Inverse of Linear Functions PDF
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This document is a lesson on finding the inverse of linear functions. It provides examples, explaining how to rewrite functions using 'y' instead of 'f(x)', switching 'x' and 'y', and solving for 'y' to find the inverse function. Graphs are included to visually demonstrate inverses as reflections over the line y = x.
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LESSON 5.4 Name Reteach Inverses of Linear Functions To find the inverse of a linear function, switch the x and the y in the equation. Find the Complete the missing information. in...
LESSON 5.4 Name Reteach Inverses of Linear Functions To find the inverse of a linear function, switch the x and the y in the equation. Find the Complete the missing information. inverse of Find the inverse of the function f(x) = −3x + 7. a linear function. First, rewrite the function using a y instead of function notation: y= x+. Switch the x and the y, and solve for y. x= - 3 +7 x - 7 = -3y x - 7 =y _ 7 = y 1 x - _ _ 1 x+_ The inverse function is g(x) = − _ 7. Graph Fill in the missing information. inverses. Graph the inverse of f(x) on the same axes as f(x). y 8 © Houghton Mifflin Harcourt Publishing Company 4 x -8 -4 0 4 8 -4 f(x) -8 The inverse of f(x) is the reflection of f(x) over the line. 1. Find the inverse of the function f(x) = __ 12 x - 4. Show your work. Algebra 1 52 Reteach You can use functional composition to determine if two functions are inverses. Show that two Complete the missing information. functions are inverses. Show that the function f(x) = 2x - 5 is the inverse of g(x) = __ 12 x + __ 52 . ( f(g(x)) = 2 _ 2 ) 5 - 5 1 x + _ 2 g(f(x)) = _ ( 1 2 x- ) 5 + _ 2 = x+ -5 =x - + =x =x Because both compositions equal x, the two functions are inverses. Show that two Complete the missing information. functions are not Show that f(x) = - __32 x + 1 and g(x) = - __23 x - 1 are not inverses. ( ) inverses. 3 - _ f(g(x)) = - _ 2 x - + 2 3 =x+ +1 =x+ Because f(g(x)) ≠ , the two functions are not inverses. 2. Show that the functions f(x) = x + 6 and g(x) = x − 6 are inverses. © Houghton Mifflin Harcourt Publishing Company 3. Show that the functions f(x) = −x + 7 and g(x) = −x − 7 are not inverses. Algebra 1 53 Reteach LESSON 8.2 Name Reteach Graph Absolute Value Functions Complete the solution by providing the missing information. Understand Translation Rule translations. h>0 Translation h units [right / left] h0 Translation k units [up / down] k0 For x < h, slope = -a Vertical stretch (a > 1) reflections. For x > h, slope = a Vertical compression (0 < a < 1) a h, slope = -a Vertical compression (-1 < a < 0) ( ) Graph g(x) = a⎜x - h⎟ + k. The vertex of g(x) = -2⎜x - 2⎟ + 4 is ,.a= and a 0. The slope of g(x) for x < is. The slope of g( x) for x > is. y 8 4 x -8 -4 0 4 8 -4 -8 Graph each function. State the vertex and the slopes. 3. 1 ⎜x⎟ g(x) = −_ 4. h(x) = 2⎜x - 3⎟ + 2 4 y y 8 8 © Houghton Mifflin Harcourt Publishing Company 4 4 x x -8 -4 0 4 8 -8 -4 0 4 8 -4 -4 -8 -8 Algebra 1 71 Reteach 10.1 Name Step It Out Learn the Math EXAMPLE 1 It is possible to write a linear y 4 inequality based on a graph. To write the inequality, determine the equation for the 2 boundary line and the inequality symbol. x The boundary line has a y-intercept at -3 and a slope -4 -2 0 2 4 of - __41 , or -4. The equation is y = -4x - 3. -2 The shading is below the line, and the boundary line is dashed. The inequality is y < -4x - 3. -4 Check this inequality using substitution. A point in the solution region A point on the boundary line A point not in the solution region should make the inequality true. should make the inequality false. should make the inequality false. Points on boundary lines that are dashed are not part of the solution region. (-2, -2)is in the solution region. ( 0, -3)is on the boundary line. ( 0, 0)is not in the solution region. (-2) < -4(-2) - 3 ( -3) < -4(0) - 3 ( 0) < -4(0) - 3 -2 < 5 -3 < - 3 0 2x + 3 2. y ≤ 2x + 3 3. y < 2x + 3 4. y ≥ 2x + 3 A. y B. y 4 4 2 2 x x -4 -2 0 2 4 -4 -2 0 2 4 -2 -2 -4 -4 C. y D. y 4 4 2 2 x x -4 -2 0 2 4 -4 -2 0 2 4 -2 -2 -4 -4 © Houghton Mifflin Harcourt Publishing Company Tell whether each ordered pair is a solution of the inequality. 5. 4x - 2y > 10; ( 1, 3) 6. x + 4y < 0; ( 0, 0) 7. x ≥ -9; ( -9, 2) 8. y < 7; ( 5, 12) 9. 3x - y ≤ 3; ( 2, -6) 10. -2x - 5y > 7; ( -1, -10) Determine whether each ordered pair is a solution of y the inequality graphed. 8 11. (2, -2) 4 12. (0, -2) x -8 -4 0 4 8 13. (-6, 0) -4 14. (6, 7) -8 Algebra 1 127 Journal and Practice Workbook 15. Open Ended Use the graph shown. y A. What inequality symbol boundary line is solid and the area below the 8 is indicated by the graph? line is shaded B. Write the linear inequality that is represented 4 by the graph. x yy -2x + 10 y -2x -2x + 10 + 10 -8 -4 0 4 8 -4 C. Describe a situation that might be represented by the graph. -8 McKayla has $10. Trading cards cost $2 each, and marbles cost $1 each. McKayla writes and graphs an inequality to determine how many trading cards and marbles she can buy with her $10. 16. Math on the Spot The Hungarian club needs to earn at least $175 from a bake sale to fund their Goulash Festival. They are selling croissants for $2.00 each and loaves of French bread for $5.00 each. Write a linear inequality to describe the situation and graph the solutions. 17. Jude’s family is planning a road trip. They are planning a budget and have $600 to spend on accommodations. They know that campsites cost $20 per night and hotel rooms cost $140 per night. A. Write a linear inequality to describe this situation. Let x represent the number of nights that Jude’s family camps and let y represent the number of nights that Jude’s family stays in a hotel. 2x+140y 20x + 140y 60 20x + 140y 60 solid line © Houghton Mifflin Harcourt Publishing Company B. Describe the boundary line of the graph of the inequality. solid line C. Interpret the point ( 16, 2). Is this a valid solution for the situation? The point (16, 2) represents 16 nights of camping and 2 nights in a hotel. D. Jude’s family wants to take the longest vacation possible. Suggest a solution. 18. Which inequality has the same graph as the inequality -6x + y > 20? A -6x + y < -20 C y > -6x + 20 B y > 6x + 20 D y < 6x + 20 19. Which ordered pair is NOT in the solution region of the inequality __ 12 x + 4y ≤ 8? (0, 2) A C (0, 16) B (16, 0) D (8, 1) Algebra 1 128 Journal and Practice Workbook Learn the Math EXAMPLE 2 Identify the vector that maps ABCD to A′B′C′D′. y A′ A 2 B′ x D′ -4 -2 0 2 4 B D C′ -2 C -4 Make a table to find the difference between the vertices of the preimage and image. Preimage Image Difference A(−3, 2) A′(0, 4) 0 − (−3) = 3, 4 − 2 = 2 B(−4, −1) B′(−1, 1) −1 − (−4) = 3, 1 − ( −1) = 2 C(−2, −3) C′(1, −1) 1 − (−2) = 3, −1 − ( −3) = 2 D(−1, −2) D′(2, 0) 2 − (−1) = 3, 0 − ( −2) = 2 The vector that maps ABCD to A′B′C′D′ is 〈 3, 2〉. Do the Math Identify the vector that maps QRST to Q′R′S′T′. © Houghton Mifflin Harcourt Publishing Company T y Preimage Image Difference 4 Q(1, 1) Q′ ( , ) −1= , −1= S′ S T′ 2 Q x ( ) R -4 -2 0 2 R(−2, 1) R′ − (−2) = −1= Q′ , , R′ -2 S(−4, 2) S′ ( , ) − (−4) = , −2= T(−1, 4) T′ ( , ) − (−1) = , −4= The vector that maps QRST to Q′R′S′T′ is. Geometry 64 Journal and Practice Workbook LESSON 5.1 Name More Practice ONLINE ONL IN E Video Tutorials and Video Tutorials and Interactive Examples Draw the figure with the given vertices. Then draw its Interactive Examples image after a translation by the given vector. 1. A(4, -2), B(1, -1), C(-3, 2); 2. A(-2, 4), B( 2, 6), C( 4, 0); vector 〈 −2, 5〉 vector 〈 3, −1〉 y y 6 6 4 4 2 2 x x -2 0 2 4 6 -4 -2 0 2 4 -2 3. A(1, 4), B(4, -3), C(-3, -2), D(-1, 3); 4. A(4, 2), B(-2, 3), C(-2, 0), vector 〈 2, −3〉 D(1, -1); vector 〈−4, 2〉 y y 4 4 2 x 2 -2 0 x 2 4 6 -2 -6 -4 -2 0 2 4 -2 -4 © Houghton Mifflin Harcourt Publishing Company -4 -6 Give the component form of a vector that maps △JKL to △J′K′L′. 5. J(−5, 2), K(−1, −3), L(1, 5) 6. J(3, 6), K(4, -1), L(6, 2) J′( −3, −1), K′( 1, −6), L′( 3, 2) J′( 6, 7), K′( 7, 0), L′( 9, 3) Preimage Image Difference J(-5, 2), J'(-3, -1) -3 -(-5)= 2, -1 - 2 = -3 K (-1, -3), K'(1, -6) 1 -(-1)= 2, -6 -(-3)= -3 L(1, 5) ,L' (3, 2) 3 - 1 = 2, 2-5= -3 Geometry 65 Journal and Practice Workbook Lesson 5.1 Step It Out Do the Math (Example 1) Preimage (x, y)→ x + ( 2 ,y− 3 ) Image X(-1, 4) ( −1, 4) → -1 + ( 2 , 4 − 3 ) ( 1 , 1 ) Y(2, 2) ( 2, 2) → ( 2 + 2 , 2 − 3 ) ( 4 , -1 ) Z(0, -1) ( 0, −1) → ( 0 + 2 , -1 − 3 ) ( 2 , -4 ) X y 4 2 Y X′ x -4 -2 0 2 4 Z Y′ -2 -4 Z′ Do the Math (Example 2) © Houghton Mifflin Harcourt Publishing Company Preimage Image Difference Q(1, 1) Q′ ( 0 , -1 ) 0 − 1 = -1 , -1 − 1 = -2 R(−2, 1) R′ ( -3 , -1 ) -3 − (−2) = -1 , -1 − 1 = -2 S(−4, 2) ( S′ -5 , 0 ) -5 − (−4) = -1 , 0 − 2 = -2 T(−1, 4 ) ( T′ -2 , 2 ) -2 − (−1) = -1 , 2 − 4 = -2 The vector that maps QRST to Q′R′S′T′ is 〈−1, −2〉. Geometry AK36 Journal and Practice Workbook Draw the figure with the given vertices. Then draw its image after a translation by the given vector. Lesson 5.1 More Practice A(4,-2), A (4, -2), BB(1,-1), (1, -1), CC(-3,2); Vector (-3, 2); The vector is 2, 5 1. C′ y Preimage (x, y) → ( x - 2, y + 5) Image 6 A(4, -2) (4, -2) → (4 - 2, -2 + 5) A'(2, 3) 4 B(1, -1) (1, -1) → (1 - 2, -1 + 5) B'(-1, 4) B′ A′ C 2 C(-3, 2) (-3, 2) → (-3 - 2, 2 + 5) C'(-5, 7) x -4 -2 0 2 4 B A(-2,4),B (2,6) , C(4,0) ; Vector < 3, -1> A 2. (x, y) → ( x + 3, y - 1) y B Preimage Image 6 B′ A(-2, 4) (x, y) → ( -2 + 3, 4 - 1) A'(1, 3) A 4 A′ B(2, 6) (x, y) → ( 2 + 3, 6 - 1) B'(5, 5) 2 C(4, 0) (x, y) → ( 4 + 3, 0 - 1) C'(7, -1) x -2 0 2 4C 6 C′ -2 A(1,4), B(4,-3) , C(-3,-2) D(-1,3) ;Vector 3. (x, y) → ( x + 2, y - 3) y Preimage Image 4 A D A(1, 4) (1, 4) → ( 1 + 2, 4 - 3) A'(3, 1) 2 A′ B(4, -3) (4, -3) → ( 4 + 2, -3 - 3) B'(6, -6) D′ x -2 0 2 4 6 C(-3, -2) ( -3, -2) → ( -3 + 2, -2 - 3) C'(-1, -5) © Houghton Mifflin Harcourt Publishing Company C -2 D(-1, 3) (-1, 3) → ( -1 + 2, 3 - 3) D'(1, 0) B -4 C′ -6 B′ A(4,2) ,B(-2,3),C( -2,0),D(1,-1); Vector 4. B′ y Preimage (x, y) → (x - 4, y + 2) Image 4 A′ B A(4, 2) (4, 2) → ( 4 - 4, 2 + 2) A'(0, 4) 2 A B(-2, 3) (-2, 3) → (-2 - 4, 3 + 2) B'(-6, 5) C′ D′ x C(-2, 0) (-2, 0) → (-2 - 4, 0 + 2) C'(-6, 2) -6 -4 C 0 2 4 D -2 D(1, -1) (1, -1) → (1 - 4, -1 + 2) D'(-3, 1) -4 Geometry AK37 Journal and Practice Workbook Give the component form of a vector that maps JKL to JKL. 5. J(5, 2), K(1, 3), J(-5,2), L(1, 5) L(1,5), K(-1,-3), J (3, 1), J' K (1, 6), LK(3,(1,-6), (-3,-1), 2) L (3,2) 5. Preimage Image Difference J(-5, 2) J' (-3, -1) -3 - (-5) = 2, -1 - 2 = -3 K(-1, -3) K' (1, -6) 1 - (-1) = 2, -6 - ( -3) = -3 L(1, 5) L' (3, 2) 3 - 1 = 2, 2 - 5 = -3 Give the component form of a vector that maps JKL to JKL. 〈2, -3〉 J(5, 2), K(1, 3), L(1, 5) J (3, 1), K (1, 6), L (3, 2) 6. Preimage Image Difference J(3, 6) J' (6, 7) 6 - 3 = 3, 7 - 6 = 1 K(4, -1) K' (7, 0) 7 - 4 = 3, 0 - (-1) = 1 L(6, 2) L' (9, 3) 9 - 6 = 3, 3 - 2 = 1 〈3, 1〉 7. Answers will vary. 8. First position: ( 7, -7) Second position: ( 7 - 14, -7) = ( -7, -7) Final position: ( -7, -7 + 21) = ( -7, 14) Subtract the first position from the final position to find the vector. -7 - 7 = -14; 14 - ( -7) = 21 〈 -14, 21〉 9. Translation vector: 〈2, 5〉 © Houghton Mifflin Harcourt Publishing Company Position of the library: ( 6 + 2, -3 + 5) = ( 8, 2) 10. Samantha is correct. Preimage (x, y) → ( x - 3, y + 3) Image F(5, -2) (x, y) → ( 5 - 3, -2 + 3) F'(2, 1) G(3, 4) (x, y) → ( 3 - 3, 4 + 3) G'(0, 7) H(-2, 3) (x, y) → ( -2 - 3, 3 + 3) H'(-5, 6) 11. Point Francis marked: ( 7, -12) Point friend marked: ( -8, 8) -8 - 7 = -15; 8 - (-12) = 20 The vector is 〈 -15, 20〉. The correct answer is D. Geometry AK38 Journal and Practice Workbook 5.2 Name Step It Out Learn the Math EXAMPLE 1 Find the center of rotation and angle of C′ A rotation of the rotation that maps △ABC onto △A′B′C′. B′ Draw a line segment between the two sets of corresponding B points. Draw the perpendicular bisectors of the points. The point of intersection of the perpendicular bisectors A′ C is the center of rotation. C′ A B′ H B A′ C Connect two corresponding vertices to the center of rotation. Measure this angle. It is the angle of rotation. C′ A 180° B′ H B A′ C © Houghton Mifflin Harcourt Publishing Company The angle of rotation is 180°. Do the Math Find the center of rotation and angle of rotation of the rotation that maps △XYZ onto △X′Y′Z′. Y′ X X′ Z′ Y Z ° The angle of rotation is. Geometry 67 Journal and Practice Workbook Learn the Math EXAMPLE 2 Draw the preimage and image of the polygon with vertices A(3, 3), B(4, 1), C(2, -1) and after a rotation counterclockwise 90°. Make a table to find the vertices of the image. Draw the preimage and image in a coordinate plane. B′ y 4 A Preimage (x, y) → ( -y, x) Image A′ 2 C′ A(3, 3) (3, 3) → ( -3, 3) A′(-3, 3) B x B(4, 1) ( 4, 1) → (-1, 4) B′(-1, 4) -4 -2 0 2 4 -2 C C(2, -1) ( 2, -1) → ( 1, 2) C′(1, 2) -4 Do the Math Draw the preimage and image of the polygon with vertices A(-4, 7), B(2, 3), C(-1, -1), and D(-5, 4) after a rotation counterclockwise 270°. Learn the Math EXAMPLE 3 Describe any rotations less than 360° that map the hexagon onto itself. Divide 360° by 6, the number of sides in the hexagon, to find the rotations that will map the hexagon onto itself. © Houghton Mifflin Harcourt Publishing Company 360° _ = 60° 6 Rotations of 60°, 120°, 180°, 240°, and 300° will map the hexagon onto itself. Do the Math Describe any rotations less than 360° that map a regular pentagon onto itself. ° ° ° ° Rotations of , , , and will map a pentagon onto itself. Geometry 68 Journal and Practice Workbook LESSON 5.2 Name More Practice ONLINE ONL IN E Video Tutorials and Video Tutorials and Interactive Examples Draw the preimage and image of each polygon under the given rotation. Interactive Examples 1. A(-1, 4), B( 3, 4), C(2, -1), D(0, 1) 2. A(-5, -6), B(-4, 1), C(1, -2), D(-2, -5) Counterclockwise 270° Counterclockwise 90° y y 4 2 2 x x -4 -2 0 2 4 6 -4 -2 0 2 4 -2 -2 -4 -4 -6 3. A(0, 3), B(3, 5), C(1, -2), D(-1, 0) 4. A(2, 4), B(3, 1), C(3, 0), D(1, -2) Counterclockwise 180° Counterclockwise 90° y y 4 4 2 2 x x -4 -2 0 2 4 -4 -2 0 2 4 -2 -2 -4 -4 © Houghton Mifflin Harcourt Publishing Company Describe any rotations less than or equal to 360° that map the polygon onto itself. 5. regular octagon 6. equilateral triangle Geometry 69 Journal and Practice Workbook 7. Use Repeated Reasoning Given the preimage of a triangle with vertices at L(4, -6), M(5, -1), and N(7, -3), rotate the image counterclockwise 810°. What are the coordinates of the vertices of the image? 8. Math on the Spot Rotate △ABC with vertices A(-1, 1), B(-4, 2), and C(-2, 5) C y by 90° counterclockwise about the origin. 4 B 2 A x -4 -2 0 2 4 -2 -4 9. Reason Cindy says that she has a regular polygon in her hands that rotates onto itself when rotated 90°. What regular polygon does Cindy have? Explain your reasoning. 10. Reason Andrea draws a figure in the first quadrant of a coordinate plane. She rotates © Houghton Mifflin Harcourt Publishing Company the figure counterclockwise. The x-coordinates of the image are the same as the y-coordinates of the preimage. The y-coordinates of the image are the opposite of the x-coordinates of the preimage. What is the smallest angle of rotation of the image? Explain your reasoning. 11. Hong is playing a board game and spins a spinner. Before spinning, the point of the spinner is at ( -2, 6). After spinning, the point is at ( -6, -2). What is the angle of rotation between the starting point and ending point of the spinner? A 90° counterclockwise C 270° counterclockwise B 180° counterclockwise D 360° counterclockwise Geometry 70 Journal and Practice Workbook Lesson 5.2 Step It Out Do the Math (Example 1) Y′ X X′ Z′ Y Z ° The angle of rotation is 90. DoPreimage ( x, y)2) the Math (Example → ( y, -x) Image A(-4, 7) (-4, 7) → ( 7, 4) A′(7, 4) B(2, 3) ( 2, 3) → ( 3, -2) B′(3, -2) C(-1, -1) ( -1, -1) → ( -1, 1) C′(-1, 1) D(-5, 4) (-5, 4) → (4, 5) D′(4, 5) A y 6 D′ 4 A′ D B © Houghton Mifflin Harcourt Publishing Company 2 C′ x -4 -2 0 2 4 6 C -2 B′ Do the Math (Example 3) 360° _ = 72° 5 ° ° ° ° Rotations of 72 , 144 , 216 , and 288 will map a pentagon onto itself. Geometry AK39 Journal and Practice Workbook Lesson 5.2 More Practice 1. y Preimage (x, y) → (y, -x) Image A 4 B A(-1, 4) (-1, 4) → ( 4, 1) A′(4, 1) 2 A′ D B(3, 4) ( 3, 4) → ( 4, -3) B′(4, -3) D′ x -4 -2 0 4 C(2, -1) ( 2, -1) → ( -1, -2) C′(-1, -2) C C′ D(0, 1) (0, 1) → ( 1, 0) D′(1, 0) B′ -4 2. y Preimage ( x, y) → ( -y, x) Image 2 A(-5, -6) (-5, -6) → ( 6, -5) A′(6, -5) B C′ x B(-4, 1) ( -4, 1) → ( -1, -4) B′(−1, −4) -4 -2 0 2 4 6 C(1, -2) ( 1, -2) → ( 2, 1) C′(2, 1) -2 C D′ D(-2, -5) (-2, -5) → ( 5, -2) D′(5, -2) B′ D -6 A′ A 3. y Preimage ( x, y) → ( -x, -y) Image B 4 A(0, 3) (0, 3) → (0, -3) A′(0, -3) A C′ B(3, 5) ( 3, 5) → ( -3, -5) B′(-3, -5) D′ x © Houghton Mifflin Harcourt Publishing Company C(1, -2) ( 1, -2) → ( -1, 2) C′(-1, 2) -4 -2 D 0 2 4 D(-1, 0) (-1, 0) → ( 1, 0) D′(1, 0) -2 C A′ -4 B′ 4. y A Preimage ( x, y) → ( -y, x) Image 4 B′ C′ A(2, 4) (2, 4) → (-4, 2) A′(-4, 2) A′ 2 B(3, 1) ( 3, 1) → (-1, 3) B′(-1, 3) D′ B x -4 -2 0 2 C4 C(3, 0) ( 3, 0) → ( 0, 3) C′(0, 3) -2 D D(1, -2) (1, -2) → ( 2, 1) D′(2, 1) -4 Geometry AK40 Journal and Practice Workbook 360° 9. Cindy has a square. 5. _ = 45° 8 360° _ = 90° Rotations of 45°, 90°, 135°, 180°, 225°, 270°, 4 315°, and 360° will map a regular octagon 10. The smallest angle of rotation is 270° because onto itself. a rotation of 270° counterclockwise produces 360° a transformation where the x-coordinates are 6. _ = 120° 3 the same as the y-coordinates of the preimage Rotations of 120°, 240°, and 360° will map an and the y-coordinates are the opposite of the equilateral triangle onto itself. x-coordinates of the preimage, (x, y) → ( y, -x). 7. A rotation of 810° counterclockwise is the same 11. 90° counterclockwise, because the transformation as a rotation of 90° counterclockwise. (−2, 6) → ( −6, −2) is ( x, y) → ( -y, x). The correct answer is A. Preimage (x, y) → ( -y, x) Image L(4, -6) (4, -6) → (6, 4) L′(6, 4) M(5, -1) (5, -1) → ( 1, 5) M′(1, 5) N(7, -3) (7, -3) → ( 3, 7) N′(3, 7) 8. The rotation of ( x, y) is ( -y, x). A(−1, 1) → A′(−1, −1) B(−4, 2) → B′(−2, −4) C(−2, 5) → C′( −5, −2) C y 4 © Houghton Mifflin Harcourt Publishing Company B 2 A x -4 -2 A′0 2 4 C′ -2 -4 B′ Geometry AK41 Journal and Practice Workbook LESSON 9.3 Name More Practice ONLINE ONL IN E Video Tutorials and Video Tutorials and Interactive Examples Interactive Examples _ _ 1. CE bisects ∠C. Find x. 2. BD bisects ∠ABC. Find x C B (5x - 2)° (3x + 18)° 36° A D x° 57° C A E B _ _ 3. CD bisects ∠C. Find x. 4. AD bisects ∠CAB. Find x A 96° A (5x + 5)° D C (2x + 2)° 36° B 86° B C D (7x - 1)° 5. If P is the incenter of _ △FGH and P W= 5, 6. Construct the angle bisectors and incenter of what is the length of PV ? the triangle below. H W A F X P © Houghton Mifflin Harcourt Publishing Company V G B C 7. If P is the incenter of △FGH and 8. If the area of the circle with center_ A is 121π m∠GPV = 65° and m∠GHF = 96°, square feet, what is the length of BA ? what is m∠PFV? D H W F X P E A B C V G F Geometry 127 Journal and Practice Workbook 9. Math on the Spot Find the angle measures in the triangle, given that point Q is the incenter of △FGH. G 100° Q 23° 5.9 F P H 10. Use Structure An architect is designing a circular window to be installed in the gable of a house. The metal window frame will be 1 foot wide and will be bordered by the roof above and a wood beam below. The distance from the center of the window to the roof is 3 feet. What will be the circumference of the circular glass pane that will be needed? 1 ft 3 ft © Houghton Mifflin Harcourt Publishing Company _ _ CD are angle bisectors of △CAB. and BD 110° A Find each measure. _ D 11. distance from D to AB 5 12. m∠DCA C E B 21° 13. m∠EDB _ bisects ∠GBC. Solve for x. 14. AB G A 41 47° B 49 A C 82 D 98 x° B 35° C Geometry 128 Journal and Practice Workbook Lesson 9.3 More Practice 1. m∠C = m∠ACE + m∠ECB ∠GFQ ≅ ∠QFP, so 9. m∠C = 36° + 36° m∠QFP = m∠GFQ = 23° m∠C = 72° m∠GHF = 180° − ( 100° + 23° + 23°) = 34° m∠A + m∠B + m∠C = 180° m∠A = x° m∠GFH = 2(23°) = 46° x° + 57° + 72° = 180° ∠GHQ ≅ ∠QHF, so x = 51 m∠GHQ = ( m∠QHF)= 2. 5x - 2 = 3x + 18 2x = 20 _ 1 (34)° = 17° 1 m( ∠GHF)= _ 2 2 x = 10 ∠QFP ≅ ∠QFG 3. 2(2x + 2)° + 96° + 36° = 180° m∠QFP + m∠FPQ + m∠FQP = 180° 2(2x + 2)° = 48° 23° + 90° + m∠FQP = 180° 2x + 2 = 24 m∠FQP = 67° 2x = 22 m∠FQH = 180° − 23° − 17° = 140° x = 11 1 (m∠FQH) = _ m∠HQP = _ 1 (140°) = 70° 4. 86° + 2(5x + 5)° + ( 7x − 1)° = 180° 2 2 10x + 10 + 7x − 1 = 94 10. The distance in feet from the center of the window to the roof is the radius of the 17x = 85 window plus the width of the frame. Therefore, x = 5 r = 3 − 1 = 2. _ _ 5. _ P is the incenter of △FGH, PW Since ≅ PV ≅ C = 2πr PX and PW = PV = PX. So, by the Substitution C ≈ 2(3.14159)(2) Property of Equality, PV = 5. C ≈ 12.6 © Houghton Mifflin Harcourt Publishing Company 6. A The circumference of the window is about 12.6 feet. _incenter of △ D 11. 5; Point D is CAB, so the distance from D to AB is the same as the distance from D to DE. B C 12. m∠ABC = 2(m∠DBE)= 2(21°) = 42° 7. m∠PGV = 180° − 90° − 65° = 25° ∠ACB = 180° − 110° − 42° = 28° m m∠HGF = 2(25°) = 50° 1 (m∠ACB)= _ DCA = _ ∠ 1 (28°) = 14° 2 2 m∠HFG = 180° − 50° − 96° = 34° 13. Since △ EDB is a right triangle, the acute angles m∠PFV = _ 1 m∠HFG = _ 34° = 17° 2 2 ∠EDB = 90° − 21° = 69° are complimentary. m _ 8. BAis a radius of the circle. 14. m∠GBC + 4 7° + 35° = 180° A = πr 2 m∠GBC = 98° πr 2 = 121π 2x° = 98°
