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20/19/02/2014 CH.7.1 Factors and Greatest Common Factors Warm Up Solve for x. 1. 16x – 3 = 12x + 13 4 2. 2x – 4 = 90 47 ABCD is a parallelogram. Find each measure. 3. CD 14 4. mC 104° Warm Up Tell whether the second number is a factor of the first number 1. 50, 6 no 2. 105, 7 yes 3. List the factors of 28. ±1, ±2, ±4, ±7, ±14, ±28 Tell whether each number is prime or composite. If the number is composite, write it as the product of two numbers. 4. 11 prime 5. 98 composite; 49 2 Classwork and Homework Classwork 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 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 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 Remember! A prime number has exactly two factors, itself and 1. The number 1 is not prime because it only has one factor. Example 1: Writing Prime Factorizations Write the prime factorization of 98. Method 1 Factor tree Method 2 Ladder diagram Choose any two factors Choose a prime factor of 98 of 98 to begin. Keep finding to begin. Keep dividing by factors until each branch prime factors until the ends in a prime factor. quotient is 1. 98 2 98 7 49 2 49 7 7 7 7 1 98 = 2 7 7 98 = 2 7 7 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 40 2 20 2 10 2 5 40 = 23 5 The prime factorization of 40 is 2 2 2 5 or 23 5. b. 33 11 33 3 33 = 3 11 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 7 7 49 = 7 7 The prime factorization of 49 is 7 7 or 72. 1 19 19 19 = 1 19 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. 26 = 2 13 52 = 2 2 13 2 13 = 26 Write the prime factorization of each number. Align the common factors. 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. factors of 12: 1, 2, 3, 4, 6, 12 List all the factors. factors of 16: 1, 2, 4, 8, 16 Circle the GCF. 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. 15 = 1 3 5 25 = 1 5 5 1 5=5 Write the prime factorization of each number. Align the common factors. The GCF of 15 and 25 is 5. 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 15x3 = 3 5 x x x 9x2 = 3 3 x x 3 Write the prime factorization of each coefficient and write powers as products. Align the common factors. 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 8x2 = 2 2 2 xx coefficient and write 7y3 = 7 y y y powers as products. Align the common factors. The GCF 8x2 and 7y3 is 1. There are no common factors other than 1. Helpful Hint If two terms contain the same variable raised to different powers, the GCF will contain that variable raised to the lower power. Check It Out! Example 3a Find the GCF of each pair of monomials. 18g2 and 27g3 18g2 = 2 3 3 27g3 = gg Write the prime factorization of each coefficient and write powers as products. 3 3 3 g g g Align the common factors. 33 gg Find the product of the common factors. The GCF of 18g2 and 27g3 is 9g2. 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 = The GCF of 16a6 and 9b is 1. 33b Align the common factors. There are no common factors other than 1. Check It Out! Example 3c Find the GCF of each pair of monomials. 8x and 7v2 8x = 2 2 2 x 7v2 = 7vv The GCF of 8x and 7v2 is 1. Write the prime factorization of each coefficient and write powers as products. Align the common factors. There are no common factors other than 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 Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Find the common factors of 18 and 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 = 3 rows 6 containers per row 24 regular milk cartons 6 containers per row = 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: 1, 2, 3, 4, 6, 9, 12, 18, 36 factors of 36 and 48. 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 12 CDs per shelf = 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 52 2. 84 22 3 7 Find the GCF of each pair of numbers. 3. 18 and 75 3 4. 20 and 36 4 Lesson Quiz: Part II Find the GCF of each pair of monomials. 5. 12x and 28x3 4x 6. 27x2 and 45x3y2 9x2 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? 17 Warm Up Solve for x. 1. x2 + 38 = 3x2 – 12 5 or –5 2. 137 + x = 180 43 3. 156 4. Find FE. 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? 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 3 Solve 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 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 10 + 22 = 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? 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 Check It Out! Example 1 Continued 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. Check It Out! Example 1 Continued 3 Solve perimeter of PQRS = Pyth. Thm. Pyth. Thm. Check It Out! Example 1 Continued Daryl needs approximately 191.3 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. Check It Out! Example 1 Continued 4 Look Back To estimate the perimeter, change the side lengths into decimals and round. , and kite is approximately . The perimeter of the 2(54) + 2 (41) = 190. So 191.3 is a reasonable answer. Example 2A: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite cons. sides ∆BCD is isos. 2 sides isos. ∆ CBF CDF isos. ∆ base s mCBF = mCDF Def. of s mBCD + mCBF + mCDF = 180°Polygon Sum Thm. Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF mBCD + mCDF + mCDF = 180° for mCBF. Substitute 52 for mBCD + 52° + 52° = 180° mCDF. mBCD = 76° Subtract 104 from both sides. Example 2B: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC ABC mADC = mABC Kite one pair opp. s Def. of s Polygon Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360° Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve. Example 2C: Using Properties of Kites In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA ABC Kite one pair opp. s mCDA = mABC Def. of s mCDF + mFDA = mABC Add. Post. 52° + mFDA = 115° mFDA = 63° Substitute. Solve. Check It Out! Example 2a In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite cons. sides ∆PQR is isos. RPQ PRQ mQPT = mQRT 2 sides isos. ∆ isos. ∆ base s Def. of s Check It Out! Example 2a Continued mPQR + mQRP + mQPR = 180° Polygon Sum Thm. 78° + mQRT + mQPT = 180°Substitute 78 for mPQR. 78° + mQRT + mQRT = 180°Substitute. 78° + 2mQRT = 180° Substitute. 2mQRT = 102° Subtract 78 from both sides. mQRT = 51° Divide by 2. Check It Out! Example 2b In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS. QPS QRS Kite one pair opp. s mQPS = mQRT + mTRS Add. Post. mQPS = mQRT + 59° Substitute. mQPS = 51° + 59° mQPS = 110° Substitute. Check It Out! Example 2c In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180°Polygon Sum Thm. mSPT = mTRS Def. of s mTRS + mTRS + mRSP = 180° Substitute. 59° + 59° + mRSP = 180°Substitute. mRSP = 62° Simplify. 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. Reading Math Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.” Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° 100 + mB = 180 Same-Side Int. s Thm. Substitute 100 for mC. mB = 80° A B Subtract 100 from both sides. Isos. trap. s base mA = mB Def. of s mA = 80° Substitute 80 for mB 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. 21.9 + 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 BKF BMJ CPCTC 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 mF. mF + mE = 180° E H mE = mH mF + 49° = 180° mF = 131° Same-Side Int. s Thm. Isos. trap. s base Def. of s Substitute 49 for mE. Simplify. Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base KM = JL JL = JN + NL Def. of segs. KM = JN + NL Substitute. Segment Add Postulate KM = 10.6 + 14.8 = 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 mS = mP 2a2 – 54 = a2 a2 Def. of s Substitute 2a2 – 54 for mS and + 27 2 a + 27 for mP. = 81 a = 9 or a = –9 Subtract a2 from both sides and add 54 to both sides. 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. AD = BC Def. of segs. Substitute 12x – 11 for AD and 12x – 11 = 9x – 2 9x – 2 for BC. 3x = 9 x=3 Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. Check It Out! Example 4 Find the value of x so that PQST is isosceles. Q S mQ = mS Trap. with pair base s isosc. trap. Def. of s 2 + 19 for mQ Substitute 2x 2x2 + 19 = 4x2 – 13 and 4x2 – 13 for mS. 32 = 2x2 x = 4 or x = –4 Subtract 2x2 and add 13 to both sides. Divide by 2 and simplify. 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. Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16.5 = 2 (25 + EH) 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? about 191.2 in. In kite HJKL, mKLP = 72°, and mHJP = 49.5°. Find each measure. 2. mLHJ 81° 3. mPKL 18° Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mWZY = 61°. Find mWXY. 119° 5. XV = 4.6, and WY = 14.2. Find VZ. 9.6 6. Find LP. 18 Warm Up 1. Find AB for A (–3, 5) and B (1, 2). 5 2. Find the slope of JK for J(–4, 4) and K(3, –3). –1 ABCD is a parallelogram. Justify each statement. 3. ABC CDA opp. s 4. AEB CED 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 mABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1. 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 6-5-1 the frame is a rectangle. Below are some conditions you can use to determine whether a parallelogram is a rhombus. Caution In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram. 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. Remember! You can also prove that a given quadrilateral is a rectangle, rhombus, or square by using the definitions of the special quadrilaterals. 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. EFGH is a parallelogram. bisecting each other 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. EFGH is a rhombus. with one pair of cons. sides 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, rhombus. , WXYZ is a 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. Check It Out! Example 3A Continued Step 2 Find KM and LN to determine if KLMN is a rectangle. Since , KMLN is a rectangle. Check It Out! Example 3A Continued 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. Check It Out! Example 3A Continued 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. Check It Out! Example 3B Continued 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. Check It Out! Example 3B Continued 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 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 JG = 2JL = 2(30.8) = 61.6 Def. of segs. Substitute and simplify. A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides. 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 13b – 9 = 3b + 4 10b = 13 b = 1.3 Def. of rhombus Substitute given values. Subtract 3b from both sides and add 9 to both sides. Divide both sides by 10. Example 2A Continued TV = XT Def. of rhombus TV = 3b + 4 Substitute 3b + 4 for XT. 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 mVTZ. mVZT = 90° 14a + 20 = 90° a=5 Rhombus diag. Substitute 14a + 20 for mVTZ. Subtract 20 from both sides and divide both sides by 14. Example 2B Continued mVTZ = mZTX Rhombus each diag. bisects opp. s mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° = 20° Substitute 5 for a and simplify. Check It Out! Example 2a CDFG is a rhombus. Find CD. CG = GF 5a = 3a + 17 a = 8.5 GF = 3a + 17 = 42.5 Def. of rhombus Substitute Simplify Substitute CD = GF Def. of rhombus CD = 42.5 Substitute Check It Out! Example 2b CDFG is a rhombus. Find the measure. mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° mGCD + mCDF = 180° b + 3 + 6b – 40 = 180° 7b = 217° b = 31° Def. of rhombus Substitute. Simplify. Divide both sides by 7. Check It Out! Example 2b Continued mGCH + mHCD = mGCD 2mGCH = mGCD Rhombus each diag. bisects opp. s 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Substitute. mGCH = 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. Helpful Hint Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms. 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. 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 = 122 so, SV TW . 1 slope of SV = 11 slope of TW = –11 SV TW Check It Out! Example 3 Continued Step 1 Show that SV and TW are congruent. Since SV = TW, Check It Out! Example 3 Continued Step 2 Show that SV and TW are perpendicular. Since Check It Out! Example 3 Continued 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: Check It Out! Example 4 Continued Statements Reasons 1. PQTS is a rhombus. 1. Given. 2. 2. Rhombus → each diag. bisects opp. s 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 2. CE 35 ft 29 ft Lesson Quiz: Part II PQRS is a rhombus. Find each measure. 3. QP 4. mQRP 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: ABE CDF