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Geometry Section 6-4 1112

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Geometry Section 6-4 1112

1. 1. SECTION 6-4 Rectangles Tuesday, April 29, 14
2. 2. ESSENTIAL QUESTIONS How do you recognize and apply properties of rectangles? How do you determine if parallelograms are rectangles? Tuesday, April 29, 14
3. 3. RECTANGLE Tuesday, April 29, 14
4. 4. RECTANGLE A parallelogram with four right angles. Tuesday, April 29, 14
5. 5. RECTANGLE A parallelogram with four right angles. Four right angles Tuesday, April 29, 14
6. 6. RECTANGLE A parallelogram with four right angles. Four right angles Opposite sides are parallel and congruent Tuesday, April 29, 14
7. 7. RECTANGLE A parallelogram with four right angles. Four right angles Opposite sides are parallel and congruent Opposite angles are congruent Tuesday, April 29, 14
8. 8. RECTANGLE A parallelogram with four right angles. Four right angles Opposite sides are parallel and congruent Opposite angles are congruent Consecutive angles are supplementary Tuesday, April 29, 14
9. 9. RECTANGLE A parallelogram with four right angles. Four right angles Opposite sides are parallel and congruent Opposite angles are congruent Consecutive angles are supplementary Diagonals bisect each other Tuesday, April 29, 14
10. 10. THEOREMS 6.13 - Diagonals of a Rectangle: 6.14 - Diagonals of a Rectangle Converse: Tuesday, April 29, 14
11. 11. THEOREMS 6.13 - Diagonals of a Rectangle: If a parallelogram is a rectangle, then its diagonals are congruent 6.14 - Diagonals of a Rectangle Converse: Tuesday, April 29, 14
12. 12. THEOREMS 6.13 - Diagonals of a Rectangle: If a parallelogram is a rectangle, then its diagonals are congruent 6.14 - Diagonals of a Rectangle Converse: If diagonals of a parallelogram are congruent, then the parallelogram is a rectangle Tuesday, April 29, 14
13. 13. EXAMPLE 1 A rectangular garden gate is reinforced with diagonal braces to prevent it from sagging. If JK = 12 feet and LN = 6.5 feet, ﬁnd KM. Tuesday, April 29, 14
14. 14. EXAMPLE 1 A rectangular garden gate is reinforced with diagonal braces to prevent it from sagging. If JK = 12 feet and LN = 6.5 feet, ﬁnd KM. Since we have a rectangle, the diagonals are congruent. Tuesday, April 29, 14
15. 15. EXAMPLE 1 A rectangular garden gate is reinforced with diagonal braces to prevent it from sagging. If JK = 12 feet and LN = 6.5 feet, ﬁnd KM. Since we have a rectangle, the diagonals are congruent. The diagonals also bisect each other, so JN = LN and KN = MN. Tuesday, April 29, 14
16. 16. EXAMPLE 1 A rectangular garden gate is reinforced with diagonal braces to prevent it from sagging. If JK = 12 feet and LN = 6.5 feet, ﬁnd KM. Since we have a rectangle, the diagonals are congruent. The diagonals also bisect each other, so JN = LN and KN = MN. So JN = LN = KN = MN = 6.5 feet and KM = KN + MN. Tuesday, April 29, 14
17. 17. EXAMPLE 1 A rectangular garden gate is reinforced with diagonal braces to prevent it from sagging. If JK = 12 feet and LN = 6.5 feet, ﬁnd KM. Since we have a rectangle, the diagonals are congruent. The diagonals also bisect each other, so JN = LN and KN = MN. So JN = LN = KN = MN = 6.5 feet and KM = KN + MN. KM = 13 feet Tuesday, April 29, 14
18. 18. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. Tuesday, April 29, 14
19. 19. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 Tuesday, April 29, 14
20. 20. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 Tuesday, April 29, 14
21. 21. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 11x + 2 = 90 Tuesday, April 29, 14
22. 22. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 11x + 2 = 90 −2 −2 Tuesday, April 29, 14
23. 23. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 11x + 2 = 90 −2 −2 11x = 88 Tuesday, April 29, 14
24. 24. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 11x + 2 = 90 −2 −2 11x = 88 11 11 Tuesday, April 29, 14
25. 25. EXAMPLE 2 Quadrilateral RSTU is a rectangle. If m∠RTU = (8x + 4)° and m∠SUR = (3x − 2)°, ﬁnd x. m∠RTU + m∠SUR = 90 8x + 4 + 3x − 2 = 90 11x + 2 = 90 −2 −2 11x = 88 11 11 x = 8 Tuesday, April 29, 14
26. 26. EXAMPLE 3 Some artists stretch their own canvas over wooden frames. This allows them to customize the size of a canvas. In order to ensure that the frame is rectangular before stretching the canvas, an artist measures the sides of the diagonals of the frame. If AB = 12 inches, BC = 35 inches, CD = 12 inches, and DA = 35 inches, how long do the lengths of the diagonals need to be? Tuesday, April 29, 14
27. 27. EXAMPLE 3 Some artists stretch their own canvas over wooden frames. This allows them to customize the size of a canvas. In order to ensure that the frame is rectangular before stretching the canvas, an artist measures the sides of the diagonals of the frame. If AB = 12 inches, BC = 35 inches, CD = 12 inches, and DA = 35 inches, how long do the lengths of the diagonals need to be? The diagonal forms a right triangle with legs of 12 and 35. We need to ﬁnd the hypotenuse. Tuesday, April 29, 14
28. 28. EXAMPLE 3 Tuesday, April 29, 14
29. 29. EXAMPLE 3 a2 + b2 = c2 Tuesday, April 29, 14
30. 30. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 Tuesday, April 29, 14
31. 31. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 144 + 1225 = c2 Tuesday, April 29, 14
32. 32. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 144 + 1225 = c2 1369 = c2 Tuesday, April 29, 14
33. 33. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 144 + 1225 = c2 1369 = c2 1369 = c2 Tuesday, April 29, 14
34. 34. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 144 + 1225 = c2 1369 = c2 1369 = c2 c = 37 Tuesday, April 29, 14
35. 35. EXAMPLE 3 a2 + b2 = c2 122 + 352 = c2 144 + 1225 = c2 1369 = c2 1369 = c2 c = 37 The diagonals must both be 37 inches Tuesday, April 29, 14
36. 36. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. Tuesday, April 29, 14
37. 37. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. Diagonals must be congruent Tuesday, April 29, 14
38. 38. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 Diagonals must be congruent Tuesday, April 29, 14
39. 39. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 Diagonals must be congruent Tuesday, April 29, 14
40. 40. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 Diagonals must be congruent Tuesday, April 29, 14
41. 41. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 Diagonals must be congruent Tuesday, April 29, 14
42. 42. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 Diagonals must be congruent Tuesday, April 29, 14
43. 43. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 Diagonals must be congruent Tuesday, April 29, 14
44. 44. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 Diagonals must be congruent Tuesday, April 29, 14
45. 45. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 Diagonals must be congruent Tuesday, April 29, 14
46. 46. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
47. 47. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
48. 48. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
49. 49. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
50. 50. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
51. 51. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
52. 52. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
53. 53. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 = −6 2 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
54. 54. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 = −6 2 = −3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
55. 55. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 = −6 2 m( JM) = −3 − 3 0 + 2 = −3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
56. 56. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 = −6 2 m( JM) = −3 − 3 0 + 2 = −6 2 = −3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
57. 57. EXAMPLE 4 Quadrilateral JKLM has vertices J(−2, 3), K(1, 4), L(3, −2), and M(0, −3). Determine whether JKLM is a rectangle by using the distance formula, then slope. JL = (−2 − 3)2 + (3 + 2)2 = (−5)2 + 52 = 25 + 25 = 50 KM = (1 − 0)2 + (4 + 3)2 = 12 + 72 = 1 + 49 = 50 m( JK) = 4 − 3 1 + 2 = 1 3 m(LM) = −3 + 2 0 − 3 = −1 −3 = 1 3 m(KL) = −2 − 4 3 − 1 = −6 2 m( JM) = −3 − 3 0 + 2 = −6 2 = −3 = −3 Diagonals must be congruent Opposite sides parallel, consecutive sides perpendicular Tuesday, April 29, 14
58. 58. PROBLEM SET Tuesday, April 29, 14
59. 59. PROBLEM SET p. 422 #1-31 odd, 41, 49, 55, 59, 61 “Character - the willingness to accept responsibility for one's own life - is the source from which self respect springs.” - Joan Didion Tuesday, April 29, 14