CP CQ, CPQ is. Q(-4, 1) O -4 - R(1, 1) 4 x Base angles and are m PQ by the Isosceles Triangle Theorem, and since a m QR = triangle can have at most right or m RS = obtuse angle, CPQ and CQP are congruent acute angles. m PS = QR, but PQ, so PQRS is a. 11. Draw a sketch. 1. Draw the net of a pentagonal pyramid = = To return, the vector components will be 13.. The vector that can be used to go directly back:,. Let cos θ = b _ c _ Let tan θ = a b b 1 1_ c = = b b + _ = 1 + ( _) 1 b + a = 1 = See page 4 of the Student Reference Guide. AF = AG AF = AF = GF = AF - GF = 15 - GF = BG = BD BG = ( ) BG = = 58 Saxon Geometry Adaptations Lesson 9

11 Practice, continued page a. A center = ( )( ) = 16. F G will be the length of FG. A outer = ( )( ) = P(center circle) = or C b., since a person would be aiming for the, it would the same probability as a random event. F G 17. Draw a sketch. SOHCAHTOA 30 = 3000 x = x The horizontal distance of the plane from its starting point is about. 18. Find the circumference of the log = π r( ) 31.4 = r r Divide the length of the hill by the circumference of the log to find the number of times the log will revolve. times 19. C O 10 π yards D The sector is of the circle. π = 0. Find the corresponding side lengths and slopes. JK= NO = KL = OP = LM = PQ = JM = NQ = The slopes of JK, KL, LM, and JM are,,, and. The slopes of NO, OP, PQ, and NQ are,,, and. J N K L M So, JKLM NOPQ. Adaptations Lesson 9 Saxon Geometry 583

12 Practice, continued page Draw a sketch. height of rider =. Reflect the figure across l 1. A C B l 1 horizontal distance of rider = Reflect the figure tan x = across l. l tan x = 3. cos θ tan θ = cos θ 4. See page 7 of the Student Reference Guide. Draw a sketch. = r 5. See page 3 of the Student Reference Guide. 6. The statements show the use of the Law of. V = 1 _ 3 ( )( ) = 7. Draw a sketch. The sides of the final image will lie on the of the original triangle, but the vertices will be different. E coincides with. F coincides with. G coincides with. 8. a. b. Continue the rotation. 9. Use inverse tangent to find θ. tan θ = tan θ = 30. tan θ = sin θ cos θ tan θ = Use inverse tangent to find θ. to θ 584 Saxon Geometry Adaptations Lesson 9

13 LESSON 93 Name Representing Solids: Orthographic Views page 607 New Concepts Math Language An orthographic drawing is a drawing that shows how a threedimensional object looks when viewed head-on from the front, top, or one side. To draw an orthographic view of a solid, imagine you are looking straight into one of its faces. Example Creating Orthographic Drawings Draw the front, top, and side orthographic views of this solid. T Draw the front face of the solid. F S Front From the top, you can see two adjacent rectangles. The solid line shows that the two parts of the object are not on the same level. From the side, you can see a square on top of a rectangle. Example Top Side Constructing from an Orthographic Drawing This orthographic drawing shows the front, top, and right-side views of a solid. Front A B C Side Top Which sketch represents the solid shown in the views? From the top, Figure A does not have the long double base seen in the top view. Figure C looks similar, but the block that comes off the back is not centered, as in the top view. The answer is B. Adaptations Lesson 93 Saxon Geometry 585

14 Lesson Practice, continued page 609 Exploration page 609 Use your textbook to complete this Exploration. Lesson Practice page 609 a. Draw the front, top, and side orthographic views of this solid. For each view, focus on the solid s visible edges. Notice their number, shape, size, and position. Front number of visible edges: Sketches pairs of congruent, parallel lines shape formed: Top number of visible edges: pairs of congruent, parallel lines shape formed: Side number of visible edges: shape formed: b. Given the three views, identify the correct sketch of the solid. In all four answer choices, the solid s base is the. In the front and side views, the top part of the solid looks like a. So it could be either a cone or a. In the top view, you see the base of the top part forms a, so the top part must be a. The correct sketch is. c. Given the three views, identify the correct sketch of the solid. Assume that the front view is angled in the top half, not vertical as shown. From the top view, the base of the solid is clearly a. It also shows that the top of the solid contains a. The front view shows that the back left side of the building is a wall. The correct sketch is. 586 Saxon Geometry Adaptations Lesson 93

15 Practice page Sketch the solid.. Write the equations in standard form. y y y Graph. 3. tan θ = 1 - cos θ cos θ tan θ = ( 1 - cos θ ) cos θ 3. = cos θ = 1 cos θ - cos θ 4. See page 3 of the Student Reference Guide. h = mm = V cylinder = π ( ) ( ) = V hole = π ( ) ( ) 3. = 1 cos θ = cos θ 5. S sphere = 4π r 1 cos θ The bowl forms a hemisphere. = V stacked discs = 6. Step 1: Step : Circle the answer. A glide reflection B glide rotation C reflection then translation D translation then reflection S = Adaptations Lesson 93 Saxon Geometry 587

16 Practice, continued page If two similar figures have a scale factor of a:b, then the ratio of their perimeters is a:b, and the ratio of their areas is a : b. 8. Draw a sketch. could be moved ratio of areas: 76:931 = : Find the prime factorization of both numbers and take their square roots. ratio of perimeters: : to make a parallelogram. Three points define sides of the parallelogram. The fourth point will define the other sides. The of that point determines whether or not there are two parallel pairs. 9. The front view shows the solid s base is shaped like the letter with two arms of length. Answer: 10. Draw a sketch. Find the side lengths and angle measures. KL = LM = KM = m KL = m LM = Use an inverse cosine function to find the other angle measures. m L = m M = 11. See page 30 of the Student Reference Guide. m KSL = 13. Draw a sketch. m K = 1. See page 35 of the Student Reference Guide. Write each equation in slope-intercept form. Answer: 14. PQ BR PB QR She is. 588 Saxon Geometry Adaptations Lesson 93

17 Practice, continued page Draw a sketch. Find the slope of a line perpendicular to the given line and that passes through (0,6). 16. total squares = slope of line through (0, 6): Use (0,6) and the slope you found to write the equation of the perpendicular line. equation: y = x + Set the two equations equal to identify the point of intersection. (, ) A = 17. What is the surface area? 18. Sketch a cube. Choose one vertex and mark the midpoints of the three edges that intersect. The cross section is a(n) S =. 19. Find the side lengths and slopes. 0. Substitute. UV = = VW = = slopes of UV, VW, WX, and XU : WX WX,,, and. and VW ; and VW UVWX is a parallelogram. 1. Draw a sketch., so b =. Use the Pythagorean theorem to find the height. Draw front, side and top views. QR = DE = EF = Label dimensions. Adaptations Lesson 93 Saxon Geometry 589

18 Practice, continued page The frictional force to the left is not affected by the force of gravity. net force left = 4. Draw a sketch. If two chords intersect in a circle, then the products of the chord segments are equal. 0, + 0, = 0, net force down = x = 5. Use the Triangle Midsegment Theorem. 6. Find the side lengths. MP = = NM = = MP and MN, so RT = MNOP is a parallelogram. 7. sin θ + cos θ = 1 cos θ = 1-8. Use solid lines to indicate two adjacent parts are not level with each other. cos θ cos θ = 1 cos θ - cos θ cos θ = 1 = + 1 = 1 cos θ - 1 cos θ Complete the transformation = = y 30. See page 5 of the Student Reference Guide. = x, x = 590 Saxon Geometry Adaptations Lesson 93

19 LESSON 94 Law of Sines page 613 Name New Concepts Although generally used to find unknown angle measures and side lengths in right triangles, trigonometric ratios can be used to find angle measures and side lengths in any triangle using the Law of Sines. The Law of Sines In any triangle ABC, a sin A = b sin B = c sin C B a C c b A Example Finding Distance with the Law of Sines Find the length of XZ. Round to the nearest hundredth. X Y m Z = = 107 XZ sin Y = XY_ sin Z XZ sin 38 = 0 sin 107 XZ(sin 107 ) = 0(sin 38 ) XZ = 0(sin 38 ) sin 107 XZ 1.88 Example Triangle Angle Sum Theorem Law of Sines Substitute. Cross multiply. Divide both sides by sin 107. Solve. Finding a Missing Angle with the Law of Sines Find the measure of L. Round to the nearest degree. LM sin N = LN sin M 1 sin N = 14 sin 75 14(sin N) = 1(sin 75 ) Law of Sines Substitute. Cross multiply. 1(sin 75 ) sin N = Divide both sides by (sin 75 ) N = sin ( 14 ) 56 m L = Triangle Angle Sum Theorem L 1 M Z N Adaptations Lesson 94 Saxon Geometry 591

20 Lesson Practice page 615 a. Find the length of TU. Round to the nearest hundredth. m T = = TU sin V = TV sin U TU sin 31 = sin TU = sin 31 = sin b. Find the measure of D and E. Round to the nearest degree. DE sin F = DF sin E sin 0 = sin E (sin E) = (sin 0 ) sin E = (sin 0 ) E = sin -1 ( = sin -1 ( (sin 0 ) ) ) 48 m D = c. Find the measure of AM. Round to the nearest foot. H 10 sin = sin M (sin M) = (sin ) sin M = sin 10 M = sin ( -1 sin ) 10 m H = ft A ft M AM sin = sin AM = sin sin ft 59 Saxon Geometry Adaptations Lesson 94

21 Practice page What is the ratio?. Humberto only found the of the. : He did not specify the. 3. When the vertex of the square meets the 4. Use the Pythagorean Theorem. road, both angles formed by the road and the sides of the square are. Each angle in a square measures. x + 90 = x = x = c = 5. Segments with equal slopes are parallel to each other. 6. Sketch the pyramid s base and use the properties of special triangles. Slope of AB = = 1 Find the slopes of the other three sides. Prove ABCD is a parallelogram = x sin 300 = sin V = 1 _ 3 Bh = 8. Find the range of values for x. x = 300 sin sin < x < Adaptations Lesson 94 Saxon Geometry 593

22 Practice, continued page Q R S T U = ( )QRSTU Q (6,6) R (1, ) S (, -6) T (, ) U (, ) Sketch Q R S T U. 10. sin = sin R sin R = R = sin sin-1 ( (sin )) 11. tan θ = θ cos θ sin θ + cos θ = 1 cos θ = 1 - tan θ = 1 - θ θ θ Q = = PR sin 13 = sin 1. Imagine you look perpendicularly at the top and see a flat shape. Draw it. Repeat this for the side and top. PR = 13 sin sin 13. Bases of a trapezoid are. 15. Circle the answer. A 0,, 0, 3 B 3, - 1 _, 1_, 5 C 1, 4,, 3 D 1, 5, -1, - 1 _ 5 V = 1 _ 3 π r 3 h 14. If an inscribed angle intercepts a semicircle, the inscribed angle is a angle. x = Substitute. h = S = π rl + π r Find l from r and h. l = S 594 Saxon Geometry Adaptations Lesson 94

23 Practice, continued page P = = A = 1 ap 17. sin θ + cos θ = sin θ = - sin θ = - Substitute into the Law of Sines. A = a_ = - b_ = - c_ Repeat the square adjacent to itself in 3 rows and 3 columns. Construct your tessellation. 19. Substitute the center and radius of the circle into the equation of a circle. (x - ) + (y - ) = 0. The angle between sides a and b can be acute or obtuse (its supplement), giving two triangles., Nadim can get 1. Set volume of a sphere equal to surface area of a sphere and solve for r. π = 4 r potentially correct answers. Because a measure. Draw the tangents. number of internal tangents: number of external tangents: total: 3. Draw a sketch. 7x - 36 = r = x = Adaptations Lesson 94 Saxon Geometry 595

24 Practice, continued page Let x = area of each small sector Draw a sketch. area of spinner = 1 _ area of large sector = 1 area of small sector = x x = 1 _ ; x = 5. triangle type: triangle marked congruent: pair(s) of pair(s) of The 6. Use the Pythagorean Theorem on each number set. Circle the answer. A 3, 4, 5 B,, 8 C,, D 7, 6, 11 Congruence Theorem 7. squares: about 8. Draw a sketch. If two chords intersect in a circle, then the of the chord segments are. number of basalt rocks per ft = equation: number of basalt rocks = c = 9. The sign is >, so the 30. Move the front and side together lining up the lines. Imagine the side always contacting the top but swinging into the paper. Move the top boundary line is above the front lining up the lines. Imagine the top always contacting the front but swinging into the paper. This is rectangular pyramid with a, and the smaller rectangular pyramid cut out of it. region it Draw the three-dimensional figure. should be shaded. Graph. 596 Saxon Geometry Adaptations Lesson 94

25 LESSON 95 Equations of Circles: Translating and Dilating page 618 Name New Concepts To translate a circle, find the center of the circle, translate the center, and write the equation for a circle with the same radius at the new center. Example Translating a Circle The equation of a circle is (x - 3) + (y + ) = 5. Translate the circle 4 units to the left and units down. Write the equation of the translated circle and sketch both circles on the coordinate plane. The center of the first circle is (3, -). The length of the radius is y 4 (3, -) x The new center is at (-1, -4). The length of the radius is still y x 4 8 The equation of the new circle is (x + 1) + (y + 4) = 5. To dilate a circle, apply the dilation to the radius and the center of the circle only. Example Dilating a Circle Centered at the Origin The equation of a circle is x + y = 49. Apply a dilation centered at the origin with a scale factor of. Write the equation of the translated circle and sketch both circles on the coordinate plane. The center will not change. The radius will be twice as long (0, 14) O y 8 (0, 7) (-1, -4) x 8 16 The given radius is 7. After the dilation, it will be 14. The new equation is x + y = 196. Adaptations Lesson 95 Saxon Geometry 597

26 Lesson Practice page 60 a. center of the first circle: (, ) y new center: (, ) length of the radius: new equation: (x ) + (y ) = -4-4 O x b. current radius: 4 y radius after dilation: new equation: = -4 - O x c. Multiply the radius by the scale factor. radius of the new circle: Apply the scale factor to each coordinate of the center y new center: (, ) = (, ) new equation: ( ) + ( ) = d. Draw a sketch. equation of siren s circle: ( + ) + ( ) = ( ) center of siren s new circle: (, ) O x radius of siren s new circle: equation of siren s new circle: + ( ) = ( ) The siren will audible at June s new house. 598 Saxon Geometry Adaptations Lesson 95

27 Practice page See page 8 of the Student Reference Guide. Draw a sketch. measure of angle = 3. Draw a sketch. 4. See page 33 of the Student Reference Guide. Let v x = horizontal speed of ball 30º = v x v x x 5. new center: (, ) radius: new equation: ( ) + = y 8 4 O x The geometric mean for positive numbers a and b is the positive number x such that a_ x = x_. b Verify that 1_ 8 is the geometric mean of 1_ and 1 3. Adaptations Lesson 95 Saxon Geometry 599

28 Practice, continued page Draw a horizontal line through C. 8. Multiply the original radius by the scale factor. original radius: A B t new radius: original center: (, ) C new center: (, ) new equation: ( ) + ( ) = 9. Use the Law of Sines to find m A. sin 108 = sin A 10. magnitude: 14 = velocity with wind: 14 = m A m B AC = 5 AC 11. Corresponding sides of similar triangles are proportional. x = 13. See page 1 of the Student Reference Guide. y = Theorem 1. original center: (, ) new center: (, ) radius: new equation: + ( ) = Draw the circles on a coordinate grid. 14. Cosine is 0 at, and tan =, so the tangent is undefined at. 600 Saxon Geometry Adaptations Lesson 95

29 Practice, continued page Use the Law of Sines. m L = LM sin 35 = See page 35 of the Student Reference Guide. i) y = x + 7 ii) y = (x - 4) iii) y - x = 7 LM Circle the answer. A i and ii C ii and iii B i and iii D i, ii, and iii 17. slope CD = slope DE = 18. See page 34 of the Student Reference Guide. slope EF = slope CF = The slopes are all, so no two sides are. CDEF be a trapezoid. x = 19. a sin a = a tan A + tan A = a = a = = 1. See page 1 of the Student Reference Guide. 0. See page 30 of the Student Reference Guide. length = width = The parallelogram is a.. The cross section is a. x = Adaptations Lesson 95 Saxon Geometry 601

30 Practice, continued page 6 3. If the pattern is rotated it will look the same. rotational symmetry? yes no order: 4. center: (, ) original radius: new radius: new equation: + = Draw the circles. 5. Draw a sketch of the circle with a line connecting the centers. Use the given values to find the lengths of the legs of a right triangle with the line that you drew as the hypotenuse. 6. After a 90 counterclockwise rotation: T: (x, y) (-y, x) Q = (, ) R = (, ) S = (, ) T = (, ) Draw the quadrilateral and the x = rotation. 7. Draw a sketch. 8. Draw the three orthographic views. r = 9. See page 30 in the Student Reference Guide. true 30. Complete the two-column proof. Statements Reasons 1. Two pairs of parallel lines 1. Given. m 1 = m. angles are congruent. 3. m = m 3 3. angles are congruent. 4. m 1 = m 4. Property of Equality 60 Saxon Geometry Adaptations Lesson 95

31 LESSON 96 New Concepts Changing Perimeter of a Polygon Name Effects of Changing Dimensions on Perimeter and Area page 64 Example A rectangle is half as tall as it is long. If its height is reduced by half its original height, what is the ratio of the new rectangle s perimeter to the original rectangle s perimeter? The length of the rectangle is x. Its height is half its length, 0.5x. P = x + x + 0.5x + 0.5x P = 3x half of original height = 1 _ (0.5x) = 0.5x P = x + x + 0.5x + 0.5x P =.5x The ratio of the new rectangle s perimeter to the original rectangle s perimeter is.5:3, or 5:6. Example Changing Area of a Polygon Find the area of the triangle. Describe how each change affects the area. Find the original area: Find the new area: A = 1 _ (y)(y) A = 1 _ y A = 1 _ (y)(y) A = y The new-to-original ratio is :1. Example Altering the Dimensions of a Circle A circle s radius is increased by a factor of 3. Find the ratio of the circle s new area and circumference to its original area and circumference. Use x as the length of the original radius. The length of the new radius will be 3x. Original Circle Changed Circle A = π x A = π (3x) = 9π x C = π x B' B y y A y A' y C = π (3x) = 6π x C C' The ratio of the new to old areas is 9:1. The ratio of the new to old circumferences is 3:1. Adaptations Lesson 96 Saxon Geometry 603

32 Lesson Practice page 67 a. One pair of opposite sides of a square is dilated by a factor of 4. The other sides stay the same. What is the ratio of the new figure s perimeter to that of the original? Let x = length of a side of original square Perimeter of Original Square Perimeter of Dilated Square P = = P = x + x + + = The ratio of the new perimeter to the old perimeter is =. b. What is the ratio of the first trapezoid s area to the second trapezoid s area? A = 1_ (b 1 + b )h First Trapezoid ( + ) ( ) = ( + ) Second Trapezoid ( ) = ( ) The ratio of the first trapezoid s area to the second trapezoid s area is =. c. The radius of a circle is x. The radius is changed by a factor of 1_. A of original circle = A of new circle = ratio of original area to new area: C of original circle = C of new circle = ratio of original circumference to new circumference: = d. A block of apartments will be built in the shape of a square. During construction, the builders learn that the length of the lot must be shortened. The new length will be nine-tenths of the original. Let s equal the side length of the original square. A = Then s is the side length of the new square. A = ratio of the lot s new area to its original area: = 604 Saxon Geometry Adaptations Lesson 96

33 Practice page m F = Use the Law of Sines to set up a proportion. sin 7 = x sin. a. Sketch the circle and chords. b. If two chords intersect in a circle, then the products of the chord segments are equal. x JN = 3. If two nonvertical lines are perpendicular, then the product of their slopes is ( ) = ( ) = 16 8 y O x y = Use the perimeter to find the side length s. s = A original : A doubled : new side length = 7. See page 3 of the Student Reference Guide. 45 = π r ( + ) 6. Copy the figure onto graph paper. Cut it out and try to fold it., because when the net is up, the two hanging squares. 8. The semicircles can be combined to make 1 circle. radius of the circle = A circle = = r Find the area of the remaining regions. A = Total area = + = Adaptations Lesson 96 Saxon Geometry 605

34 Practice, continued page T: (x, y) (, ) The point (x, y) is rotated 10., the Law of Sines be used because it requires about the origin. that at least one triangle given, and its side be known. In the such pair exists. 11. π r = π 1. Write and solve a proportion. r = So r = A = k = 13. equation for shifted circle: ( ) + = ( ) 14. Leif translated the figure instead of and y 8 4 O x See page 3 of the Student Reference Guide. Find the perimeter: P = translating. 16. Draw a sketch. Write and solve a proportion to find the length x of the man s shadow. Circle the answer. A ft B 11,110 ft C ft D 5555 ft x = 606 Saxon Geometry Adaptations Lesson 96

35 Practice, continued page Complete the sketch of the solid. 18. Sketch two cylinders with the same approximate diameter. Use dotted lines for the cross sections. The cross section will have the larger area because the cross section in the The solid is a. direction it slants forming an. 19. See page 4 of the Student Reference Guide. What piece of information is needed? 0. number of feet in a yard: number of feet in a square yard: 10 yd = 10 ft = multiply by 1. Beginning with the right triangle with shorter leg z, find and label the length of its hypotenuse.. Find the area A m of a face of the model. A m = 1 _ ( )( ) = length of the 1st hypotenuse = Use the length of the 1st hypotenuse to find the length of the shorter leg of the nd triangle and then use that length to find the length of the nd hypotenuse. length of the nd hypotenuse = Proceed in the same manner through the 3rd and 4th triangles to find the length of hypotenuse x in terms of z. x = If two similar figures have a scale factor of a:b, then the ratio of their areas is a :b. Let A = number of square feet of glass needed for each pane 41 A = = A Adaptations Lesson 96 Saxon Geometry 607

36 Practice, continued page Classify the solid: 4. See page 3 of the Student Reference Guide. vertices: edges: bases: S = 5. There are in a circle. = 1 x x = 6. Sketch the triangles on a coordinate grid and find their areas. A of ABC = A of A B C = 7. (x ) + (y ) = 8 y 8. See page 3 of the Student Reference Guide. Let n = number of sides O 4 8 x 9. ( a_ tan θ b) sin θ = _ ( _) b) ( _ 1 _ = ( ) -4-8 n = 30. Let r = radius of the replica replica s circumference = clock tower circumference = ratio: = : = = 1 b_ c = 608 Saxon Geometry Adaptations Lesson 96

37 LESSON 97 Concentric Circles page 630 Name New Concepts Math Language Concentric circles are coplanar circles with the same center. Example Determining When Two Circles are Concentric Determine if the circles in each diagram are concentric. Explain your reasoning. The circles are not concentric. The circles are coplanar, but they intersect at one point, so they cannot have the same center. r 3 B r r A The circles with center A are concentric but B is not concentric with either of them. Given equations of circles in the form (x - h) + (y - k) = r, concentric circles have the same h and k values but different r values. Example Equations of Concentric Circles Write the equations for the concentric circles centered at (-, 3). Describe how the circles are alike and different. Both circles have center (-, 3). So, one side of each equation is (x + ) + (y - 3). smaller circle: r = 3 r = 9 larger circle: r = 9 r = 81 Write the equation for each circle. (x + ) + (y - 3) = 9 (x + ) + (y - 3) = 81 The circles are coplanar and they share the same center. They have different radii. The larger circle is the smaller circle dilated by a scale factor of y -4 O 4-4 x Adaptations Lesson 97 Saxon Geometry 609

38 Lesson Practice page 63 a. Concentric circles are coplanar and have the same center., they are, but do not have the same. b. Write the equations for these concentric circles. Tell how they are alike and different. The space between grid lines is units. 1 8 y The center of both circles is (, ). The radius of the smaller circle is. The radius of the larger circle is. 4-4 V(8, 4) 8 1 x The equation for the smaller circle is (x ) + (y ) =. The equation for the larger circle is (x ) + (y ) =. The circles coplanar. The circles have the same center. The circles have the same radius. The larger circle is the smaller circle dilated by a scale factor of. c. Find the area of the annulus. Express the answer in terms of π. The annulus is the region between two concentric circles. To find its area, subtract the area of the smaller circle from the area of the larger circle. radius of smaller circle = ft A smaller circle = radius of larger circle = ft A larger circle = A annulus = A larger circle - A smaller circle = - = d. Look at Example 4 on page 63. Give the probability a randomly thrown dart will hit the board outside the bull s eye and dark-shaded region. Express answers in terms of π. Given: radius of the central circle = in. Given: width of each annulus = 1 in. The value of r for the entire target is. The area A 1 of the entire target = The value of r for the central circle through circle 30 is. For the circular region of the board through 30, the area A =. 7 ft 4 ft The area of the target region is A 1 - A =. P(outside bull s eye and darker region) = area of the outer two rings = area of the entire target 610 Saxon Geometry Adaptations Lesson 97

39 Practice page He thought that the sine of an angle was. T : (x, y) (y, x) the of the lengths of the to the, A : (, ) B : (, ) C : (, ) D : (, ) E : (, ) instead of the other way around; the y 4 correct sine is O 4 x Subtract the area of the smaller circle from the area of the larger circle. Express answers in terms of π. 4. h = k = (x ) + (y ) = ( ) equation of the circle: A annulus = 5. Use the properties of a right triangle. 6. Draw a sketch. 16 = d 1 36 = d Find the difference in the distances. horizontal component: They have moved closer. Adaptations Lesson 97 Saxon Geometry 611

40 Practice, continued page To find the area of the annulus, subtract the area of the smaller circle from the area of the larger circle. Express answers in terms of π. P(annulus) = area of annulus area of the figure = = 8. D 0, k (x, y) (kx, ky); k = D 0, (3, 3) = (, ) D 0, (5, 1) = (, ) D 0, (1, 1) = (, ) For a dilation centered at (1, ), add 1 to each x-coordinate and to each y-coordinate. M(6, ) = M (, ) N(, ) = N (, ) P(, ) = P (, ) 9. Circle the answer y A 104 mm B 117 mm C 73 mm D 130 mm -4 - O 4 x See page 3 in the Student Reference Guide. r = L one can = Area of the sheet of metal = ft = in Divide to find the number of lateral surfaces. = lateral surfaces 13. Multiply each side by the scale factor. 1. For four points not to form a quadrilateral, at least must be collinear. So point must lie on the line segment connecting of the other points. 14. See page 3 in the Student Reference Guide. dilated perimeter: 3( ) + 3( ) + 3( ) = Multiply the by. 61 Saxon Geometry Adaptations Lesson 97

41 Practice, continued page To find the area of the annulus, subtract the area of the smaller circle from the area of the larger circle. Express answers in terms of π. 16. Taking the inverse of any function as the argument of that function will result in the argument of the = function; in this case,. 17. For any triangle, ABC: sin 17 = sin θ 1 a = b = c. sin A sin B sin C 18. The rhombus has side lengths, the product of the of adjacent sides of the rectangle is, and the slopes of the sides of the trapezium area all from θ 1 = θ = each other. 19. Assume that the m ABC is 10. Both of the rotations are about the original position of the central fountain. The figure formed by AB and BC is a. The perimeter is 700 = 1. a. Each is to one pair of sides and has a magnitude to the length of those sides. b. of rotation = center: c. Any reflection the 0. Use the Pythagorean Theorem to find the other side length. A = = = A community center A community center = cm Divide by 100 to find the area in m. parallelograms so that they parallelogram. match the original = Adaptations Lesson 97 Saxon Geometry 613

42 Practice, continued page Plot the given points on a coordinate plane. Connect the points to form the sides of the polygon. Then use the distance formula to find the length of each side. 4. equations: (x ) + (y ) = (x ) + (y ) = The circles have the center at (, ). r small circle = The polygon is a. r large circle = P = 5. x + (y ) = y 6. See page 30 in the Student Reference Guide. m NSM = O 8 16 x Converse: This is. 9. See page 8 in the Student Reference Guide. 8. The lengths of corresponding sides of similar triangles are proportional. w = 30. Conditional statements have the form If p, then q.,. x = 614 Saxon Geometry Adaptations Lesson 97

43 LESSON 98 Law of Cosines page 636 Name New Concepts Math Language The Law of Cosines relates the sides and angles of any triangle. Use the Law of Cosines when two side lengths and the included angle measure are given or all three side lengths of a triangle are given. In any triangle ABC: a = b + c - bc (cos A) b = a + c - ac (cos B) c = a + b - ab (cos C ) Example The Law of Cosines Finding Distance with the Law of Cosines Find c. Round your answer to the nearest tenth. A b C c a B Apply the Law of Cosines. c = a + b - ab(cos V ) c = (80)(60)(cos ) c c 33. U c T V Example Finding a Missing Angle with the Law of Cosines Find m C. Round your answer to the nearest degree. Use the Law of Cosines. c = a + b - ab(cos C ) 6 = (3)(57)(cos C ) -49 = -3648(cos C ) E cos C Use inverse cosine to find m C. m C 83. Example Using the Law of Cosines with a Right Triangle Find y. Round your answer to the nearest tenth. y = r + t - rt(cos S ) y = (1)(8)(cos 90 ) y = y = y Since cos 90 = 0, the third term of the equation is eliminated. Notice that after the term is eliminated, the Law of Cosines is identical to the Pythagorean Theorem. T 6 y 1 C D R 8 S Adaptations Lesson 98 Saxon Geometry 615

44 Lesson Practice page 638 a. Find c. Round your answer to the nearest whole number. Use c = a + b - ab(cos C). In the formula, C represents P. Substitute: m P = b = a = c = + - ( )( )(cos ) c = c b. Find m X. Round to the nearest degree. Use c = a + b - ab (cos C). Substitute: a = b = c = 7 7 = + - ( )( )(cos X ) cos X Use inverse cosine to find m X. m X, which rounded is c. Find a using the Law of Cosines. Round your answer to the nearest tenth. Then, suggest another way to find a. Use a = b + c - bc(cos A). In the formula A represents N. a = + - ( )( )(cos ) = + - a =, when rounded to the nearest tenth. Since LMN is a right triangle, you could also solve using. d. How far is Strom from Milan? Round to the nearest hundredth yard. Let P be the starting point. Let p be Strom s distance from Milan. Draw a sketch. Use p = b + c - bc(cos P ). p = + - ( )( )(cos P) p = + - p, which to the nearest hundredth of a yard =. 616 Saxon Geometry Adaptations Lesson 98

45 Practice page Use c = a + b - ab(cos C)., their images are the (3x) = ( ) + ( ) - ( )( )cos θ = + - cos θ = cos θ θ. It does not matter if the dilation or the translation occurs first, because the dilation changes only the of the circle, leaving the center unmoved, while the translation changes only the location of the of the circle, without changing the. The equation of the new circle is ( ) + ( ) =. 3. Subtract the area of the smaller circle from the larger circle to find the area of the annulus. 4. See page 8 in the Student Reference Guide. distance = A annulus = 5. To find the magnitude of a vector, place its initial point on the origin and use the distance formula. d = 7. sin θ = 0.6 sin θ = 0.6 cos θ sin θ cos θ = 0.6 cos θ = 0.6 cos θ = θ 6. See page 8 in the Student Reference Guide. 8. y = x = SX SY since they are both of the same circle. By the definition of triangles, SYX is isosceles. By the Isosceles Triangle Theorem, SXY Angles Theorem,. By the SXY and SYX. By the Property of Congruence, 1. Adaptations Lesson 98 Saxon Geometry 617

46 Practice, continued page Since the second floor is halfway up, it is a midsection of each triangular face. 10. Sketch u. Then draw as many vectors as you can that are perpendicular to u. side length of the second floor: A = There are angles of unique vectors; with above and below the positive x-axis. 11. Use c = a + b - ab(cos C ). 1 = cos θ = θ + - ( )( )(cos θ) 1. Redraw PQR so it has the same orientation as JKL. JKL and PQR are triangles. JK acute K JKL PQR by the Congruence Theorem _ 6 6 = = new height 14. Draw a sketch. P new = The widths of both figures make up 1_ of the perimeter of each figure. Apply the scale factor to find what portion the heights are of the new perimeter. P new = P + ( 1 _ P) 15. See page 30 in the Student Reference Guide. If the diagonals of a parallelogram are, it is a. other endpoint: (, ) 16. Draw a sketch of a right triangle. Let the shorter leg = 1 unit and then find the length of the other two sides. sin 60 = y = cos = y x = 618 Saxon Geometry Adaptations Lesson 98

47 Practice, continued page equation of larger circle: ( ) + = equation of smaller circle: 18. Use c = a + b - ab(cos C ). When m C = 90, cos C = means the entire expression, which ( ) + = The circles are with centers at of the smaller circle is circle is.. The radius and the larger c = a + b, is the = 0. The remaining equation,. 19. The angle is not the included angle. The angles opposite the 18-unit side 0. Use AAS Triangle Congruence, CPCTC, the Reflexive Property and the Hinge Theorem to complete the two-column proof. can either be information, it is or. From the given Statements 1. JMK LMH 1. Reasons to tell which.. LHK JKH. 1. Draw a sketch. Use the properties of a right triangle to find the exact length l of a leg. l = 3. Use FOIL to find (sin θ + cos θ ). (sin θ + cos θ ) = 3. JK HL 4. JKM Given. Use c = a + b - ab(cos C ). Let c = distance between Adrian and Shannon c = ( ) + ( ) - ( )( )cos c 4. See page 8 in the Student Reference Guide. m H = 87 The square root of sin θ + cos θ equal sin θ + cos θ. x = Adaptations Lesson 98 Saxon Geometry 619

48 Practice, continued page Convert percent to decimals and multiply by 360. Circle the answer. 6. tangents: common tangents: A red: 10 ; blue: 10 ; green: 10 B red: 145 ; blue: 15 ; green: 90 C red: 10 ; blue: 105 ; green: 75 D red: 160 ; blue: 60 ; green: Graph the given coordinates. The base of the square is dilated by 3, so it becomes a. P image = = A image = b h = = 9. If two similar figures have a scale factor of a:b, then the ratio of their perimeters is a:b, and the ratio of their areas is a : b. The larger circle will have a circumference times that of the smaller circle and an area that is 8. r entire figure = A entire figure = A r=5 = A r=1 = Find the sum of the areas of the two shaded regions. P = P = target regions area of entire circle 30. Trace, fold, copy. B m A t C times the area of the smaller circle. 60 Saxon Geometry Adaptations Lesson 98

49 LESSON 99 Volume Ratios of Similar Solids page 64 Name New Concepts Theorem 99-1 If two similar solids have a scale factor of a:b, then, 1. the ratio of the perimeters of their corresponding faces is a:b.. the ratio of the areas of their corresponding faces is a :b. 3. the ratio of their volumes is a 3 :b 3. Example Corresponding Perimeters of Similar Solids Two similar rectangular pyramids have a scale factor of 3:. Determine the perimeter of the smaller pyramid s base. First, determine the perimeter of the smaller pyramid s base. P = = 1 3 in. 7.5 in. By Theorem 99-1, the perimeter of the smaller pyramid s base is in a 3: ratio with the perimeter of the larger pyramid s base. 1 P = 3 _ Example P = 14 inches Finding Surface Areas of Similar Solids The surface area of the smaller pyramid is 54 cm. What is the surface area of the larger pyramid? Since the pyramids are similar, apply Theorem From the dimensions given, the ratio of the solid s sides is 15:6 or 5:. Solve for A. 5 = A 54 A = cm 6 cm The surface area of the larger pyramid is square centimeters. Adaptations Lesson 99 Saxon Geometry 61

50 New Concepts, continued page 643 Example Finding Volumes of Similar Solids The two cylinders shown are similar. The volume of the smaller cylinder is 38 cubic feet. What is the volume of the larger cylinder? Looking at the radii, the scale factor is 3:. By Theorem 99-1, the ratio of the volumes will be 33. Write a 3 proportion. 3 3 = V 3 38 V = ft The volume of the larger cylinder is 18.5 cubic feet. ft Lesson Practice page 644 a. Find the perimeter of the larger prism s base. For similar solids with a scale factor of a:b, the ratio of the perimeters of their corresponding faces is a:b. The formula for the perimeter of a triangle is P =. So the smaller prism s base is = in. 3_ 4 = 40 P 3P = P larger = in. b. The surface area of the smaller prism is 18 in. Find the surface area A of the larger prism. For similar solids with a scale factor of a:b, the ratio of the areas of their corresponding faces is a :b. 4 = A 9A = 18 A = in c. The volume of the smaller prism is 33 in 3. Find the volume V of the larger prism. For similar solids with a scale factor of a:b, the ratio of their volumes is a 3 : b 3. The volume of the smaller prism is. 3 4 = 33 3 V = 4 (33) Vlarger = in 3 d. Find the volume of the actual booster. Round to the nearest cubic meter. V = π r h For the model, r =, and h = 59. For the model, V = π. 1 3 V actual booster 3 = π V = π cm 3 To convert cm 3 to m 3, divide by V = m 3 6 Saxon Geometry Adaptations Lesson 99

51 Practice page c = a + b - ab(cos C) Substitute. x = + - ( )( )cos 78 x. See page 34 in the Student Reference Guide. length shorter leg = length hypotenuse = P = 3. Finish the sketch. 4. If two similar solids have a scale factor of a:b, the ratio of the areas of their corresponding faces is a : b. Find the area of each face of P. Use the ratio 3 to find the area of each 4 corresponding face of Q. 5. P pyramid P = 6. a. Substitute 0 for a. 4_ 5 = P pyramid P P pyramid Q P pyramid Q = base length 4 _ 5 = 6 _ l l = width 4 _ 5 = 4 w w = sloping edge 4 _ 5 = 10 s s = 7. Use the Law of Sines, AB 5 sin 108 = 0 sin θ = CB. sin C sin A x = b. Isolate y. Identify the slope and the y-intercept. m = b = - 1 Describe what happens to the slopes of the lines and their y-intercepts as a becomes larger. 8. Since only measure of the trapezoid is changing in the ratio :1, the ratio of the areas is changing by the ratio of: :. θ Adaptations Lesson 99 Saxon Geometry 63

52 Practice, continued page Use the Law of Cosines to find the angle at the fountain and the angle at the park. c = a + b - ab(cos C ). 110 = ( )(15)cos C 95 = (110)( )cos C Subtract the sum of the angles from 180 to find the angle at the art museum. 10. See page 9 in the Student Reference Guide. ratio of the smaller to larger perimeter: : ratio of the smaller to larger area: : fountain = park = art museum = 11. Find the following. 1. See page 3 in the Student Reference Guide. A entire disc = π A central hole = π = = r = A recordable region = A entire disc - A central hole A shaded region = ( )π - A central hole Find the ratio of the recordable area to the shaded region. Then use it to find the number of minutes of music that the entire disc can hold. time = 13. See page 9 in the Student Reference Guide. Use the Pythagorean Theorem to find AB and BC. AB AD BC CD L = 14. Sketch and label the triangle. Use c = a + b - ab(cos C). Find the measure of each angle. = A = B = C 64 Saxon Geometry Adaptations Lesson 99

53 Practice, continued page Let x = geometric mean 16. a. P = b. A = A triangle + A rectangle A = + = x = 17. sin θ + cos θ = sin θ = sin θ = - cos θ - cos θ Divide both sides of the equation by cos θ. θ = - cos θ 18. Each face of the triangular pyramid is an equilateral triangle. Draw a face of a triangular pyramid. Add an altitude. If the side length of each face is a, then, because each face can be split into two triangles, the area of each face is. Let the side length of each face of the other triangle = b. Similarly, the area of each face on the other pyramid is. So the ratio of the areas of 19. The larger the angle, the the opposite side. The smallest angle is C = 5, so the shortest side is, followed by, and then longest side. 1. See page 30 in the Student Reference Guide. length right shorter leg = their faces is :. 0. Find the radius of the entire bagel including its center hole. r = Subtract the smaller circle s area from the larger circle s area to find the area of the annulus. A = π - π =. fourth interior angle = Use the Pythagorean Theorem to find PW. h = Adaptations Lesson 99 Saxon Geometry 65

54 Practice, continued page She 4 by 4 to get instead of 4 and Let x = price of milk and let y = price of eggs y 4 y -4 - O 4 x Each of the four faces of A and B are right triangles. 6. Draw a sketch. Find the area of each triangular face, including the base. Add the areas to find S. S A = To find the surface area of B, set up a proportion: = S B Circle the answer. S B = 7. (x ) + (y ) = Draw the original circle and the translated circle on a coordinate grid. 9. Draw a sketch. Equilateral triangles have sides. Since one side coincides with a side of the hexagon, which is inches, all the side A kite C square B parallelogram D none of these 8. Use the Law of Sines to find m S. Then use the Law of Sines to find S. Circle the answer. A 4 B 14 C 1 D m PQ = m RS = PS, and PQ So PQRS is a trapezoid. m QR = m PS = lengths will be inches. 66 Saxon Geometry Adaptations Lesson 99

55 LESSON 100 Transformation Matrices page 648 Name New Concepts Corresponding numbers in matrices are in the same row and column in each matrix: and 7 are corresponding. To add matrices, add the corresponding numbers. Example Adding Matrices Math Language Add the two matrices. A matrix is an ordered, rectangular arrangement of numbers. Matrices are used to describe transformations on the coordinate plane. A point matrix is a matrix that represents coordinates. These can be coordinates of a line segment or a polygon. The first row of a point matrix contains x- values and the next row contains y-values Add the components of each matrix Example 4 0 = + (-1) (-1) + 0 = 1 3 Adding a Matrix to a Point Matrix Write a point matrix for the line segment AB. Add the point matrix to the matrix given below Finally, graph the new segment next to the original segment to show the transformation. The point matrix for AB is Add the matrices = = O y 7-1 A 4 B 6 x The translation is one unit to the right and up one unit. Adaptations Lesson 100 Saxon Geometry 67

56 Lesson Practice page 650 a. Add the matrices = + = = 1 + (-1) = 0 The sum of the matrices is 0. b. Find a point matrix for AB. Row 1 contains x-values. Row contains y-values Add the transformation matrix O - -4 y B A x 4 Add (-1) to each number: 1 3 Point A = (, 1); Point B = (4, ) Describe the change. The matrix translates the line down unit(s) and unit(s). c. A line segment is translated 3 units left and 1 unit down. Write the transformation in a matrix. Add (-3) to each x-value; add (-1) to each y-value. matrix: -3-1 d. Ferris wheel seats are always horizontal. A rotation of the wheel translates a car 6 meters right and 5 meters down. Write a translation matrix to represent this transformation. The translation matrix is: 68 Saxon Geometry Adaptations Lesson 100

57 Practice page R larger circle = + in. A larger circle = π ( ) = π A smaller circle = π ( ) = π A annulus = A larger circle - A smaller circle =. To find the magnitude of a vector, place its initial point on the origin and use the distance formula. d = (x - x 1 ) + (y - y 1 ) = = (60 - ) + ( ) 3. A transformation across a line so that the image is congruent to the preimage but has a different orientation than the preimage is a. 4. If two similar solids have a scale factor of a:b, the ratio of the areas of their corresponding faces is a : b. S A = 3 S B 5, so S B = S A. Circle the answer: A S B = _ 3 S 5 A B S B = 5 S 9 A C S B = 9 5 S A D S B = 15 7 S A 5. x-values in first row, y-values in second row 6. Solve for sin θ in sin θ + cos θ = 1. horizontal change: vertical change: L M N -4-1 unit(s) unit(s) 7. What is the measure of JK if v = 3.? sin θ + cos θ = 1 sin θ = 1 - cos θ sin θ = sin θ = sin θ 8. Corresponding parts of the segments are proportional. Draw a sketch. Write a proportion to solve for y and x. m JK = y = x = Adaptations Lesson 100 Saxon Geometry 69

58 Practice, continued page See page 7 in the Student Reference Guide. 10. x = y = Sketch a right triangle. Draw altitudes. The orthocenter is at the of the matrix. 11. Look at the front, top, and side views to eliminate choices. Circle the answer. A B 1. The equation of a circle is. When the circle is centered at the origin, h and k =. Since r is given as.5, the equation is. C D 14. cos = V x V x = 13. a:b = : _ radius: 3 = r = _ height: 3 = h = 15. Sketch the net of a pentagonal pyramid. = sin = V y, 16. Use the Law of Cosines. = m P = = m R = + + V y = - ( )( )(cos P ) - ( )( )(cos R ) Use the Triangle Angle Sum Theorem to find Q. m Q = 630 Saxon Geometry Adaptations Lesson 100

59 Practice, continued page The diagonals of a square or are perpendicular. The diagonals of a square are length. The quadrilateral that has diagonals of different lengths is a. 18. If two solids have a scale factor of a:b, the ratio of areas of corresponding faces is a : b. A square pyramid has triangular faces. Its base is a. Find the areas of the triangles and base. 3 8 = A small base A A large base = large base 3 8 = A small face A A large face = large face 19. There is(are) common tangent(s). 0. a Law of Sines: sin A = b sin B = c sin C Law of Cosines: a = b + c - bc cos θ In Law of Sines, number of sides involved ; number of angles involved. In Law of Cosines, number of sides involved ; number of angles involved. In a triangle with given angles, number of sides involved ; number of angles involved. The law that can be used to find a side with the given triangle is. 1. S = L + B; for the area of the bases, use the Pythagorean theorem or triangle.. Use formula for lateral surface area of a pyramid and calculate area of base. area of 1 face = perimeter = area of all faces = lateral surface area = area of 1 equilateral base = 1 _ area of all equilateral bases = area of base = total area = + total surface area = + Adaptations Lesson 100 Saxon Geometry 631

60 Practice, continued page x-values in first row; y-values in second row Donna confused columns and. The correct matrix is Alter (y + 1) = x to make its graph symmetrical about the x-axis. Translate the equation one unit by subtracting from y + 1. The equation would be. 5. Sketch QRST and JKLM. 6. (x - h) + (y - k) = r r is increased by a factor of. ST = equation of second circle: 7. Translation matrix = final matrix - initial matrix Translation matrix: x 1 y 1 + x - (x - 1) - x 1 y 1 8. Find r, then use the formula for the surface area of a sphere. 8 = r r = = 9. Simplify and cross multiply = 33 y y = 30. Statements 1. is the center of the circle.. PA PC 3. APB S sphere = 4π 1. Reasons. Radii of the same circle are. 3. Theorem 4. PAB 4. SAS Theorem 63 Saxon Geometry Adaptations Lesson 100

61 INVESTIGATION 10 Name Fractals page 653 Math Language Iteration is the repeated application of the same rule. A fractal is a figure generated by iteration. Fractals are created using combinations of transformations. One famous fractal is the Sierpinski triangle. It is an iteration of an equilateral triangle in which an inverted equilateral triangle is removed from the interior of the original equilateral triangle. This is done by drawing the original triangle s midsegments and then removing the central triangle. This fractal is a transformation that consists of a rotation and a dilation of the original equilateral triangle. 1. Connect the midsegments of the shaded triangles. Shade the upright triangles. The result is the second iteration. Math Language A self-similar figure can be divided into parts, each of which is similar to the entire figure. Copy the new figure and repeat this process. The result is the third iteration.. The parts of the Sierpinski triangle similar to the entire figure. It self-similar figure. 3. A Sierpinski triangle is a fractal formed from a(n) by removing triangles with vertices at the of the of each remaining. Adaptations Investigation 10 Saxon Geometry 633