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Introduction As an honors student you are expected to complete the following packet the summer before beginning the Geometry Honors course. This packet consists of topics and problem sets from Chapter 1 of the Geometry textbook as well as important prerequisite Algebra 1 knowledge. This packet will be assigned to students on move-up day and extra copies will be provided to students during the first week of school if they were unable to attend move-up day. All packets must be completed and handed in the second week of school. The date is to be determined by individual teachers. This summer assignment will count as a test. A short review of the packet will take place and then an in-class test will be given on all the material in this packet. Algebra Topic 1- Solving Equations An equation is a statement that two expressions are equal. Examples of Equations – 2x 3y 2 7 x 4 y 2 2x y 8 4 w 13 3w 5w 6w 8 Addition and Subtraction Properties of Equality If a b then a c b c and a c b c . We can add or subtract the same quantity from both sides of an equation without changing the truth of the statement. Multiplication and Division Properties of Equality a b If a b , then ac bc and (if c 0) . c c We can multiply or divide each side of an equation by the same quantity (as long as it isn’t zero) without changing the truth of the statement. Zero Product Property If ab 0 , then a 0 or b 0 . If the product of two or more numbers is zero, at least one of the numbers must be zero. Example: Solve the equation 3x 7 2 x 4 for x. Subtract 2x from each side Add 7 to each side 3x 7 2 x 4 2x 2x x7 4 7 7 x 11 Ordered Pair – An ordered pair is written as x, y . They are frequently listed in tables and are solutions to equations. Examples: (2, 3) 5, 7 3, 4 1 , 15 2 Problem Set Topic 1 1. Refer to the figure. a) Write a formula for the perimeter P of the figure. b) Find x if P 64. x 10 B C x2 x 11 A 2. The numbers between – 3 and 5 have been shaded on the number line. 5 -3 a) Is 30 shaded? b) Is 22 shaded? c) Is 22 shaded? 18 d) Is shaded? 5 14 e) Is shaded? 3 f) Is – 3 shaded? In problems 4 and 5, solve each equation. 3. 53x 2 50 4. 3x 6 9 4 5. Billie’s bank contains nickels, dimes and quarters. a) If Billie has eight nickels, five dimes and seven quarters, what is the total value of the coins in the bank? b) Suppose Billie has n nickels, d dimes, and q quarters. Write a formula that can be used to find V, the total value of the coins. 6. If w 2 x y , what is a) The value of w for which 3, 4 is a solution of the equation? b) The value of x for which w 8 and x, 5 is a solution of the equation? 7. The perimeter of the rectangle is 62 centimeters. Find the area of the rectangle. (Hint: First find the value of x). 2x 3 x5 8. Two hot air balloons are flying over a field. The height h, in feet, of the first balloon t seconds from now can be found with the formula ht 50 34t . The height g, in feet, of the second balloon t seconds from now can be found with the formula g t 400 16t . a. In how many seconds will the balloons be at the same height? b. How high will they be at that time? 9. Solve 3x 9 y 12 z for x. 10. Solve 2 x 3 y 10 y 6 x for y. 11. Express a in terms of x. 4x a 11 2x 12. If y 9 4 x and z 6 x 11 , for what value of x is y = z? In problems 13 – 14, solve each equation for x and simplify each expression. 2 13. x 3 7 6 x 2 3 14. abc ac ab bab bc ac 15. Refer to the figure. a. Write an expression for the surface area of the box. b. Find x if the surface area is 261. 12 x 6 16. Refer to the diagram. a. Find the coordinates of point R. b. Find the lengths of PR and RQ . c. Find the area of PQR. y P 0, 3 d. Use the Pythagorean Theorem to find the length of PQ . x R 17. If s 5 x 7 and t x 10 , for what value of x is s 2t 4 ? Q12,2 Algebra Topic 2 - Exponents and Radicals The Product Rule – If a is a real number and m and n are integers, then a m * a n a m n Rule – If x is a real number and n is a positive integer, then Example 3 * 3 (3 * 3 * 3)(3 * 3) 3 * 3 * 3 * 3 * 3 35 3 2 The Power Rule – If a is a real number and m and n are integers, then a m 5 3 5 n a mn Example - 53 * 53 * 53 * 53 * 53 533333 515 x n n x whenever x is defined. n n Example - (4 16 ) 4 16 Rule – If x and y are nonnegative real numbers, then x x x y xy , where y 0 y y Example - 5 2 10 and 5 2 5 2 The Quotient Rule – If a is a real number and m and am n are integers, then n a mn a Standard Radical Form – Radicals written in the form a b where b is an integer have no perfect square factors. Example 35 3 * 3 * 3 * 3 * 3 3 * 3 * 3 * 3 * 3 1* 33 33 or 2 3*3 3*3 3 5 3 352 33 2 3 Example - Rule – If a is a real number, and a 0 , and n is an 1 integer, then a n n . a 1 Example - 5 3 3 5 The Distributive Property of Exponentiation over Multiplication – If x and y are real numbers and n is an integer, then x n * y n ( x * y) n Example - 6 5 * 2 5 (6 * 2) 5 The Distributive Property of Exponentiation over Division – If x and y are real numbers y 0 , and n is an integer, then xn x y n y n 5 Example - 65 6 35 5 2 2 Rule – If a is a real number and a 0 then a 0 1 Example - 5 0 1 252 9 * 28 3 4 7 6 7 Conjugates – Expressions in the form a b c and a b c Example - 2 7 and 2 7 are conjugates. Problem Set –Topic 2 In problems 1-2 simplify each expression. 3 2 1. 2b 3 * 3 b 4 2. 12 w 4 15 w12 * 8w8 (6 w) 2 In problems 3-5, indicate whether each statement is True or False. ( x 0) 4. x a * x b x ab 3. x 2a ( x a ) 2 5. xa x1 a x 6. Evaluate the following a. 5 2 3 b. 53 2 c. 5 5 d. 5 2.5 7. Find an expression for the sum of the volumes of the cube and the box. x2 x2 x x x2 x2 3 8. Evaluate each expression. a. 6 2 b. 32 1 d. 2 1 31 1 1 1 c. 2 3 1 1 In problems 9-10 rewrite each expression without parenthesis using only negative exponents. x 3 y 2 9. 4 w 5 x 4 y 3 10. 4 w 5 2 11. Write an expression, in simplified form, for the area of the trapezoid. x4 y9 4x 2 y 5 3x 4 y 9 12. Simplify x * x 2 2 y * 4 y 2 1 13. Solve 5 5 x 5 3 xy 14. Evaluate x y 2 18 19 2 *y 1 1 (5 z ) 3 w 3 for w, x, y, and z. x4 * 4 y 5 for (x, y) = (5.2, -2). In problems 15-17 rewrite each expression without parentheses, using only positive exponents. x 6 15. 4 y 5 x3 16. 2 x 5 17. x 3 y 4 5 In problems 18-20 simplify each expression. 18. 20 xy 3 z 2 24 x 4 y 2 z 19. x 2 ( x 2 x 2) x 3 ( x 1) 20. xy ( x 2 y 2 ) 2 4 In problems 21-24 simplify each expression. 21. xy 2 x y xy 23. 3x y * 5x 15 x y 2 22. 4 4 xy 5 18 3 5 3 10 x 3 20 x 2 15 x 4 30 x 3 24. xy 17 Algebra Topic 3–Factoring & Solving Quadratic Equations Multiplication of a monomial and polynomial example – Find the product 2 x 3 ( x 3 3x 2 2 x 5) 2 x 3 ( x 3 3x 2 2 x 5 Write product 2 x 3 ( x 3 ) 2 x 3 (3x 2 ) 2 x 3 (2 x) 2 x 3 (5) Distributive Property 2 x 6 6 x 5 4 x 4 10 x 3 Product of powers property Multiplication of binomials using FOIL example Find the product 3a 4a 2 (3a)( a) (3a)( 2) (4)( a) (4)( 2) Write product of terms 3a 2 (6a) 4a (8) Multiply 3a 2 2a 8 Combine like terms Factoring Trinomials ( x 2 bx c ) Rule - x 2 bx c =(x + p) (x + q) provided p + q = b and p*q=c. Example when b and c are positive – Factor x 2 11x 18 Find two positive factors of 18 whose sum is 11. 18, 1 9, 2 6, 3 The factors 9 & 2 have a product of 18 and a sum of 11, so they are the correct value of p and q. You can check your solution by distribution. Does ( x 9)( x 2) x 2 11x 18 ? Example when b is negative and c is positive – Factor n 2 6n 8 Because b is negative and c is positive, p and q must both be negative. Make a list of the negative factors of 18 -8, -1 -4, -2 The factors -4 & -2 have a product of 18 and a sum of -6, so they are the correct value of p and q. You can check your solution by distribution. Does (n 4)( n 2) n 2 6n 8 ? Example when b is positive and c is negative – Factor y 2 2 y 15 Because c is negative, p and q must have different signs. Make a list of the factors of -15 -15, 1 15, -1 -5, 3 5, -3 The factors 5 & -3 have a product of -15 and a sum of 2, so they are the correct value of p and q. You can check your solution by distribution. Does ( y 5)( y 3) y 2 2 y 15 Example when b is negative and c is negative – Factor s 2 2s 24 Because both b and c are negative, p and q must have different signs. Make a list of the factors of -24 1, -24 -1, 24 2, -12 -2, 12 3, - 8 -3, 8 4, -6 -4, 6 The factors 4 & -6 have a product of -24 and a sum of -2, so they are the correct value of p and q. You can check your solution by distribution. Does ( s 4)( s 6) s 2 2s 24 Solving Quadratic Equations Solving by Factoring – Once you have factored the equation you set each parenthesis equal to zero to determine the values of x. Example – Solve the quadratic equation for x using factoring m 2 5m 6 0 Step 1 – Factor (m + 2)(m + 3) = 0 Step 2 – Set each parenthesis equal to zero (m + 2) = 0 and (m + 3) = 0 Step 3 – Solve each equation for x. m+2=0 m = -2 m+3=0 m = -3 Solving by the Quadratic Formula - Sometimes quadratic equations can not be factored so we must use the quadratic formula. If ax 2 bx c 0 then x b b 2 4ac 2a Example – Solve the quadratic equation 3 x 2 5 x 8 0 for x. Step 1 – Determine the value of a, b & c. a = 3 b = 5 c = -8 Step 2 – Substitute the values for a, b, & c into the quadratic formula. x 5 (5) 2 4(3)( 8) 2*3 Step 3 – Simplify. 5 (5) 2 4(3)( 8) 5 25 96 5 121 5 11 x 2*3 6 6 6 Step 4 – Break expression into two parts. x 5 11 6 1 and 6 6 x 5 11 16 2.6 6 6 Problem Set - Topic 3 Directions – Simplify the following expressions. 1. x 82x 5 2. x 6 3. 5 x 2 (3x 2 6 x 12) 4. 3xx 4x 2 5. 3x 2 6 2 x 2 7 2 Directions – Factor the following expressions. 6. x 2 8 x 20 7. x 2 16 x 64 8. 2 x 2 8 x 24 9. x 3 11x 2 30 x 10. y 2 8 y 48 Directions – Solve the following quadratic equations using either factoring or the quadratic formula. 11. x 2 2 x 63 0 12. 5 x 2 13x 6 13. n 2 n 6 14. 3 x 2 7 x 4 0 15. 3x 2 4 11x 16. 6 x 2 2 x 28 0 17. x 2 17 x 18 Algebra Topic 4 – System of Equations Linear Systems by Substitution - the strategy is to obtain an equation in only one variable. The solution is the point of intersection of two lines. Example - 1 y x 5 Solve the system 2 2 x y 7 The first equation is solved for y, so we can substitute equation. Then 2 x y 7 becomes 2 x 1 x 5 for the value of y in the second 2 1 x5 7. 2 2.5 x 5 7 2.5 x 2 Now this equation has only one variable And we can solve for x. x 0.8 To find the y-coordinate, substitute 0.8 for the value of x in the first equation. y 1 (0.8) 5 2 0.4 5 5.4 y 1 x 5 becomes 2 By this method, the solution is , x, y 0.8,5.4 Linear Systems by Elimination - There are two types of elimination. The first is called the addition/subtraction method. The second requires you to multiply before you can eliminate with addition and subtraction. Example Addition & Subtraction Method – Step 1 – Add or Subtract the equations to eliminate one variable Step 2 – Solve the resulting equation for the other variable Step 3 – Substitute in either original equation to find the value of the eliminated variable 2 x 3 y 11 2 x 5 y 13 Solve the linear system Step 1 – Add the equations to eliminate the variable. Step 2 – Solve for y. 2 x 3 y 11 2 x 5 y 13 Step 3 – Substitute 3 for y in either equation and solve for x. 2 x 3 y 11 2 x 3(3) 11 x 1 Write Equation 1 Substitute 3 for y Solve for x 8 y 24 y3 The solution is (1, 3) Example Multiplication First Method Sometimes in a linear system variables can not be eliminated by using addition and subtraction. First we must multiply one or both of the equations by a constant so that adding or subtracting the equations will eliminate a variable. 6 x 5 y 19 2 x 3 y 5 Solve the linear system Equation 1 Equation 2 Solution – Step 1 – Multiply Eqn 2 by -3 so that the coefficients of x are opposites. Step 2 – Add the equations 6 x 5 y 19 6 x 9 y 15 4y 4 Step 3 – Solve for y y 1 6 x 5 y 19 2 x 3 y 5 *(-3) Step 4 – Substitute -1 for y in either of the original equations and solve for x. 2x 3 y 5 2 x 3(1) 5 2 x (3) 5 2x 8 x4 The solution is 4,1 Write Equation 2 Substitute -1 for y Multiply Subtract -3 from each side Divide each side by 2. Problem Set Topic 4 Directions – Solve each system of equations for the missing values using the indicated method. Write your solution as a point of intersection. Check your solution in both equations. Show all work. 4 x 3 y 27 y 2x 1 1. 5 x y 13 4 x 3 y 18 2. Substitution Method Add/Sub/Mult. Method 3x 4 y 26 x 2 y 2 Substitution Method 7 x 3 y 23 x 2 y 13 Add/Sub/Mult. Method x 3 y 13 x y 5 Either Method 3 x 7 y 22 2 x 8 y 2 Either Method 3. 4. 5. 6. Chapter 1 –Introduction to Geometry Geometry Topic 1 - 1.1 Getting Started Points are represented by dots. We use capital letters to name points. B A E C D Lines are made up of points and are straight. The arrows at the end of a line show that the lines extend infinitely far in both directions. E B m l C D F The line on the left is called line m. We can name a line by any two points on it, so the line in the middle can be called one of the following; BD , BC , CD , CB , DB , DC The line on the right can be called by any of three names; line l, EF , FE In Algebra 1 you learned that a number line is formed when a numerical value is assigned to each point on a line. B A The coordinate for A is -2. -4 -3-2-1 0 1 2 3 4 The coordinate for B is 1 1 2 A line segment or segment are made up of points and are straight. A segment has a definite beginning and end. S R P X Q A segment is named in terms of its endpoints. The segment on the left can be called either RS or SR . In the second figure there are two segments. The vertical one can be called either PX or XP . The horizontal segment can be named as either XQ or QX . A ray is like a line and segment because it is made up of points and it is straight. It is different from a line or segment because it begins at an endpoint and then extends infinitely far in only one direction. B C D E A When we name a ray, we name the endpoint first so that it is clear where the ray begins. The ray on the left is called AB . The ray on the right is called CD or CE . As long as we use the endpoint first any other point on the ray can be used in its name. An angle is made up of two rays with a common endpoint. This endpoint is called the vertex of the angle. The rays are called the sides of the angle. A Y P B C 3 1 O 2 R X The angle of the left is called 3 . The 3 placed inside the angle near the vertex names it. The middle angle can be called by any of these three names; In the figure on the right there are three angles. BAC o o o CAB A 1 can be called POY or YOP 2 can be called YOR , YOX , ROY or XOY . The other angle can be named POR . See if you can find three other names for this angle. A triangle has three segments as its sides and three angles even though they do have endpoints. Triangle ABC = ABC A B C The triangle is the union of three segments ABC AB BC AC . The intersection of any two sides is a vertex of the triangle. AB BC B . Problem Set 2.1 (pg. 7 #1-6, 8-14) l 1. What are three possible names for the line shown? B A C 2. What are four possible names for the angle shown? 7 E D 3. Can the ray shown be called XY ? Y X 4. Name the sides of RST . R S T 5. a. AB BC C D b.EC EA c. AC DB E d .DC AB e AC EC f .BA BC A g .EC CB BE 6. a. Name OPR in all possible ways. B S T O b. What is the vertex of TOS . c. How many angles have vertex R? d. Name TSP in all other possible ways. P R e. How many triangles are there in the figure? 7a. A line is named up of b. An angle is the union of two . with a common . 8. Draw a number line and label points F, G, H and J with the coordinates 4 2 , 2, 5, and 3.5 3 respectively. One of these points is the midpoint (the halfway point) between two others. Which is it? 8.6 cm 9. Given a rectangle with sides 2.5 cm and 8.6 cm long, find a. The rectangle’s area 2.5 cm b. The rectangle’s perimeter (the distance around it) H 10a. In HJK , HJ is twice as long as JK and exactly as long as HK . If the length of HJ is 15, find the perimeter of HJK J b. If the length of HJ were 4x, the length of HK were K 3x, the length of JK were 2x, and the perimeter of HJK were 63, what would the length of HJ be? 11. Draw a diagram in which AB CD CB. 12. Draw a diagram in which the intersection of AEF and DPC is ED. 13. a What percentage of the triangles in the diagram T E have CT as a side? A b. What percentage have AC as a side? X C