To the Student...4. Part Part - PDF Free Download

Table of Contents To the Student.......................................................4 Session I: Pretest...........................................5 Part 1............................................................6 Part 2...........................................................13 Session II: Lessons..........................................21 Lesson 1: Identify and Use Spatial Relationships.......................22 A. Define Terms Related to Spatial Relationships................22 B. Describe Properties of Spatial Relationships.................27 Lesson 2: Identify and Use Angle Relationships........................31 A. Define Terms Related to Angle Relationships.................31 B. Describe Angle Relationships with Parallel Lines and Transversals......................................35 C. Use Angle Relationships to Solve Problems.................38 D. Use Algebra to Solve Problems Related to Angle Relationships...41 Lesson 3: Identify and Use Triangle Properties........................44 A. Define Terms Related to Triangle Relationships...............44 B. Describe Properties of Triangle Relationships................48 C. Describe Inequality Properties of Triangles...................52 D. Use Algebra to Solve Problems Related to Triangle Properties....55 Lesson 4: Solve Triangle Problems Using the Pythagorean Theorem.......59 A. Find the Missing Measures of Right Triangles................59 Lesson 5: Use Congruence of Triangles..............................64 A. Recognize Congruence Relationships in Triangles.............64 B. Use Congruence of Triangles to Solve Problems..............69 Lesson 6: Use Coordinate Geometry................................72 A. Find Information about Triangles in Coordinate Planes..........72 B. Use the Distance Formula to Find Measures and Classify Triangles..................................75 C. Find Midpoints of Segments.............................78 Session III: Posttest..........................................81 Part 1...........................................................82 Part 2...........................................................89 3

Lesson 4 Solve Triangle Problems Using the Pythagorean Theorem A Find the Missing Measures of Right Triangles In a right triangle, the side opposite the right angle is called the hypotenuse of the triangle. The sides opposite the acute angles are called the legs of the triangle. In this triangle, the hypotenuse is A B and the legs are A C and B C. b A c In a right triangle, the lengths of the sides are related. This relationship is given by the Pythagorean Theorem. C a B The Pythagorean Theorem states that in any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. For triangle ABC above, this means that: c 2 a 2 b 2 Given the lengths of any two sides of a right triangle, the Pythagorean Theorem can be used to find the length of the missing side. 59

Skills Coach, Geometry: Streamline to Proficiency Triangles Coached Practice 1 In JKL, L is the right angle, j 8, and k 6. Find the length of the hypotenuse. J k l L j K STRATEGY: Use the Pythagorean Theorem to find l. The given side lengths, k and j, are the lengths of the legs of the triangle. The missing side length, l, is the length of the hypotenuse. Use the given lengths of the legs and the Pythagorean Theorem to find l. l 2 j 2 k 2 l 2 8 2 6 2 l 2 64 36 l 2 100 l 100 l 10 SOLUTION: The length of the hypotenuse l is 10. 60

Lesson 4: Solve Triangle Problems Using the Pythagorean Theorem Coached Practice 2 In MNP, M is a right angle, m 13, and n 12. What is p? M p n N m P STRATEGY: Use the Pythagorean Theorem to find p. The missing length p is the length of a leg of the right triangle. To find p, use the given lengths and the Pythagorean Theorem to solve for p. m 2 n 2 p 2 13 2 12 2 p 2 169 144 p 2 25 p 2 25 p 5 p SOLUTION: The length of the leg p is 5. 61

Skills Coach, Geometry: Streamline to Proficiency Triangles Coached Practice 3 A 20-foot long ladder is leaning against a house. The distance from the base of the ladder to the base of the house is 6 feet. To the nearest foot, how far up the side of the house does the ladder reach? A b c 20 C a 6 B STRATEGY: Use the Pythagorean Theorem. The length of the ladder is the hypotenuse of a right triangle. Both the distance from the base of the ladder to the base of the house and the distance the ladder reaches up the side of the house are legs of the right triangle. Substitute the known distances in the Pythagorean Theorem, and solve for the missing side length. c 2 a 2 b 2 20 2 6 2 b 2 400 36 b 2 364 b 2 364 b You can use a calculator to find that 364 19.08. SOLUTION: The ladder reaches about 19 feet up the side of the house. 62

Lesson 4: Solve Triangle Problems Using the Pythagorean Theorem Independent Practice Use the Pythagorean Theorem to solve the following problems. Show all your work. 1. In ABC, AB 15 and AD 3. B 4. A model of a wheelchair ramp is designed so that the ramp rises by 1 inch for every foot the ramp covers. 1 in. ramp 1 ft C D A a. To the nearest hundredth of an inch, how many inches long is the model ramp? What is BD? Round your answer to the nearest tenth. 2. One leg of a right triangle has a length of 33 cm, and the hypotenuse has a length of 65 cm. What is the length of the other leg? b. The above ramp is a model of an actual ramp that has a rise of 6 inches and a run of 5 feet. To the nearest hundredth of an inch, how long is the actual ramp? 3. The Great Pyramid of Giza has a square base with sides that are 229 meters long. 229 m 229 m What is the length in meters of a diagonal of the square base? Round your answer to the nearest whole number. 63

Table of Contents To the Student.......................................................5 Session I: Pretest...........................................7 Part 1............................................................8 Part 2...........................................................14 Session II: Lessons..........................................21 Lesson 1: Identify and Use Polygon Properties........................22 A. Define Terms Related to Polygonal Relationships..............22 B. Classify Two-Dimensional Geometric Figures.................26 C. Use Congruence of Polygons to Solve Problems..............31 D. Use Sums of Interior Angles of Triangles to Find Sums of Angles of Polygons..................................34 E. Use Characteristics of Polygons to Solve Problems............37 Lesson 2: Identify and Use Quadrilateral Properties....................43 A. Classify Special Quadrilaterals by Characteristics.............43 B. Find Missing Measures of Special Quadrilaterals..............51 C. Describe Properties of Quadrilateral Relationships.............56 Lesson 3: Use Coordinate Geometry................................60 A. Use the Distance Formula to Find Measures and Classify Quadrilaterals..............................60 B. Know How to Calculate the Slope of a Line..................66 C. Know That Slopes of Parallel Lines Are Equal................71 D. Know that Slopes of Perpendicular Lines Are Opposite Reciprocals..................................75 Session III: Posttest..........................................81 Part 1...........................................................82 Part 2...........................................................88 3

Lesson 1 Identify and Use Polygon Properties A Define Terms Related to Polygonal Relationships A polygon is a closed two-dimensional geometric figure formed by three or more line segments that meet at their endpoints. These line segments are called sides. Each point where two sides of a polygon meet is called a vertex. A polygon always has as many sides as it has vertices (plural of vertex). The figure below shows a polygon that has six sides and six vertices. side vertex A line segment connecting two nonconsecutive vertices of a polygon is called a diagonal. The dotted segments in the polygons below are diagonals. Convex Concave 22

Lesson 1: Identify and Use Polygon Properties Look at the polygon on the left of the previous page. All the diagonals of that polygon lie inside the polygon. In fact, any line segment drawn between any two points of that polygon would lie inside the polygon. That is because the polygon on the left is convex. Notice that the polygon on the right looks a little caved in. That is because it is a concave polygon. In a concave polygon, it is possible to draw a line segment between two points of the polygon that lies partially or entirely outside the polygon. Most of the important facts and properties about polygons apply to convex polygons. So, in this book, polygon will mean convex polygon unless otherwise noted. You should also know about the angles of a polygon. An angle that is formed when two sides of a polygon meet at a vertex is called an interior angle of the polygon, or simply an angle of the polygon. An angle formed at the point where a side and an extension of an adjacent side meet is called an exterior angle. interior angle exterior angle If all the sides of a polygon are equal in length, the polygon is equilateral. If all of the interior angles have equal measures, the polygon is equiangular. If a polygon s sides are all equal in length and its angles are also all equal in measure, then the polygon is called a regular polygon. Equilateral Equiangular Regular (Equilateral and Equiangular) 23

Skills Coach, Geometry: Streamline to Proficiency Polygons Coached Practice Use all appropriate vocabulary words from the lesson to describe each of the following: figure ABCD, C D, A C, point D, D, 1, DAB A 1 B D C STRATEGY: SOLUTION: Review the meanings of the bold vocabulary words in the lesson, and use what you know about geometry symbols. For example, C D refers to line segment CD. Figure ABCD is a polygon. All four of its sides are equal in length, so figure ABCD is an equilateral polygon. Since any line segment drawn between any two points inside this polygon will lie inside the polygon, figure ABCD is convex. C D is a side of the polygon. A C is a diagonal of polygon ABCD because it connects vertices A and C, which are not consecutive. Point D is a vertex of the polygon. D is an interior angle of the polygon because it is formed by sides D A and D C. 1 is an exterior angle of the polygon because it is formed by side A B and an extension of the adjacent side B C. DAB is an interior angle of polygon ABCD because it is formed by sides A D and A B. Three letters are used to name DAB because there is more than one angle with vertex A. 24

Lesson 1: Identify and Use Polygon Properties Independent Practice 1. a. List all the vocabulary words from the lesson that describe figure PQRST. P T Q b. What terms can you use to describe point S? Q R? TPQ? S R 2. Look at polygon ABCDE. a. Name each exterior angle shown, using three letters to name each. A B L b. How many diagonals can you draw from vertex A? Name them. E D C M 3. Is it possible for a polygon to be regular if it is not equilateral? Explain your answer. 4. A triangle contains a 75 angle. Is the triangle equiangular? Explain your answer. 5. Will two diagonals of a convex polygon always intersect? Make a sketch to illustrate your answer. 25

Skills Coach, Geometry: Streamline to Proficiency Polygons B Classify Two-Dimensional Geometric Figures The table at the right shows how some polygons are classified based on the number of sides. Recall that in any polygon, the number of sides equals the number of vertices, so you could also classify a polygon by counting its vertices. Number of sides 3 4 5 6 7 8 9 10 Classification Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon In addition, polygons can be classified by the lengths of their sides. A polygon with all sides congruent, or equal in length, is equilateral. Triangles can be further classified as shown below. Equilateral (all sides congruent) Another way to classify polygons is by angle measure. A polygon with all angles congruent, or equal in measure, is equiangular. Triangles can be classified by the measures of their angles. Recall that an acute angle measures less than 90, a right angle measures exactly 90, and an obtuse angle measures greater than 90 but less than 180. The examples below show how triangles can be classified based on the types of angles they have. Equiangular (all angles congruent) Isosceles (at least two sides congruent) Acute (all angles acute) Right (one right angle) Scalene (no sides congruent) Obtuse (one obtuse angle) 26

Lesson 1: Identify and Use Polygon Properties Remember, if a polygon is both equilateral and equiangular, then it is a regular polygon. A regular polygon that is also a triangle is called an equilateral triangle. An equilateral triangle has three congruent sides and three congruent 60 angles. When classifying polygons, it is also helpful to remember that a polygon can be classified as convex or concave. Remember, if a line segment drawn between two points of a polygon lies partially or entirely outside the polygon, it is concave. If no such line segment can be drawn, it is convex. How can you classify each polygon? Coached Practice 1. 2. 3. 4. STRATEGY: Here are some of the questions you can ask yourself: How many sides does the polygon have? Is it equilateral? equiangular? regular? If it is a triangle, what special names apply? Is it concave? 27

Skills Coach, Geometry: Streamline to Proficiency Polygons SOLUTION: 1. This is a pentagon because it has five sides. It is concave because a line segment connecting two of its points can be drawn outside the pentagon as shown by this diagram. 2. The figure for question 2 is an isosceles triangle because it has exactly two congruent sides. It is an acute triangle because all three angles are acute. So, it is an isosceles acute triangle. 3. This is a scalene triangle because no sides are congruent. It is obtuse because it has one angle that is greater than 90. So, it is a scalene obtuse triangle. 4. This is an octagon because it has 8 sides. It is equiangular because all its interior angles are congruent. So, it is an equiangular octagon. It is not a regular octagon because its sides do not appear to have equal lengths. 28

Lesson 1: Identify and Use Polygon Properties 1. How can you classify this polygon? Independent Practice 2. How can you classify this polygon? 3. What are all the classifications that apply to both of these polygons? 60º 60º 60º 29

Skills Coach, Geometry: Streamline to Proficiency Polygons 4. What is the difference between these two polygons? 5. Draw an example of each polygon described below. Use marks to show congruent angles or congruent sides. If any of these polygons cannot be drawn, explain why. a. an isosceles obtuse triangle b. an isosceles scalene triangle c. a scalene right triangle d. an obtuse right triangle e. a pentagon with exactly two sides congruent f. a regular quadrilateral 30

Table of Contents To the Student.......................................................5 Session I: Pretest...........................................7 Part 1............................................................8 Part 2...........................................................15 Session II: Lessons..........................................23 Lesson 1: Reflections............................................24 A. Sketch Reflections of Figures............................24 B. Reflect Figures in a Coordinate Plane......................28 C. Use Reflections in Real-World Situations....................34 D. Describe Which Properties of a Figure Do Not Change After a Reflection......................................37 E. Describe Which Properties of a Figure Change After a Reflection......................................40 Lesson 2: Translations............................................44 A. Sketch Translations of Figures............................44 B. Translate Figures in a Coordinate Plane.....................47 C. Use Translations in Real-World Situations...................51 D. Describe Which Properties of a Figure Do Not Change After a Translation.....................................54 E. Describe Which Properties of a Figure Change After a Translation.....................................58 Lesson 3: Rotations..............................................61 A. Sketch Rotations of Figures..............................61 B. Rotate Figures in a Coordinate Plane.......................65 C. Use Rotations in Real-World Situations.....................72 D. Describe Which Properties of a Figure Do Not Change After a Rotation.......................................75 E. Describe Which Properties of a Figure Change After a Rotation.......................................78 3

Lesson 4: Dilations..............................................81 A. Sketch Dilations of Figures..............................81 B. Dilate Figures in a Coordinate Plane Using (0, 0) as a Center of Dilation...........................................85 C. Dilate Figures in a Coordinate Plane Using Any Point as a Center of Dilation.....................................89 D. Describe Which Properties of a Figure Do Not Change After a Dilation........................................93 E. Describe Which Properties of a Figure Change After a Dilation........................................96 Session III: Posttest.........................................99 Part 1..........................................................100 Part 2..........................................................106

Lesson 2 Translations A Sketch Translations of Figures A translation is a type of transformation that moves each point of a figure in a plane the same distance in the same direction. A translation can also be referred to as a shift or a slide. To translate a point P in a plane, move it a certain distance along a straight line in any direction. In the figure below, point P was translated to form point P. P P To translate a figure, move each point of the figure the same distance in the same direction. In the figure below, triangle F G H is a translation of triangle FGH. F F G H G H 44

Lesson 2: Translations Draw a translation of Figure A below. Coached Practice STRATEGY: SOLUTION: Choose a direction and a distance to translate the figure. Then move each point of Figure A that distance in that direction. One possible translation of the figure is shown below. 45

Skills Coach, Geometry: Streamline to Proficiency Transformations Independent Practice 1. Complete Figure X so that it represents a translation of Figure X. Figure X Figure X 2. Draw a translation of Figure ABCD below. A B D C 3. Translate the figure below so that point P is the translation of point P. P P 46

Lesson 2: Translations B Translate Figures in a Coordinate Plane In a coordinate plane, a translation can be described by the number of units a figure moves left, right, up, or down. Given the coordinates of a point, you can find the coordinates of the image of the point by adding or subtracting the number of units the point moves in either direction. If point P(x, y) is translated h units to the right and k units up, the image is point P (x h, y k). y P (x h, y k) This can be represented symbolically as P(x, y) P (x h, y k). Translations to the right or up are represented by positive values of h and k. Translations to the left or down are represented by negative values of h and k. P(x, y) h units k units x 47

Skills Coach, Geometry: Streamline to Proficiency Transformations Coached Practice 1 Find the coordinates of the image of point P( 1, 4) after it is translated 6 units to the right and 2 units down. Draw the image of the point on the coordinate grid. y P 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 2 3 4 5 6 x STRATEGY: Add or subtract the number of units the point will be translated from each of the coordinates. The image of the point will be 6 units to the right of P, so add 6 to the x-coordinate. The image of the point will be 2 units down from P, so subtract 2 from the y-coordinate. P( 1, 4) P ( 1 6, 4 2) P (5, 2) SOLUTION: The image of point P( 1, 4) is P (5, 2). y P 6 5 4 3 2 P 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 2 3 4 5 6 x 48

Lesson 2: Translations Coached Practice 2 Translate quadrilateral MNPQ by (x, y) (x 3, y 4). Sketch the image of the figure and give the coordinates of its vertices. y 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 M N 6 5 4 3 2 1 2 3 4 5 6 Q P x STRATEGY: To translate a figure, translate each of the vertices and then draw the image of the figure. Subtract 3 from the x-coordinate of each vertex, and add 4 to the y-coordinate of each vertex. M( 2, 1) M ( 2 3, 1 4) M ( 5, 3) N( 1, 1) N ( 1 3, 1 4) N ( 4, 5) P(3, 2) P (3 3, 2 4) P (0, 6) Q(2, 2) Q (2 3, 2 4) Q ( 1, 2) SOLUTION: The image of quadrilateral M N P Q is shown at the right. The coordinates of the vertices of M N P Q are: M ( 5, 3) N ( 4, 5) P (0, 6) Q ( 1, 2) N M N y 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 M Q 2 3 4 5 6 P Q P x 49

Skills Coach, Geometry: Streamline to Proficiency Transformations Independent Practice Let P be the point (3, 1). Give the coordinates of the image point P after each of the following translations. y 5. Translate the figure below by (x, y) (x 4, y 7). y 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 2 3 4 5 6 x 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 2 3 4 5 6 x 1. 2 units up 2. 2 units to the right and 1 unit up 3. (x, y) (x 2, y 3) 4. (x, y) (x 2, y 3) 50

Lesson 2: Translations C Use Translations in Real-World Situations Many of the movements of figures and objects in real life can be described by translations. For example, if a car travels in a straight line, the car is being translated, along with anything inside the car. For the car shown below, the movement is forward and in a straight line. Translations can be used to make patterns, such as those seen in floor or ceiling tiles and wallpaper designs. This tile pattern is made from horizontal and vertical translations of a single tile. By translating the tile in different directions and different distances, the entire pattern can be extended indefinitely. 51

Skills Coach, Geometry: Streamline to Proficiency Transformations Coached Practice Use translations to create a tiling pattern with the figure shown. STRATEGY: SOLUTION: Translate the figure several times to create a tiling pattern. Each time, be sure that the image of the figure does not overlap any other figures. One possible pattern is shown at the right. 52

Lesson 2: Translations Independent Practice 1. The hexagon below can be used to create a tiling pattern such as you would see in a floor design. To see the pattern created by this hexagon, translate the hexagon so that the point labeled P falls on each given point. P 2. The figure below shows the flight path of an airplane. Does this path represent a translation? Explain your answer. 53

Skills Coach, Geometry: Streamline to Proficiency Transformations D Describe Which Properties of a Figure Do Not Change After a Translation Remember that a translation in a plane shifts a figure by a certain distance in a certain direction. Because the distance and direction are fixed, the properties of a figure do not change after a translation. In the figure to the right, triangle ABC was translated from triangle A B C. A C A B B C The shape and size of the image are the same as the preimage. So, the image and the preimage are congruent. Therefore, in the figure above ABC A B C. Because the image and the preimage are congruent, the following statements are true: 1) The corresponding sides have the same lengths. 2) The corresponding angles have the same measures. The orientation of the figure does not change after a reflection. Notice that the vertices of square ABCD and square A B C D are both labeled in a clockwise direction; the orientation of the figure remains the same. A D B C A D B C 54

Lesson 2: Translations Coached Practice Pentagon JKLMN on the left was translated to form the pentagon on the right, as shown below. J K N L M Label the vertices of the image of pentagon J K L M N. Then identify the congruent sides and the congruent angles. STRATEGY: Remember that the orientation of the figure does not change. Label the vertices of the image in the same direction as the vertices of pentagon JKLMN. Also, remember that the figure and its image are congruent, so corresponding sides and corresponding angles are congruent. 55

Skills Coach, Geometry: Streamline to Proficiency Transformations SOLUTION: The vertices of the image figure should be labeled counterclockwise, as shown below. J K N J L M K N L M The following sides are congruent: J K J K K L K L L M L M M N M N N J N J The following angles are congruent: J J K K L L M M N N 56

Lesson 2: Translations Independent Practice 1. Circle A was translated to create circle A. A r = 3 in. 2. Triangle RST was translated to create triangle R S T. S A R 70º T 70º R S T What is the radius of circle A? What is m T? (Remember that the sum of the measures of the interior angles of any triangle is 180.) 3. Quadrilateral A B C D is a translation of the quadrilateral to its left. C B D A Label the vertices of the quadrilateral on the left. Then identify the congruent sides and the congruent angles. 57

Skills Coach, Geometry: Streamline to Proficiency Transformations E Describe Which Properties of a Figure Change After a Translation You have seen that translations preserve many properties of a figure. In fact, a translation changes only the location of a figure. The location of a figure changes. This figure was translated 2 inches to the right. A A E E B D B D C C The image of the figure is identical to the preimage; only the location has changed. Every part of the image moved the same distance in the same direction from the preimage. When a point in a coordinate plane is translated, its coordinates change. If a point is translated If a point is translated If a point is translated horizontally by h units, vertically by k units, horizontally by h units the translation is the translation is and vertically by k units, (x, y) (x h, y). (x, y) (x, y k). the translation is (x, y) (x h, y k). y (x, y)(x+h, y) x y (x,y+k) (x,y) x y (x, y) (x+h, y+k) x 58

Lesson 2: Translations Coached Practice Triangle ABC was translated to form triangle A B C. Describe the translation. y A C 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 2 3 B 4 5 6 A C B x STRATEGY: Choose a vertex and its image, and find the change in the x- and y-coordinates. The vertices of triangle ABC are: The vertices of triangle A B C are: A(3, 2) A ( 2, 4) B(4, 5) B ( 1, 3) C(2, 2) C ( 3, 0) Remember that in a translation, every point in the figure is translated the same distance in the same direction. To find the change in the x- and y-coordinates, choose any vertex and its image, such as A and A. Subtract the x-coordinate of A from the x-coordinate of A, and then subtract the y-coordinate of A from the y-coordinate of A x-coordinates y-coordinates 2 3 5 4 2 2 Point A is 5 units to the left and 2 units up from point A. SOLUTION: Each of the vertices of the triangle moved 5 units to the left and 2 units up. The translation can be described as (x, y) (x 5, y 2). 59

Skills Coach, Geometry: Streamline to Proficiency Transformations Independent Practice 1. The point ( 5, 4) was translated to (1, 1). Describe this translation. 2. The point (3, 4) was translated to (3, 1). Describe this translation. 3. The point (6, 2) was translated to ( 5, 4). Describe this translation. 4. Quadrilateral ABCD was translated to form quadrilateral A B C D. Describe the translation. y A D 6 5 4 3 2 1 6 5 4 3 2 1 0 1 2 3 4 5 6 7 1 C B 2 3 4 5 6 D B A C x 60

Lesson 1 Solve Problems Using Similarity A Recognize Similar Figures Similar figures are figures that have the same shape, but not necessarily the same size. In similar figures, the corresponding angles are congruent and the corresponding side lengths are proportional. The ratio of the corresponding side lengths is called the similarity ratio. The symbol for is similar to is. When naming a pair of similar figures, it is important to name the vertices in the same order. The order in which the vertices are named tells which parts correspond to each other. In the figure at the right, ABCD EFGH. The F angles are marked to show which pairs are G congruent. The following statements describe the relationships between ABCD and EFGH: B C AB BC CD DA EF FG GH HE A D E H A E B F C G D H 22

Lesson 1: Solve Problems Using Similarity Coached Practice Decide whether each pair of figures is similar. Explain your answer. 1. 6 3. L 6 64º 4 3 6 4 8 8 10 K 85º 4 120º J 3 6 91º M T 2. E B 2z z x A 2x D y 2y C F S 89º 8 12 59º 12 113º R 6 99º U STRATEGY: Remember the two conditions for figures to be similar: The corresponding angles are congruent. The corresponding side lengths are proportional. Determine whether each pair of figures satisfies both of these conditions. 23

Skills Coach, Geometry: Streamline to Proficiency Similarity and Trigonometry SOLUTION: 1. The corresponding angles are congruent. However 4 8 6, so the corresponding side 10 lengths are not proportional. These figures are not similar. 2x 2y 2z 2. The corresponding angles are congruent. Because 2, the x y z corresponding side lengths are proportional. ABC DEF 3. Because 6 3 8 4 1 2 1 2 2, the corresponding side lengths are proportional. 6 6 However, the corresponding angles are not congruent. These figures are not similar. 24

Lesson 1: Solve Problems Using Similarity Independent Practice For exercises 1 4, decide whether each pair of figures is similar. Explain your answer. 1. F 3.5 68º 3.5 A 4 132º 4 E 112º 112º G D 48º 48º B 3.5 68º 3.5 4 132º 4 C H 2. 10 5 120º 5 120º 120º 5 10 120º 120º 10 120º 5 5 120º 120º 120º 5 10 10 120º 120º 10 25

Skills Coach, Geometry: Streamline to Proficiency Similarity and Trigonometry 3. A B W X D C Z Y 4. 53º 15 20 127º 8 10 65º 115º 12 9 26

Lesson 1: Solve Problems Using Similarity 5. In the figure below, BAC ADC. B D 5 4.6 11.1 A 12 C a. Name the three pairs of congruent angles in the similar triangles. b. What is the ratio of the side lengths of BAC to the side lengths of ADC, rounded to the nearest tenth? 6. Remember that in congruent figures, all corresponding angles are congruent and all corresponding sides are congruent. Explain why congruent figures are also similar. 27

Skills Coach, Geometry: Streamline to Proficiency Similarity and Trigonometry B Use Proportions from Sides of Similar Triangles to Find Missing Side Measures A proportion is a statement of equality between two ratios, such as a d c. One way to b solve a proportion is to cross multiply; for the proportion a d c, ad bc. b The side lengths of similar triangles are proportional, so you can use proportions to find lengths of missing sides in similar triangles. Because the similarity ratio is the same for any pair of corresponding sides, it does not matter which pair of corresponding sides are used in the proportion. In the figure below, ABC DEF. Coached Practice A 6 D Find DF. 4 C B F E 3 28

Lesson 1: Solve Problems Using Similarity STRATEGY: Write and solve a proportion to find DF. In these triangles, A B corresponds to D E and A C corresponds to D F. Because each of these side lengths is known except for DF, you can write and solve a proportion to find DF. AB DE AC DF SOLUTION: DF 2 6 3 4 DF Substitute the known values into the proportion. 6 DF 3 4 Cross multiply. 6 DF 12 Simplify. DF 2 Divide both sides by 6. 29

Skills Coach, Geometry: Streamline to Proficiency Similarity and Trigonometry Independent Practice 1. In the figure below, BLK BHJ. B 3. In the figure below, LKJ MKL and LKJ MLJ. L m 12 6.4 L 1.75 K J 1 K 11.5 M a. What is m, rounded to the nearest tenth? What is BL? H 3.5 J b. What is JL, rounded to the nearest tenth? 2. In the figure below, ABC EFG. B 2.5 C F 4 G 3 A Find x. 5 E x 30

Lesson 1: Solve Problems Using Similarity C Use Properties of Similar Triangles to Find Missing Measures in Real-World Situations Similar triangles can be used to solve many real-world problems. Many of these problems involve indirect measurement, in which lengths can be found without actually measuring them. At a certain time of day, a 5-foot tall man casts a shadow that is 7 feet long. At the same time, a tree casts a shadow that is 21 feet long. How tall is the tree? Coached Practice 5 ft 7 ft 21 ft STRATEGY: SOLUTION: The man and his shadow and the tree and its shadow form two similar right triangles. Write and solve a proportion to find the height of the tree. Let h represent the height of the tree. 5 h 7 21 5 21 7 h Cross multiply. 5 21 h 7 7 Divide both sides by 7. 105 h 7 15 h The tree is 15 feet tall. Simplify. 31

Table of Contents To the Student.......................................................5 Session I: Pretest...........................................7 Part 1............................................................8 Part 2...........................................................15 Session II: Lessons..........................................23 Lesson 1: Find Geometric Probabilities..............................24 A. Use Area Formulas to Find Geometric Probabilities...........24 Lesson 2: Classify and Describe Three-Dimensional Figures.............31 A. Classify Three-Dimensional Figures........................31 B. Describe the Intersection of a Plane with a Three-Dimensional Figure...............................40 Lesson 3: Find Surface Area.......................................46 A. Find the Surface Area of Rectangular Prisms Using Formulas....46 B. Find the Surface Area of Pyramids Using Formulas............50 C. Find the Surface Area of Cylinders Using Formulas............54 D. Find the Surface Area of Cones Using Formulas..............58 E. Find the Surface Area of Spheres Using Formulas.............62 Lesson 4: Find Volume............................................65 A. Find the Volume of Rectangular Prisms Using Formulas........65 B. Find the Volume of Pyramids Using Formulas................70 C. Find the Volume of Cylinders Using Formulas................74 D. Find the Volume of Cones Using Formulas..................78 E. Find the Volume of Spheres Using Formulas.................81 F. Describe How the Change in One Dimension of a Geometric Shape Affects Perimeter, Area, Surface Area, or Volume................................84 G. Describe How the Change in More Than One Dimension of a Geometric Shape Affects Perimeter, Area, Surface Area, or Volume.......................................91 Session III: Posttest.........................................99 Part 1..........................................................100 Part 2..........................................................105 3

Lesson 3 Find Surface Area A Find the Surface Area of Rectangular Prisms Using Formulas Recall that a rectangular prism has six faces three pairs of congruent rectangles. 6 inches 10 inches 4 inches The figures below show the faces of this rectangular prism. Front and back Top and bottom 10 inches 6 inches Sides 6 inches 4 inches 10 inches 4 inches 46

Lesson 3: Find Surface Area To find the surface area of the rectangular prism on the previous page, you can find the areas of all six faces and then add: Front and back faces: Each face is (6)(10), so both faces are (2)(6)(10). Top and bottom faces: Each face is (4)(10), so both faces are (2)(4)(10). Side faces: Each face is (4)(6), so both faces are (2)(4)(6). Surface area: SA (2)(6)(10) (2)(4)(10) (2)(4)(6) 120 80 48 248 square inches Another way to find the surface area of a rectangular prism is to use the following formula: Surface Area of a Rectangular Prism SA 2B Ph, where B is the area of the base, P is the perimeter of the base, and h is the height of the prism. So, you can use the formula to find the area of the prism on the previous page. Recall that any face of a rectangular prism can be used as the base. You can use the 10 in.-by-4 in. rectangle as the base and 6 in. as the height as follows: B (10)(4) 40 square inches P 10 4 10 4 28 inches h 6 inches SA 2B Ph (2)(40) (28)(6) 80 168 248 square inches A cube is a special rectangular prism. All of the faces of a cube are congruent squares. In the figure at the right, each face is a square that measures 5 inches by 5 inches. 5 inches To find the surface area of this cube, you can multiply the area of one face by 6: Area of each face: Surface area: (5)(5) 25 square inches SA (6)(25) 150 square inches 5 inches Because a cube is a rectangular prism, you can also use the formula to find the surface area. Use one of the 5 in.-by-5 in. squares as the base and 5 in. as the height. B (5)(5) 25 square inches P 5 5 5 5 20 inches h 5 inches SA 2B Ph (2)(25) (20)(5) 50 100 150 square inches 5 inches 47

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume Coached Practice Find the surface area of the rectangular box shown below. 6 inches 3 inches 4 inches STRATEGY: Use the formula for the surface area of a rectangular prism. Choose one of the faces to use as the base. The base used here is the 3 in.-by-4 in. rectangle, so the height is 6 in. B (3)(4) 12 square inches P 3 4 3 4 14 inches h 6 inches SA 2B Ph (2)(12) (14)(6) 24 84 108 square inches SOLUTION: The surface area is 108 square inches, or 108 in. 2 48

Lesson 3: Find Surface Area Independent Practice For problems 1 3, find the surface area of each rectangular prism. 1. 4 feet 3. 6 cm 12 feet 3 feet 24 cm 8 cm 2. 8 feet 4. Adrienne wants to wrap a gift box that measures 10 inches on each edge. What is the minimum number of square feet of paper needed to cover all the surfaces of the box? 2 feet 2 feet 10 in. 5. Roberto built a concrete patio measuring 12 feet by 24 feet by 6 inches. He plans to paint the top surface and all of the sides of the patio. 24 feet What is the surface area of the part he wants to paint? 12 feet 6 inches 49

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume B Find the Surface Area of Pyramids Using Formulas A regular pyramid has a base that is a regular polygon. All of its lateral faces are congruent isosceles triangles. In the regular pyramid below, the base is a square that measures 6 inches by 6 inches. The slant height of the pyramid is 10 inches. 10 inches 6 inches 6 inches The base and one triangular face of the pyramid are shown below. Each face is an isosceles triangle; the base of the triangle is 6 inches and the height is 10 inches. The height of each triangle is the slant height of the pyramid. Triangular Face Square Base 6 inches 6 inches 6 inches 10 inches 50

Lesson 3: Find Surface Area To find the surface area of the triangle on the previous page, you can find the areas of the square base and the four triangular faces and then add. Recall that the area of a triangle is 1 bh, where b is the base of the triangle and h is the height. 2 Square base: (6)(6) Triangular faces: 1 1 Each triangle is (6)(10), so all 4 triangles are (4) 2 2 (6)(10). Surface area: 1 SA (6)(6) 4 2 (6)(10) 36 120 156 square inches You can also use the following formula to find the surface area. This formula applies to any regular pyramid. Surface Area of a Regular Pyramid SA B 1 Pl, where 2 B is the area of the base, P is the perimeter of the base, and l is the slant height of the pyramid. For the pyramid above, B (6)(6) 36 square inches P 6 6 6 6 24 inches l 10 inches SA B 1 1 Pl 36 (24)(10) 36 120 156 square inches 2 2 51

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume Coached Practice Find the surface area of the square pyramid below. 36 cm 20 cm 20 cm STRATEGY: Use the formula for the surface area of a regular pyramid. B (20)(20) 400 cm 2 P 20 20 20 20 80 cm l 36 cm 1 SA B 1 Pl 400 (80)(36) 400 1440 1840 cm 2 2 2 SOLUTION: The surface area is 1840 square centimeters, or 1840 cm 2. 52

Lesson 3: Find Surface Area Independent Practice For problems 1 and 2, find the surface area of each square pyramid. 1. 5 ft 3. Laura is making pyramid paper weights for gifts. Each pyramid has a square base that measures 2 inches by 2 inches and a slant height of 3 inches. What is the total surface area of each pyramid? 4 ft 4 ft 3 in. 2. 8.4 m 2 in. 2 in. 6 m 6 m 4. Mei-Lin s roof is in the shape of a pyramid. What is the total area of the surfaces of the roof? (Hint: Do not include the area of the base.) 30 ft 17 ft 30 ft 53

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume C Find the Surface Area of Cylinders Using Formulas A cylinder has a curved lateral surface and two congruent circular bases. The cylinder shown below has radius r and height h. r h The radius of a cylinder is the same as the radius of its circular base. To work with cylinders, you will need the following formulas for circles: Circumference of a circle: C 2 r Area of a circle: A r 2 If the lateral surface of a cylinder is unrolled and flattened to form a two-dimensional pattern called a net, it becomes a rectangle. This rectangle has a length of 2 r, the circumference of the circular base, and a width of h, the height of the cylinder. The net below shows the circular bases and the lateral surface of a cylinder. r Base Lateral Surface Base r 2 r h 54

Lesson 3: Find Surface Area To find the surface area of a cylinder, add the areas of the two circular bases to the area of the lateral surface (the rectangle). SA 2 area of base circumference height SA 2 r 2 2 r h So, the formula is: Surface Area of a Cylinder SA 2 r 2 2 rh, where r is the radius of the cylinder, and h is the height of the cylinder. To find the lateral surface area of a cylinder (the area of the surfaces, not including the bases), use 2 rh. Being able to find the lateral surface area comes in handy if you need to find the area of a label for a soup can or if a problem asks you about painting only the curved surfaces of a cylinder. Coached Practice Find the surface area of the cylinder shown below. 3 m 10 m STRATEGY: Use the formula for the surface area of a cylinder. Note that the diameter of the cylinder is 3 m, so the radius is: 1 3 1.5 m. The height of this cylinder is its length, 10 m. 2 For, either approximate it as 3.14 or use the key on a calculator. SA 2 r 2 2 rh 2 (1.5) 2 2 (1.5)(10) 4.5 30 34.5 34.5 3.14 108.3 The surface area is 34.5 square meters, which is approximately 108.3 square meters. SOLUTION: The surface area is 34.5 m 2, which is approximately 108.3 m 2. 55

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume Independent Practice For problems 1 3, find the surface area of each cylinder. Give your answer in terms of and rounded to the nearest tenth of an inch. 1. 30 inches 30 inches 2. 1.4 in. 3.4 in. 3. 20 in. 2 in. 56

Lesson 3: Find Surface Area 4. The outside lateral surface of the cylindrical water tank shown below must be painted. Neither of the bases will be painted. Each gallon of paint covers 300 square feet. 30 ft 60 ft a. What is the approximate area to be painted? b. If paint is sold in 1-gallon cans, how many cans of paint will be needed? 5. What is the outside surface area of a cylindrical garden hose that has a diameter of 1 inch and a length of 25 feet? a. Find your answer in square inches. b. Find your answer in square feet. (Hint: 1 square foot 144 square inches) 57

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume D Find the Surface Area of Cones Using Formulas A cone has a curved lateral surface and one circular base. The cone below has a slant height, l, and a base with radius, r. The slant height is the length of a line segment from the vertex to the outer boundary of the circular base. The radius of the circular base is also the radius of the cone. l r Use the following formula to find the surface area of a cone. Surface Area of a Cone SA r 2 rl, where r is the radius of the base of the cone, and l is the slant height of the cone. To find only the lateral surface area, use rl. To find only the area of the circular base, use r 2. 58

Lesson 3: Find Surface Area Coached Practice Find the surface area of the cone shown below. 9 cm 6 cm STRATEGY: Use the formula for the surface area of a cone. 1 The diameter of the base is 6 cm, so the radius is: 6 3 cm. 2 The slant height is 9 cm. Remember that you can either approximate as 3.14 or use the key on your calculator. SA r 2 rl (3) 2 (3)(9) 9 27 36 36 3.14 113 SOLUTION: The surface area is 36 cm 2, which is approximately 113 cm 2. 59

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume Independent Practice For problems 1 3, find the surface area of each cone. Give your answer in terms of and rounded to the nearest tenth of a meter. 1. 2 m 10 m 2. 8 m 10 m 3. 15 m 5 m 60

Lesson 3: Find Surface Area 4. An ice cream cone has a diameter of 3 inches at its open circular end and a slant height of 5 inches. Yolanda will wrap a piece of paper around the lateral surface of the cone to catch any ice cream that drips down. Approximately what area of the cone will the paper cover? 3 in. 5 in. 5. The lamp shade below is part of a cone. The diameter of the bottom opening is 20 inches. The diameter of the top opening is 10 inches (not shown on diagram). Find the area of the fabric that forms the lamp shade. (Hint: Imagine a cone with a slant height of 13 inches and a base diameter of 10 inches. Then imagine a cone with a slant height of 26 inches and a base diameter of 20 inches. Use those cones to find the amount of fabric in the lamp shade.) 13 in. 20 in. 13 in. 61

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume E Find the Surface Area of Spheres Using Formulas A sphere is the set of all points that are a given distance from a point. That point is the center of the sphere and the distance is the radius r. r Use the following formula to find the surface area of a sphere. Surface Area of a Sphere SA 4 r 2, where r is the radius of the sphere. 62

Lesson 3: Find Surface Area Find the surface area of the sphere shown at the right. Coached Practice 10 in. STRATEGY: Use the formula for the surface area of a sphere. 1 The diameter is 10 inches, so the radius is: 10 5 inches. 2 Use either 3.14 as an approximation for or use the key on your calculator. SA 4 r 2 4 (5) 2 4 (25) 100 100 3.14 314 SOLUTION: The surface area is 100 in. 2, which is approximately 314 in. 2 63

Skills Coach, Geometry: Streamline to Proficiency Surface Area and Volume Independent Practice For problems 1 and 2, find the surface area of each sphere. Give your answer in terms of and rounded to the nearest tenth of a centimeter. 1. 4. Earth is nearly spherical, with a radius of approximately 3960 miles. Find the approximate surface area of Earth, to the nearest thousand square miles. 20 cm 2. 1.2 cm 3960 miles 5. Find the approximate surface area, in square inches, of each of the following. Use 3.14 to approximate. (The diameters given are approximate in some cases.) 3. Find the surface area of the hemisphere (half a sphere, not including the circular base) shown below. 3 ft a. basketball, diameter of 9 inches b. baseball, diameter of 3 inches c. tennis ball, diameter of 2.6 inches d. golf ball, diameter of 1.7 inches 64

Skills Coach, Geometry: Streamline to Proficiency Similarity and Trigonometry Independent Practice 1. Karen is 6 feet tall. At a certain time of day, she casts a shadow that is 15 feet long. At the same time, a telephone pole casts a shadow that is 40 feet long. 6 ft 15 ft 40 ft How tall is the telephone pole? 2. A rock climber wants to estimate the height of a cliff. The climber is 5 feet tall, and casts a shadow that is 9 feet long. At the same time, the cliff casts a shadow that is 58.5 feet long. 5 ft 9 ft 58.5 ft Note: Figure is not drawn to scale. How tall is the cliff? 32

Lesson 1: Solve Problems Using Similarity 3. Ellen is 5.5 feet tall. She walked towards a flagpole that is 25 feet tall, and stopped when she reached the end of the flagpole s shadow. At that time, she measured the length of her shadow as 3 feet. 25 ft 5.5 ft 3 ft How far is Ellen from the base of the flagpole, rounded to the nearest tenth of a foot? 33