20/02/2014 CH.7.2 Factoring by GCF.

20/02/2014 CH.7.2 Factoring by GCF

Warm Up Solve for x. 1. 16x – 3 = 12x + 13 2. 2x – 4 = 90 ABCD is a parallelogram. Find each measure. 3. CD 4. mC 4 47 14 104°

Warm Up 1. 2(w + 1) 2. 3x(x2 – 4) 2w + 2 3x3 – 12x 3. 4h2 and 6h 2h
Simplify. 2w + 2 3x3 – 12x Find the GCF of each pair of monomials. 3. 4h2 and 6h 2h 4. 13p and 26p5 13p

Objective Factor polynomials by using the greatest common factor.

Recall that the Distributive Property states that ab + ac =a(b + c)
Recall that the Distributive Property states that ab + ac =a(b + c). The Distributive Property allows you to “factor” out the GCF of the terms in a polynomial to write a factored form of the polynomial. A polynomial is in its factored form when it is written as a product of monomials and polynomials that cannot be factored further. The polynomial 2(3x – 4x) is not fully factored because the terms in the parentheses have a common factor of x.

Example 1A: Factoring by Using the GCF
Factor each polynomial. Check your answer. 2x2 – 4 2x2 = 2  x  x Find the GCF. 4 = 2  2 2 The GCF of 2x2 and 4 is 2. Write terms as products using the GCF as a factor. 2x2 – (2  2) 2(x2 – 2) Use the Distributive Property to factor out the GCF. Multiply to check your answer. Check 2(x2 – 2) The product is the original polynomial.  2x2 – 4

Aligning common factors can help you find the greatest common factor of two or more terms.
Writing Math

Example 1B: Factoring by Using the GCF
Factor each polynomial. Check your answer. 8x3 – 4x2 – 16x 8x3 = 2  2  2  x  x  x Find the GCF. 4x2 = 2  2  x  x 16x = 2  2  2  2  x The GCF of 8x3, 4x2, and 16x is 4x. 2  2  x = 4x Write terms as products using the GCF as a factor. 2x2(4x) – x(4x) – 4(4x) Use the Distributive Property to factor out the GCF. 4x(2x2 – x – 4) Check 4x(2x2 – x – 4) Multiply to check your answer. The product is the original polynomials. 8x3 – 4x2 – 16x 

Example 1C: Factoring by Using the GCF
Factor each polynomial. Check your answer. –14x – 12x2 – 1(14x + 12x2) Both coefficients are negative. Factor out –1. 14x = 2   x 12x2 = 2  2  3  x  x Find the GCF. The GCF of 14x and 12x2 is 2x. 2  x = 2x –1[7(2x) + 6x(2x)] Write each term as a product using the GCF. –1[2x(7 + 6x)] Use the Distributive Property to factor out the GCF. –2x(7 + 6x)

 Example 1C: Continued Factor each polynomial. Check your answer.
–14x – 12x2 Check –2x(7 + 6x) Multiply to check your answer. –14x – 12x2  The product is the original polynomial.

When you factor out –1 as the first step, be sure to include it in all the other steps as well.
Caution!

Example 1D: Factoring by Using the GCF
Factor each polynomial. Check your answer. 3x3 + 2x2 – 10 3x3 =  x  x  x Find the GCF. 2x2 =  x  x 10 =  5 There are no common factors other than 1. 3x3 + 2x2 – 10 The polynomial cannot be factored further.

 Check It Out! Example 1a Factor each polynomial. Check your answer.
5b + 9b3 5b = 5  b Find the GCF. 9b = 3  3  b  b  b The GCF of 5b and 9b3 is b. b Write terms as products using the GCF as a factor. 5(b) + 9b2(b) Use the Distributive Property to factor out the GCF. b(5 + 9b2) Multiply to check your answer. b(5 + 9b2) Check The product is the original polynomial. 5b + 9b3 

Check It Out! Example 1b Factor each polynomial. Check your answer. 9d2 – 82 9d2 = 3  3  d  d Find the GCF. 82 =  2  2  2  2  2 There are no common factors other than 1. 9d2 – 82 The polynomial cannot be factored further.

Check It Out! Example 1c Factor each polynomial. Check your answer. –18y3 – 7y2 – 1(18y3 + 7y2) Both coefficients are negative. Factor out –1. 18y3 = 2  3  3  y  y  y Find the GCF. 7y2 = 7  y  y y  y = y2 The GCF of 18y3 and 7y2 is y2. Write each term as a product using the GCF. –1[18y(y2) + 7(y2)] –1[y2(18y + 7)] Use the Distributive Property to factor out the GCF.. –y2(18y + 7)

Factor each polynomial. Check your answer.
Check It Out! Example 1d Factor each polynomial. Check your answer. 8x4 + 4x3 – 2x2 8x4 = 2  2  2  x  x  x  x 4x3 = 2  2  x  x  x Find the GCF. 2x2 = 2  x  x 2  x  x = 2x2 The GCF of 8x4, 4x3 and –2x2 is 2x2. Write terms as products using the GCF as a factor. 4x2(2x2) + 2x(2x2) –1(2x2) 2x2(4x2 + 2x – 1) Use the Distributive Property to factor out the GCF. Check 2x2(4x2 + 2x – 1) Multiply to check your answer. 8x4 + 4x3 – 2x2 The product is the original polynomial.

To write expressions for the length and width of a rectangle with area expressed by a polynomial, you need to write the polynomial as a product. You can write a polynomial as a product by factoring it.

Example 2: Application The area of a court for the game squash is (9x2 + 6x) square meters. Factor this polynomial to find possible expressions for the dimensions of the squash court. A = 9x2 + 6x The GCF of 9x2 and 6x is 3x. = 3x(3x) + 2(3x) Write each term as a product using the GCF as a factor. = 3x(3x + 2) Use the Distributive Property to factor out the GCF. Possible expressions for the dimensions of the squash court are 3x m and (3x + 2) m.

Check It Out! Example 2 What if…? The area of the solar panel on another calculator is (2x2 + 4x) cm2. Factor this polynomial to find possible expressions for the dimensions of the solar panel. A = 2x2 + 4x The GCF of 2x2 and 4x is 2x. = x(2x) + 2(2x) Write each term as a product using the GCF as a factor. = 2x(x + 2) Use the Distributive Property to factor out the GCF. Possible expressions for the dimensions of the solar panel are 2x cm, and (x + 2) cm.

Sometimes the GCF of terms is a binomial
Sometimes the GCF of terms is a binomial. This GCF is called a common binomial factor. You factor out a common binomial factor the same way you factor out a monomial factor.

Example 3: Factoring Out a Common Binomial Factor
Factor each expression. A. 5(x + 2) + 3x(x + 2) The terms have a common binomial factor of (x + 2). 5(x + 2) + 3x(x + 2) (x + 2)(5 + 3x) Factor out (x + 2). B. –2b(b2 + 1)+ (b2 + 1) The terms have a common binomial factor of (b2 + 1). –2b(b2 + 1) + (b2 + 1) –2b(b2 + 1) + 1(b2 + 1) (b2 + 1) = 1(b2 + 1) (b2 + 1)(–2b + 1) Factor out (b2 + 1).

Example 3: Factoring Out a Common Binomial Factor
Factor each expression. C. 4z(z2 – 7) + 9(2z3 + 1) There are no common factors. 4z(z2 – 7) + 9(2z3 + 1) The expression cannot be factored.

Check It Out! Example 3 Factor each expression. a. 4s(s + 6) – 5(s + 6) The terms have a common binomial factor of (s + 6). 4s(s + 6) – 5(s + 6) (4s – 5)(s + 6) Factor out (s + 6). b. 7x(2x + 3) + (2x + 3) 7x(2x + 3) + (2x + 3) The terms have a common binomial factor of (2x + 3). 7x(2x + 3) + 1(2x + 3) (2x + 3) = 1(2x + 3) (2x + 3)(7x + 1) Factor out (2x + 3).

Check It Out! Example 3 : Continued
Factor each expression. c. 3x(y + 4) – 2y(x + 4) There are no common factors. 3x(y + 4) – 2y(x + 4) The expression cannot be factored. d. 5x(5x – 2) – 2(5x – 2) The terms have a common binomial factor of (5x – 2 ). 5x(5x – 2) – 2(5x – 2) (5x – 2)(5x – 2) (5x – 2)2 (5x – 2)(5x – 2) = (5x – 2)2

You may be able to factor a polynomial by grouping
You may be able to factor a polynomial by grouping. When a polynomial has four terms, you can make two groups and factor out the GCF from each group.

Example 4A: Factoring by Grouping
Factor each polynomial by grouping. Check your answer. 6h4 – 4h3 + 12h – 8 Group terms that have a common number or variable as a factor. (6h4 – 4h3) + (12h – 8) 2h3(3h – 2) + 4(3h – 2) Factor out the GCF of each group. 2h3(3h – 2) + 4(3h – 2) (3h – 2) is another common factor. (3h – 2)(2h3 + 4) Factor out (3h – 2).

Example 4A Continued Factor each polynomial by grouping. Check your answer. Check (3h – 2)(2h3 + 4) Multiply to check your solution. 3h(2h3) + 3h(4) – 2(2h3) – 2(4) 6h4 + 12h – 4h3 – 8 6h4 – 4h3 + 12h – 8  The product is the original polynomial.

Example 4B: Factoring by Grouping
Factor each polynomial by grouping. Check your answer. 5y4 – 15y3 + y2 – 3y (5y4 – 15y3) + (y2 – 3y) Group terms. Factor out the GCF of each group. 5y3(y – 3) + y(y – 3) 5y3(y – 3) + y(y – 3) (y – 3) is a common factor. (y – 3)(5y3 + y) Factor out (y – 3).

Example 4B Continued Factor each polynomial by grouping. Check your answer. 5y4 – 15y3 + y2 – 3y Check (y – 3)(5y3 + y) Multiply to check your solution. y(5y3) + y(y) – 3(5y3) – 3(y) 5y4 + y2 – 15y3 – 3y 5y4 – 15y3 + y2 – 3y  The product is the original polynomial.

Check It Out! Example 4a Factor each polynomial by grouping. Check your answer. 6b3 + 8b2 + 9b + 12 (6b3 + 8b2) + (9b + 12) Group terms. 2b2(3b + 4) + 3(3b + 4) Factor out the GCF of each group. (3b + 4) is a common factor. 2b2(3b + 4) + 3(3b + 4) (3b + 4)(2b2 + 3) Factor out (3b + 4).

Check It Out! Example 4a Continued
Factor each polynomial by grouping. Check your answer. 6b3 + 8b2 + 9b + 12 Multiply to check your solution. Check (3b + 4)(2b2 + 3) 3b(2b2) + 3b(3)+ (4)(2b2) + (4)(3) 6b3 + 9b+ 8b2 + 12 The product is the original polynomial. 6b3 + 8b2 + 9b + 12 

Check It Out! Example 4b Factor each polynomial by grouping. Check your answer. 4r3 + 24r + r2 + 6 (4r3 + 24r) + (r2 + 6) Group terms. 4r(r2 + 6) + 1(r2 + 6) Factor out the GCF of each group. 4r(r2 + 6) + 1(r2 + 6) (r2 + 6) is a common factor. (r2 + 6)(4r + 1) Factor out (r2 + 6).

Check It Out! Example 4b Continued
Factor each polynomial by grouping. Check your answer. Check (4r + 1)(r2 + 6) Multiply to check your solution. 4r(r2) + 4r(6) +1(r2) + 1(6) 4r3 + 24r +r2 + 6 4r3 + 24r + r2 + 6  The product is the original polynomial.

If two quantities are opposites, their sum is 0.
(5 – x) + (x – 5) 5 – x + x – 5 – x + x + 5 – 5 0 + 0 Helpful Hint

Recognizing opposite binomials can help you factor polynomials
Recognizing opposite binomials can help you factor polynomials. The binomials (5 – x) and (x – 5) are opposites. Notice (5 – x) can be written as –1(x – 5). –1(x – 5) = (–1)(x) + (–1)(–5) Distributive Property. = –x + 5 Simplify. = 5 – x Commutative Property of Addition. So, (5 – x) = –1(x – 5)

Example 5: Factoring with Opposites
Factor 2x3 – 12x – 3x by grouping. 2x3 – 12x – 3x (2x3 – 12x2) + (18 – 3x) Group terms. 2x2(x – 6) + 3(6 – x) Factor out the GCF of each group. 2x2(x – 6) + 3(–1)(x – 6) Write (6 – x) as –1(x – 6). 2x2(x – 6) – 3(x – 6) Simplify. (x – 6) is a common factor. (x – 6)(2x2 – 3) Factor out (x – 6).

Check It Out! Example 5a Factor each polynomial by grouping. 15x2 – 10x3 + 8x – 12 (15x2 – 10x3) + (8x – 12) Group terms. Factor out the GCF of each group. 5x2(3 – 2x) + 4(2x – 3) 5x2(3 – 2x) + 4(–1)(3 – 2x) Write (2x – 3) as –1(3 – 2x). Simplify. (3 – 2x) is a common factor. 5x2(3 – 2x) – 4(3 – 2x) (3 – 2x)(5x2 – 4) Factor out (3 – 2x).

Check It Out! Example 5b Factor each polynomial by grouping. 8y – 8 – x + xy (8y – 8) + (–x + xy) Group terms. 8(y – 1)+ (x)(–1 + y) Factor out the GCF of each group. 8(y – 1)+ (x)(y – 1) (y – 1) is a common factor. (y – 1)(8 + x) Factor out (y – 1) .

Lesson Quiz: Part I Factor each polynomial. Check your answer. 1. 16x + 20x3 2. 4m4 – 12m2 + 8m Factor each expression. 3. 7k(k – 3) + 4(k – 3) 4. 3y(2y + 3) – 5(2y + 3) 4x(4 + 5x2) 4m(m3 – 3m + 2) (k – 3)(7k + 4) (2y + 3)(3y – 5)

Lesson Quiz: Part II Factor each polynomial by grouping. Check your answer. 5. 2x3 + x2 – 6x – 3 6. 7p4 – 2p p – 18 7. A rocket is fired vertically into the air at 40 m/s. The expression –5t t + 20 gives the rocket’s height after t seconds. Factor this expression. (2x + 1)(x2 – 3) (7p – 2)(p3 + 9) –5(t2 – 8t – 4)

Warm Up 1. 50, 6 2. 105, 7 3. List the factors of 28. no yes
1. 50, , 7 3. List the factors of 28. Tell whether each number is prime or composite. If the number is composite, write it as the product of two numbers. Tell whether the second number is a factor of the first number no yes ±1, ±2, ±4, ±7, ±14, ±28 4. 11 prime 5. 98 composite; 49  2

Classwork and Homework
7.1 (Pages 459 to 461) Exercises 1, 17 to 30, 31, 32 to 35, 36, 37, 38, 47 to 55, 57, 58, 60 to 68. Homework Homework booklet Ch. 7.1

Objectives Write the prime factorization of numbers.
Find the GCF of monomials.

Vocabulary prime factorization greatest common factor

The whole numbers that are multiplied to find a product are called factors of that product. A number is divisible by its factors. You can use the factors of a number to write the number as a product. The number 12 can be factored several ways. Factorizations of 12 1 12  2 6 3 4

The order of factors does not change the product, but there is only one example below that cannot be factored further. The circled factorization is the prime factorization because all the factors are prime numbers. The prime factors can be written in any order, and except for changes in the order, there is only one way to write the prime factorization of a number. Factorizations of 12 1 12  2 6 3 4

A prime number has exactly two factors, itself and 1
A prime number has exactly two factors, itself and 1. The number 1 is not prime because it only has one factor. Remember!

Example 1: Writing Prime Factorizations
Write the prime factorization of 98. Method 1 Factor tree Method 2 Ladder diagram Choose any two factors of 98 to begin. Keep finding factors until each branch ends in a prime factor. Choose a prime factor of 98 to begin. Keep dividing by prime factors until the quotient is 1. 98  98 49 7 1 2 98 =  98 =  The prime factorization of 98 is 2  7  7 or 2  72.

Check It Out! Example 1 Write the prime factorization of each number. a. 40 b. 33 40  2 5 33 3 11 40 = 23  5 33 = 3  11 The prime factorization of 40 is 2  2  2  5 or 23  5. The prime factorization of 33 is 3  11.

Check It Out! Example 1 Write the prime factorization of each number. c. 49 d. 19 49  19 1 49 = 7  7 19 = 1  19 The prime factorization of 49 is 7  7 or 72. The prime factorization of 19 is 1  19.

Factors that are shared by two or more whole numbers are called common factors. The greatest of these common factors is called the greatest common factor, or GCF. Factors of 12: 1, 2, 3, 4, 6, 12 Factors of 32: 1, 2, 4, 8, 16, 32 Common factors: 1, 2, 4 The greatest of the common factors is 4.

Example 2A: Finding the GCF of Numbers
Find the GCF of each pair of numbers. 100 and 60 Method 1 List the factors. factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100 List all the factors. factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 Circle the GCF. The GCF of 100 and 60 is 20.

Example 2B: Finding the GCF of Numbers
Find the GCF of each pair of numbers. 26 and 52 Method 2 Prime factorization. Write the prime factorization of each number. 26 =  13 52 = 2  2  13 Align the common factors. 2  13 = 26 The GCF of 26 and 52 is 26.

Check It Out! Example 2a Find the GCF of each pair of numbers. 12 and 16 Method 1 List the factors. List all the factors. factors of 12: 1, 2, 3, 4, 6, 12 Circle the GCF. factors of 16: 1, 2, 4, 8, 16 The GCF of 12 and 16 is 4.

Check It Out! Example 2b Find the GCF of each pair of numbers. 15 and 25 Method 2 Prime factorization. Write the prime factorization of each number. 15 = 1  3  5 25 = 1  5  5 Align the common factors. 1  = 5 The GCF of 15 and 25 is 5.

You can also find the GCF of monomials that include variables
You can also find the GCF of monomials that include variables. To find the GCF of monomials, write the prime factorization of each coefficient and write all powers of variables as products. Then find the product of the common factors.

Example 3A: Finding the GCF of Monomials
Find the GCF of each pair of monomials. 15x3 and 9x2 Write the prime factorization of each coefficient and write powers as products. 15x3 = 3  5  x  x  x 9x2 = 3  3  x  x Align the common factors. 3  x  x = 3x2 Find the product of the common factors. The GCF of 15x3 and 9x2 is 3x2.

Example 3B: Finding the GCF of Monomials
Find the GCF of each pair of monomials. 8x2 and 7y3 Write the prime factorization of each coefficient and write powers as products. 8x2 = 2  2  2  x  x 7y3 =  y  y  y Align the common factors. There are no common factors other than 1. The GCF 8x2 and 7y3 is 1.

If two terms contain the same variable raised to different powers, the GCF will contain that variable raised to the lower power. Helpful Hint

Check It Out! Example 3a Find the GCF of each pair of monomials. 18g2 and 27g3 Write the prime factorization of each coefficient and write powers as products. 18g2 = 2  3  3  g  g 27g3 =  3  3  g  g  g Align the common factors. 3  3  g  g Find the product of the common factors. The GCF of 18g2 and 27g3 is 9g2.

Find the GCF of each pair of monomials.
Check It Out! Example 3b Find the GCF of each pair of monomials. Write the prime factorization of each coefficient and write powers as products. 16a6 and 9b 16a6 = 2  2  2  2  a  a  a  a  a  a 9b =  3  b Align the common factors. The GCF of 16a6 and 9b is 1. There are no common factors other than 1.

Check It Out! Example 3c Find the GCF of each pair of monomials. 8x and 7v2 Write the prime factorization of each coefficient and write powers as products. 8x = 2  2  2  x 7v2 =  v  v Align the common factors. There are no common factors other than 1. The GCF of 8x and 7v2 is 1.

Example 4: Application A cafeteria has 18 chocolate-milk cartons and 24 regular-milk cartons. The cook wants to arrange the cartons with the same number of cartons in each row. Chocolate and regular milk will not be in the same row. How many rows will there be if the cook puts the greatest possible number of cartons in each row? The 18 chocolate and 24 regular milk cartons must be divided into groups of equal size. The number of cartons in each row must be a common factor of 18 and 24.

Example 4 Continued Find the common factors of 18 and 24. Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 The GCF of 18 and 24 is 6. The greatest possible number of milk cartons in each row is 6. Find the number of rows of each type of milk when the cook puts the greatest number of cartons in each row.

Example 4 Continued 18 chocolate milk cartons 6 containers per row = 3 rows 24 regular milk cartons = 4 rows When the greatest possible number of types of milk is in each row, there are 7 rows in total.

Check It Out! Example 4 Adrianne is shopping for a CD storage unit. She has 36 CDs by pop music artists and 48 CDs by country music artists. She wants to put the same number of CDs on each shelf without putting pop music and country music CDs on the same shelf. If Adrianne puts the greatest possible number of CDs on each shelf, how many shelves does her storage unit need? The 36 pop and 48 country CDs must be divided into groups of equal size. The number of CDs in each row must be a common factor of 36 and 48.

Check It Out! Example 4 Continued
Find the common factors of 36 and 48. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 The GCF of 36 and 48 is 12. The greatest possible number of CDs on each shelf is 12. Find the number of shelves of each type of CDs when Adrianne puts the greatest number of CDs on each shelf.

36 pop CDs 12 CDs per shelf = 3 shelves 48 country CDs = 4 shelves When the greatest possible number of CD types are on each shelf, there are 7 shelves in total.

Lesson Quiz: Part I Write the prime factorization of each number. 1. 50 2. 84 Find the GCF of each pair of numbers. 3. 18 and 75 4. 20 and 36 2  52 22  3  7 3 4

Lesson Quiz: Part II Find the GCF of each pair of monomials. 5. 12x and 28x3 6. 27x2 and 45x3y2 7. Cindi is planting a rectangular flower bed with 40 orange flower and 28 yellow flowers. She wants to plant them so that each row will have the same number of plants but of only one color. How many rows will Cindi need if she puts the greatest possible number of plants in each row? 4x 9x2 17

Warm Up Solve for x. 1. x = 3x2 – 12 x = 180 3. 4. Find FE. 5 or –5 43 156

Classwork/Homework Classwork (Pages 444 to 448 ) Exercises 1, 14 to 25, 26, 27 to 32, 34 to 36, 40, 42, 47, 48, 49. Homework Homework Booklet Chapter: 6.6

Objectives Use properties of kites to solve problems.
Use properties of trapezoids to solve problems.

Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid
base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid

A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

Example 1: Problem-Solving Application
Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along . She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?

Understand the Problem
Example 1 Continued 1 Understand the Problem The answer will be the amount of wood Lucy has left after cutting the dowel. 2 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find , and Add these lengths to find the length of .

Example 1 Continued Solve N bisects JM. Pythagorean Thm.
3 N bisects JM. Pythagorean Thm. Pythagorean Thm.

Example 1 Continued Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is, 36 – 32.4  3.6 cm Lucy will have 3.6 cm of wood left over after the cut.

Example 1 Continued Look Back
4 Look Back To estimate the length of the diagonal, change the side length into decimals and round , and The length of the diagonal is approximately = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer.

Check It Out! Example 1 What if...? Daryl is going to make a kite by doubling all the measures in the kite. What is the total amount of binding needed to cover the edges of his kite? How many packages of binding must Daryl buy?

Understand the Problem
Check It Out! Example 1 Continued 1 Understand the Problem The answer has two parts. • the total length of binding Daryl needs • the number of packages of binding Daryl must buy

2 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Use the Pythagorean Theorem and the properties of kites to find the unknown side lengths. Add these lengths to find the perimeter of the kite.

Solve 3 Pyth. Thm. Pyth. Thm. perimeter of PQRS =

Daryl needs approximately inches of binding. One package of binding contains 2 yards, or 72 inches. packages of binding In order to have enough, Daryl must buy 3 packages of binding.

4 Look Back To estimate the perimeter, change the side lengths into decimals and round. , and The perimeter of the kite is approximately 2(54) + 2 (41) = 190. So is a reasonable answer.

Example 2A: Using Properties of Kites
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite  cons. sides  ∆BCD is isos. 2  sides isos. ∆ CBF  CDF isos. ∆ base s  mCBF = mCDF Def. of   s mBCD + mCBF + mCDF = 180° Polygon  Sum Thm.

Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCDF + mCDF = 180° Substitute 52 for mCDF. mBCD + 52° + 52° = 180° Subtract 104 from both sides. mBCD = 76°

Example 2B: Using Properties of Kites
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC Kite  one pair opp. s  mADC = mABC Def. of  s Polygon  Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360°

Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

Example 2C: Using Properties of Kites
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA  ABC Kite  one pair opp. s  mCDA = mABC Def. of  s mCDF + mFDA = mABC  Add. Post. 52° + mFDA = 115° Substitute. mFDA = 63° Solve.

Check It Out! Example 2a In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite  cons. sides  ∆PQR is isos. 2  sides  isos. ∆ RPQ  PRQ isos. ∆  base s  mQPT = mQRT Def. of  s

Check It Out! Example 2a Continued
mPQR + mQRP + mQPR = 180° Polygon  Sum Thm. Substitute 78 for mPQR. 78° + mQRT + mQPT = 180° 78° + mQRT + mQRT = 180° Substitute. 78° + 2mQRT = 180° Substitute. Subtract 78 from both sides. 2mQRT = 102° mQRT = 51° Divide by 2.

Check It Out! Example 2b In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS. QPS  QRS Kite  one pair opp. s  mQPS = mQRT + mTRS  Add. Post. mQPS = mQRT + 59° Substitute. mQPS = 51° + 59° Substitute. mQPS = 110°

Check It Out! Example 2c In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180° Polygon  Sum Thm. mSPT = mTRS Def. of  s mTRS + mTRS + mRSP = 180° Substitute. 59° + 59° + mRSP = 180° Substitute. Simplify. mRSP = 62°

A trapezoid is a quadrilateral with exactly one pair of parallel sides
A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base.

If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

Theorem 6-6-5 is a biconditional statement
Theorem is a biconditional statement. So it is true both “forward” and “backward.” Reading Math

Example 3A: Using Properties of Isosceles Trapezoids
Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A  B Isos. trap. s base  mA = mB Def. of  s mA = 80° Substitute 80 for mB

Example 3B: Using Properties of Isosceles Trapezoids
KB = 21.9 and MF = 32.7. Find FB. Isos.  trap. s base  KJ = FM Def. of  segs. KJ = 32.7 Substitute 32.7 for FM. KB + BJ = KJ Seg. Add. Post. BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.

Example 3B Continued Same line. KFJ  MJF Isos. trap.  s base  Isos. trap.  legs  ∆FKJ  ∆JMF SAS CPCTC BKF  BMJ FBK  JBM Vert. s 

Example 3B Continued Isos. trap.  legs  ∆FBK  ∆JBM AAS CPCTC FB = JB Def. of  segs. FB = 10.8 Substitute 10.8 for JB.

Check It Out! Example 3a Find mF. mF + mE = 180° Same-Side Int. s Thm. E  H Isos. trap. s base  mE = mH Def. of  s mF + 49° = 180° Substitute 49 for mE. mF = 131° Simplify.

Check It Out! Example 3b JN = 10.6, and NL = Find KM. Isos. trap. s base  KM = JL Def. of  segs. JL = JN + NL Segment Add Postulate KM = JN + NL Substitute. KM = = 25.4 Substitute and simplify.

Example 4A: Applying Conditions for Isosceles Trapezoids
Find the value of a so that PQRS is isosceles. Trap. with pair base s   isosc. trap. S  P mS = mP Def. of  s Substitute 2a2 – 54 for mS and a for mP. 2a2 – 54 = a2 + 27 Subtract a2 from both sides and add 54 to both sides. a2 = 81 a = 9 or a = –9 Find the square root of both sides.

Example 4B: Applying Conditions for Isosceles Trapezoids
AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles. Diags.   isosc. trap. Def. of  segs. AD = BC Substitute 12x – 11 for AD and 9x – 2 for BC. 12x – 11 = 9x – 2 Subtract 9x from both sides and add 11 to both sides. 3x = 9 x = 3 Divide both sides by 3.

Check It Out! Example 4 Find the value of x so that PQST is isosceles. Trap. with pair base s   isosc. trap. Q  S mQ = mS Def. of  s Substitute 2x for mQ and 4x2 – 13 for mS. 2x = 4x2 – 13 Subtract 2x2 and add 13 to both sides. 32 = 2x2 Divide by 2 and simplify. x = 4 or x = –4

The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.

Example 5: Finding Lengths Using Midsegments
Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.

Substitute the given values.
Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = (25 + EH) 2 Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides.

Lesson Quiz: Part I 1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite? In kite HJKL, mKLP = 72°, and mHJP = 49.5°. Find each measure. 2. mLHJ 3. mPKL about in. 81° 18°

Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mWZY = 61°. Find mWXY. 5. XV = 4.6, and WY = Find VZ. 6. Find LP. 119° 9.6 18

Warm Up 5 –1 1. Find AB for A (–3, 5) and B (1, 2).
2. Find the slope of JK for J(–4, 4) and K(3, –3). ABCD is a parallelogram. Justify each statement. 3. ABC  CDA 4. AEB  CED 5 –1  opp. s  Vert. s Thm.

Objective Prove that a given quadrilateral is a rectangle, rhombus, or square.

When you are given a parallelogram with certain
properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

Example 1: Carpentry Application
A manufacture builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem

Check It Out! Example 1 A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle? Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one  of WXYZ is a right . If one angle is a right , then by Theorem the frame is a rectangle.

Below are some conditions you can use to determine whether a parallelogram is a rhombus.

In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.
Caution To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43.

You can also prove that a given quadrilateral is a
rectangle, rhombus, or square by using the definitions of the special quadrilaterals. Remember!

Example 2A: Applying Conditions for Special Parallelograms
Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a rhombus. The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

Example 2B: Applying Conditions for Special Parallelograms
Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a square. Step 1 Determine if EFGH is a parallelogram. Given Quad. with diags. bisecting each other  EFGH is a parallelogram.

Example 2B Continued Step 2 Determine if EFGH is a rectangle. Given. EFGH is a rectangle. with diags.   rect. Step 3 Determine if EFGH is a rhombus. with one pair of cons. sides   rhombus EFGH is a rhombus.

Example 2B Continued Step 4 Determine is EFGH is a square. Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition. The conclusion is valid.

Check It Out! Example 2 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: ABC is a right angle. Conclusion: ABCD is a rectangle. The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram .

Example 3A: Identifying Special Parallelograms in the Coordinate Plane
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)

Example 3A Continued Step 1 Graph PQRS.

Example 3A Continued Step 2 Find PR and QS to determine if PQRS is a rectangle. Since , the diagonals are congruent. PQRS is a rectangle.

Example 3A Continued Step 3 Determine if PQRS is a rhombus. Since , PQRS is a rhombus. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.

Example 3B: Identifying Special Parallelograms in the Coordinate Plane
Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3) Step 1 Graph WXYZ.

Example 3B Continued Step 2 Find WY and XZ to determine if WXYZ is a rectangle. Since , WXYZ is not a rectangle. Thus WXYZ is not a square.

Example 3B Continued Step 3 Determine if WXYZ is a rhombus. Since (–1)(1) = –1, , WXYZ is a rhombus.

Check It Out! Example 3A Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

Check It Out! Example 3A Continued
Step 1 Graph KLMN.

Step 2 Find KM and LN to determine if KLMN is a rectangle. Since , KMLN is a rectangle.

Step 3 Determine if KLMN is a rhombus. Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus.

Step 4 Determine if KLMN is a square. Since KLMN is a rectangle and a rhombus, it has four right angles and four congruent sides. So KLMN is a square by definition.

Check It Out! Example 3B Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)

Check It Out! Example 3B Continued
Step 1 Graph PQRS.

Step 2 Find PR and QS to determine if PQRS is a rectangle. Since , PQRS is not a rectangle. Thus PQRS is not a square.

Step 3 Determine if PQRS is a rhombus. Since (–1)(1) = –1, are perpendicular and congruent. PQRS is a rhombus.

Lesson Quiz: Part I 1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

Lesson Quiz: Part II 2. Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: PQRS and PQNM are parallelograms. Conclusion: MNRS is a rhombus. valid

Lesson Quiz: Part III 3. Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply. AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD = 1, so AC  BD. ABCD is a rhombus.

Objectives Prove and apply properties of rectangles, rhombuses, and squares. Use properties of rectangles, rhombuses, and squares to solve problems.

Vocabulary rectangle rhombus square

A second type of special quadrilateral is a rectangle
A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

Example 1: Craft Application
A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect.  diags.  KM = JL = 86 Def. of  segs.  diags. bisect each other Substitute and simplify.

Check It Out! Example 1a Carpentry The rectangular gate has diagonal braces. Find HJ. Rect.  diags.  HJ = GK = 48 Def. of  segs.

Check It Out! Example 1b Carpentry The rectangular gate has diagonal braces. Find HK. Rect.  diags.  Rect.  diagonals bisect each other JL = LG Def. of  segs. JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

A rhombus is another special quadrilateral
A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

Like a rectangle, a rhombus is a parallelogram
Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

Example 2A: Using Properties of Rhombuses to Find Measures
TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9 = 3b + 4 Substitute given values. 10b = 13 Subtract 3b from both sides and add 9 to both sides. b = 1.3 Divide both sides by 10.

Example 2A Continued TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV = 3b + 4 TV = 3(1.3) + 4 = 7.9 Substitute 1.3 for b and simplify.

Example 2B: Using Properties of Rhombuses to Find Measures
TVWX is a rhombus. Find mVTZ. mVZT = 90° Rhombus  diag.  14a + 20 = 90° Substitute 14a + 20 for mVTZ. Subtract 20 from both sides and divide both sides by 14. a = 5

Example 2B Continued Rhombus  each diag. bisects opp. s mVTZ = mZTX mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° = 20° Substitute 5 for a and simplify.

Check It Out! Example 2a CDFG is a rhombus. Find CD. CG = GF Def. of rhombus 5a = 3a + 17 Substitute a = 8.5 Simplify GF = 3a + 17 = 42.5 Substitute CD = GF Def. of rhombus CD = 42.5 Substitute

Check It Out! Example 2b CDFG is a rhombus. Find the measure. mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° mGCD + mCDF = 180° Def. of rhombus b b – 40 = 180° Substitute. 7b = 217° Simplify. b = 31° Divide both sides by 7.

Check It Out! Example 2b Continued
mGCH + mHCD = mGCD Rhombus  each diag. bisects opp. s 2mGCH = mGCD 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Substitute. mGCH = 17° Simplify and divide both sides by 2.

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms.
Helpful Hint

Example 3: Verifying Properties of Squares
Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

Example 3 Continued Step 1 Show that EG and FH are congruent. Since EG = FH,

Example 3 Continued Step 2 Show that EG and FH are perpendicular. Since ,

Example 3 Continued Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other.

SV ^ TW SV = TW = 122 so, SV @ TW . slope of TW = –11 1 slope of SV =
Check It Out! Example 3 The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other. SV = TW = so, TW . 1 11 slope of SV = slope of TW = –11 SV ^ TW

Step 1 Show that SV and TW are congruent. Since SV = TW,

Step 2 Show that SV and TW are perpendicular. Since

Step 3 Show that SV and TW bisect each other. Since SV and TW have the same midpoint, they bisect each other. The diagonals are congruent perpendicular bisectors of each other.

Example 4: Using Properties of Special Parallelograms in Proofs
Given: ABCD is a rhombus. E is the midpoint of , and F is the midpoint of . Prove: AEFD is a parallelogram.

Example 4 Continued ||

Check It Out! Example 4 Given: PQTS is a rhombus with diagonal Prove:

Statements Reasons 1. PQTS is a rhombus. 1. Given. 2. Rhombus → each diag. bisects opp. s 2. 3. QPR  SPR 3. Def. of  bisector. 4. 4. Def. of rhombus. 5. 5. Reflex. Prop. of  6. 6. SAS 7. 7. CPCTC

Lesson Quiz: Part I A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length. 1. TR CE 35 ft 29 ft

Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP mQRP 42 51°

Lesson Quiz: Part III 5. The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

Lesson Quiz: Part IV 6. Given: ABCD is a rhombus. Prove: 

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