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Special Parallelograms

m 1 = 62, m 2 = 62, m 3 = 56. m 1 = 90, m 2 = 20, m 3 = 20, m 4 = 70. Special Parallelograms. Lesson 6-4. Lesson Quiz. 1. The diagonals of a rectangle have lengths 4 + 2 x and 6 x – 20. Find x and the length of each diagonal.

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Special Parallelograms

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  1. m 1 = 62, m 2 = 62, m 3 = 56 m 1 = 90, m 2 = 20, m 3 = 20, m 4 = 70 Special Parallelograms Lesson 6-4 Lesson Quiz 1. The diagonals of a rectangle have lengths 4 + 2x and 6x – 20. Find x and the length of each diagonal. 6; each diagonal has length 16. Find the measures of the numbered angles in each rhombus. 3. 2. Determine whether the quadrilateral can be a parallelogram. If not, write impossible. Explain. 4. Each diagonal is 15 cm long, and one angle of the quadrilateral has measure 45. 5. The diagonals are congruent, perpendicular, and bisect each other. Impossible; if diagonals of a parallelogram are congruent, the quadrilateral is a rectangle, but a rectangle has four right angles. Yes; if diagonals of a parallelogram are congruent, the quadrilateral is a rectangle, and if diagonals of a parallelogram are perpendicular, the quadrilateral is a rhombus, and a rectangle that is a rhombus is a square. 6-5

  2. Trapezoids and Kites Lesson 6-5 Notes The parallel sides of a trapezoid are its bases. The nonparallel sides are its legs.Two angles that share a base of a trapezoid are base angles of the trapezoid. 6-5

  3. Trapezoids and Kites Lesson 6-5 Notes 6-5

  4. Trapezoids and Kites Lesson 6-5 Notes 6-5

  5. Trapezoids and Kites Lesson 6-5 Notes 6-5

  6. Trapezoids and Kites Lesson 6-5 Notes 6-5

  7. mX + mW = 180 Two angles that share a leg of a trapezoid are supplementary. 156 + mW = 180 Substitute. mW = 24 Subtract 156 from each side. Trapezoids and Kites Lesson 6-5 Additional Examples Finding Angle Measures in Trapezoids XYZW is an isosceles trapezoid, and mX = 156. Find mY, mZ, and mW. Because the base angles of an isosceles trapezoid are congruent, mY = mX = 156 and mZ = mW = 24. Quick Check 6-5

  8. Trapezoid ABDC is part of an isosceles triangle whose vertex angle is at the center of the spider web. Because there are 6 adjacent congruent vertex angles at the center of the web, together forming a straight angle, each vertex angle measures , or 30. 180 6 Trapezoids and Kites Lesson 6-5 Additional Examples Real-World Connection Half of a spider’s web is shown below, formed by layers of congruent isosceles trapezoids. Find the measures of the angles in ABDC. By the Triangle Angle-Sum Theorem, mA + mB + 30 = 180, so mA + mB = 150. Because ABDC is part of an isosceles triangle, mA = mB, so 2mA = 150 and mA = mB = 75. 6-5

  9. Another way to find the measure of each acute angle is to divide the difference of 180 and the measure of the vertex angle by 2: 180 – 30 2 = 75 Because the bases of a trapezoid are parallel, the two angles that share a leg are supplementary, so m C = m D = 180 – 75 = 105. Trapezoids and Kites Lesson 6-5 Additional Examples (continued) Quick Check 6-5

  10. Diagonals of a kite are perpendicular. m 2 = 90 RU = RS Definition of a kite m 1 = 72 Isosceles Triangle Theorem m 3 + mRDU + 72 = 180 Triangle Angle–Sum Theorem m RDU = 90 Diagonals of a kite are perpendicular. m 3 + 90 + 72 = 180 Substitute. m 3 + 162 = 180 Simplify. m 3 = 18 Subtract 162 from each side. Trapezoids and Kites Lesson 6-5 Additional Examples Finding Angle Measures in Kites Quick Check Find m 1, m 2, and m 3 in the kite. 6-5

  11. m B = 45, m C = m D = 135 Trapezoids and Kites Lesson 6-5 Lesson Quiz Use isosceles trapezoid ABCD for Exercises 1 and 2. 1. If m A = 45, find m B, m C, and m D. 2. If AC = 3x – 16 and BD = 10x – 86, find x. 10 Use kite GHIJ for Exercises 3–6. 3. Find m 1. 4. Find m 2. 5. Find m 3. 6. Find m 4. 90 9 81 40 6-5

  12. Trapezoids and Kites Lesson 6-5 Check Skills You’ll Need (For help, go to Lesson 6-1.) Find the values of the variables. Then find the lengths of the sides. 3. 1. 2. Check Skills You’ll Need 6-5

  13. Trapezoids and Kites Lesson 6-5 Check Skills You’ll Need (For help, go to Lesson 6-1.) Solutions 1.a – 1.4 = 2a – 7. Solve for a to get a = 5.6. Substitute 5.6 for a in the expression a – 1.4: 5.6 – 1.4 = 4.2. The lengths of the sides are 4.5, 4.2, 4.2, and 4.5. 2. 4y + 6 = 7y – 3. Solve for y to get y = 3. Substitute 3 for y in the expression 4y + 6: 4(3) + 6 = 18. Substitute 3 for y in the expression 5y + 1.4: 5(3) + 1.4 = 16.4. The lengths of the sides are 18, 4.8, 18, and 16.4. 3.n = 3m and n + 6 = 7m – 14. Substitute 3m for n in n + 6 = 7m – 14: 3m + 6 = 7m – 14. Solve for m to get m = 5. Substitute 5 for m in n = 3m to get n = 15. Substitute 5 for m in 7m – 14: 7(5) – 14 = 21. The lengths of the sides are 15, 15, 21, and 21. 6-5

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