3 = 1, a c, a d, b c, and b d.

Transcription

1 hapter 7 Maintaining Mathematical Proficienc (p. 357) 1. (7 x) = 16 (7 x) = 16 7 x = 7 7 x = 3 x 1 = 3 1 x = 3. 7(1 x) + = 19 7(1 x) = 1 7(1 x) = x = x = x 1 = 1 x = 3. 3(x 5) + 8(x 5) = 11(x 5) = 11(x 5) 11 = 11 x 5 = x = 7. Slope of line a: ( ) = + 6 = 6 = 3 ( ) Slope of line b: = + 0 ( 3) 3 = 3 = 3 Slope of line c: ( 3) = = 3 6 = 1 Slope of line d: = = ecause the slopes of line a and line b are equal, a b. 1 ecause ( () = 1, c d. ) 5. Slope of line a: 1 ( 3) = Slope of line b: 1 ( 3) 0 ( 3) = Slope of line c: 1 = 3 = 3 Slope of line d: () = 6 = 3 = 3 + = 6 8 = 3 ecause the slopes of line a and line b are equal, a b. ecause the slopes of line c and line d are equal, c d. ecause ( 3) ( 3 = 1, a c, a d, b c, and b d. ) ( ) 6. Slope of line a: = + ( ) + = 6 = 1 3 Slope of line b: 3 ( ) = 3 + = 1 = Slope of line c: = 1 = 3 Slope of line d: 0 () = 1 = 1 1 ecause ( = 1, b d and c d. ) ecause the slopes of line b and line c are equal, b c. 7. You can follow the order of operations with all of the other operations in the equation and treat the operations in the expression separatel. hapter 7 Mathematical Practices (p. 358) 1. false; There is no overlap between the set of trapezoids and the set of kites.. true; There is no overlap between the set of kites and the set of parallelograms. 3. false; There is area inside the parallelogram circle for rhombuses and other parallelograms that do not fall inside the circle for rectangles.. true; ll squares are inside the categor of quadrilaterals. 5. Sample answer: ll kites are quadrilaterals. No trapezoids are kites. ll squares are parallelograms. ll squares are rectangles. Some rectangles are squares. No rhombuses are trapezoids. 6. Quadrilaterals Parallelograms Trapezoids Rhombuses Squares Rectangles clic Quadrilaterals Kites opright ig Ideas Learning, LL Geometr 31

2 7.1 xplorations (p. 359) 1. a. The sum of the measures of the interior angles of quadrilateral is =. The sum of the measures of the interior angles of pentagon FGHI is = 50. c. Number of sides, n Sum of angle measures, S Measure of one interior angle F G H 3. The sum S of the measure of the interior angles of a polgon is given b the equation S = (n )180, where n is the number of sides of the polgon. b. Number of sides, n Sum of angle measures, S c I (9, 160) (8, 1080) (7, 900) (6, 70) (5, 50) (, 360) (3, 180) x d. S = (n ) 180; Let n represent the number of sides of the polgon. If ou subtract and multipl the difference b 180, then ou get the sum of the measures of the interior angles of a polgon.. a. n equation used to determine the measure of one angle in a regular polgon is S = (n ) 180. n b. S = (5 ) = = 50 5 = 108 The measure of one interior angle of a regular pentagon is (n )180. S = n (1 ) 180 = 1 = (10)(180) = = 150 The measure of one interior angle in a regular dodecagon is Monitoring Progress (pp ) 1. (n ) 180 = (11 ) 180 = = 160 The sum of the measures of the interior angles of the 11-gon is (n ) 180 = 10 (n ) 180 = n = 8 n = 10 The polgon has 10 sides, so it is a decagon. 3. x + 3x + 5x + 7x = 16x = 16x 16 = 16 x =.5 3x = 3.5 = x = 5.5 = x = 7.5 = The angle measures are.5, 67.5, 11.5, and Geometr opright ig Ideas Learning, LL

3 . The total sum of the angle measures of the polgon with 5 sides is x = x = 50 x = 06 x = 103 m S = m T = Pentagon is equilateral but not equiangular. 6. The sum of the measures of the exterior angles is. The sum of the known exterior angle measures is = 83. So, the measure of the exterior angle at the sixth vertex is 83 = You can find the measure of each exterior angle b subtracting the measure of the interior angle from 180. In xample 6, the measure of each exterior angle is = xercises (pp ) Vocabular and ore oncept heck 1. segment connecting consecutive vertices is a side of the polgon, not a diagonal.. The one that does not belong is the sum of the measures of the interior angles of a pentagon. This sum is 50, but in each of the other three statements the sum is. Monitoring Progress and Modeling with Mathematics 3. (n ) 180 = (9 ) 180 = = 160 The sum of the measures of the interior angles in a nonagon is (n ) 180 = (1 ) 180 = = 160 The sum of the measures of the interior angles in a 1-gon is (n ) 180 = (16 ) 180 = = 50 The sum of the measures of the interior angles in a 16-gon is (n ) 180 = 70 (n ) = n = n = 6 The polgon has 6 sides, so it is a hexagon. 8. (n ) 180 = 1080 (n ) 180 = n = 6 n = 8 The polgon has 8 sides, so it is an octagon. 9. (n ) 180 = 50 (n ) 180 = n = 1 n = 16 The polgon has 16 sides, so it is a 16-gon. 10. (n ) 180 = 30 (n ) 180 = n = 18 n = 0 The polgon has 0 sides, so it is a 0-gon. 11. XYZW is a quadrilateral, therefore the sum of the measures of the interior angles is ( ) 180 = x = 96 + x = 360 x = 6 1. HJKG is a quadrilateral, therefore the sum of the measures of the interior angles is ( ) 180 = x = 9 + x = 360 x = KLMN is a quadrilateral, therefore the sum of the measures of the interior angles is ( ) 180 = x + 9 = 71 + x = x = (n ) 180 = (0 ) 180 = = 30 The sum of the measures of the interior angles in a 0-gon is 30. opright ig Ideas Learning, LL Geometr 33

4 1. is a quadrilateral, therefore the sum of the measures of the interior angles is ( ) 180 =. x = x + 61 = 360 x = The polgon has 6 sides, therefore the sum of the measures of the interior angles is (6 ) 180 = x = x = 70 x = The polgon has 5 sides, therefore the sum of the measures of the interior angles is (5 ) 180 = x = x = 50 x = The polgon has 6 sides, therefore the sum of the measures of the interior angles is (6 ) 180 = x = x = 70 x = The polgon has 8 sides, therefore the sum of the measures of the interior angles is (8 ) 180 = x x = x = x = 65 1 x = The polgon has 5 sides, therefore the sum of the measures of the interior angles is (5 ) 180 = 50. x + x = 50 x = 50 x = 18 m X = m Y = 9 x = 9 0. The polgon has 5 sides, therefore the sum of the measures of the interior angles is (5 ) 180 = 50. x x = 50 x + 56 = 50 x = 8 m X = m Y = 1 x = 1 1. The polgon has 6 sides, therefore the sum of the measures of the interior angles is (6 ) 180 = x + x = 70 x = 70 x = 01 m X = m Y = x = The polgon has 6 sides, therefore the sum of the measures of the interior angles is (6 ) 180 = x x = 70 x + 50 = 70 x = 70 m X = m Y = x = x + 9 = 360 x = 111 x = x + x = x + 5 = 360 x = x + x = 5x + 00 = 360 5x = 160 x = x x + 5 = 3x + 16 = 360 3x = 198 x = 66 x = The sum of the measures of the interior angles of a pentagon is (5 ) 180 = = 50. ach interior angle: 50 5 = 108 ach exterior angle: 5 = 7 The measure of each interior angle of a regular pentagon is 108 and the measure of each exterior angle is The sum of the measures of the interior angles of an 18-gon is (18 ) 180 = = 880. ach interior angle: = 160 ach exterior angle: 18 = 0 The measure of each interior angle of a regular 18-gon is 160 and the measure of each exterior angle is 0. 3 Geometr opright ig Ideas Learning, LL

5 9. The sum of the measures of the interior angles of a 5-gon is (5 ) 180 = = 770. ach interior angle: = 17 ach exterior angle: 5 = 8 The measure of each interior angle of a regular 5-gon is 17 and the measure of each exterior angle is The sum of the measures of the interior angles of a 90-gon is (90 ) 180 = = 15,80. ach interior angle: 15,80 = ach exterior angle: 90 = The measure of each interior angle of a regular 90-gon is 176 and the measure of each exterior angle is. 31. The measure of one interior angle of a regular pentagon was found, but the measure of one exterior angle should be found b dividing b the number of angles. The correct response should be 5 = There is onl one exterior angle at each vertex. So, the measure of one exterior angle is found b dividing b the number of vertices. The correct response should be 5 = regular hexagon has 6 sides. The sum of the measures of the interior angles is (6 ) 180 = 70. The measure of each interior angle is 70 6 = regular decagon has 10 sides. The sum of the measures of the interior angles is (10 ) 180 = 10. The measure of each interior angle is 10 = 1. The measure of each 10 exterior angle is 10 = (n ) 180 = x n (n ) 180 = nx n = nx n = nx n 180 = nx n 180 nx = 360 n(180 x) = 360 n(180 x) 180 x = x 360 n = 180 x The formula to find the number of sides n in a regular polgon given the measure of one interior angle x 360 is: n = 180 x. 36. x = n n x = n 360 n n x = 360 n x = 360 x x n = 360 x The formula to find the number of sides n in a regular polgon given the measure of one exterior angle x is: n = 360 x. 37. n = = = 15 The number of sides of a polgon where each interior angle has a measure of 156 is n = = 15 = The number of sides of a polgon where each interior angle has a measure of 165 is. 39. n = 9 = 0 The number of sides of a polgon where each exterior angle has a measure of 9 is n = = 60 6 The number of sides of a polgon where each exterior angle has a measure of 6 is , ;. n = = 18 = 0. n = = 9 = 0. n = = n = = Solving the equation found in xercise 35 for n ields a positive integer greater than or equal to 3 for and, but not for and.. In a pentagon, when all the diagonals from one vertex are drawn, the polgon is divided into three triangles. ecause the sum of the measures of the interior angles of each triangle is 180, the sum of the measures of the interior angles of the pentagon is (5 ) 180 = = In a quadrilateral, when all the diagonals from one vertex are drawn, the polgon is divided into two triangles. ecause the sum of the measures of the interior angles of each triangle is 180, the sum of the measures of the interior angles of the quadrilateral is 180 =. opright ig Ideas Learning, LL Geometr 35

6 . es; ecause an interior angle and an adjacent exterior angle of a polgon form a linear pair, ou can use the Polgon xterior ngles Theorem (Thm. 7.) to find the measure of the exterior angles, and then ou can subtract this value from 180 to find the interior angle measures of a regular polgon. 5. hexagon has 6 angles. If of the exterior angles have a measure of x, the other two each have a measure of (x + 8), and the total exterior angle sum is, then: x + [(x + 8)] = x + [x + 96] = 360 x + x + 19 = 360 8x + 19 = 360 8x = 168 x = 1 (x + 8) = (1 + 8) = 69 = 138 So, the measures of the exterior angles are 1, 1, 1, 1, 138, and es; The measure of the angle where the polgon caves in is greater than 180 but less than. 7. ivide the quadrilateral into two triangles. The sum of the measures of the interior angles of each triangle is 180. Therefore, the total sum of the interior angle measurements for this quadrilateral is 180 =. ivide the heptagon into five triangles. The sum of the measures of the interior angles of each triangle is 180. Therefore, the total sum of the interior angle measurements for this heptagon is = 900. When diagonals are drawn from the vertex of the concave angle as shown, the polgon is divided into n triangles whose interior angle measures have the same total as the sum of the interior angle measures of the original polgon. So, an expression to find the sum of the measures of the interior angles for a concave polgon is (n ) The base angles of P are congruent exterior angles of the regular octagon, each with a measure of 5. So, m P = 180 (5 ) = 90. G H F P ivide the pentagon into three triangles. The sum of the measures of the interior angles of each triangle is 180. Therefore, the total sum of the interior angle measurements for this pentagon is = 50. ivide the hexagon into four triangles. The sum of the measures of the interior angles of each triangle is 180. Therefore, the total sum of the interior angle measurements for this hexagon is 180 = a. The formula for the number of sides n in a regular polgon, where h(n) is the measure of an interior angle is h(n) = (n ) 180. n b. h(9) = (9 ) 180 = = 160 = c. 150 = (n ) 180 n 150n = (n ) n = 180n n = n 30 = n = 1 When h(n) = 150, n = Geometr opright ig Ideas Learning, LL

7 d. Number of sides h(n) Measure of each interior angle (8, 135) (7, 18.6) (6, 10) (5, 108) (, 90) (3, 60) n The value of h(n) increases on a curve that gets less steep as n increases. 50. no; The interior angles are supplements of the adjacent exterior angles, and because the exterior angles have different values, the supplements will be different as well. 51. In a convex n-gon, the sum of the measures of the n interior angles is (n ) 180 using the Polgon Interior ngles Theorem (Thm. 7.1). ecause each of the n interior angles forms a linear pair with its corresponding exterior angle, ou know that the sum of the measures of the n interior and exterior angles is 180n. Subtracting the sum of the interior angle measures from the sum of the measures of the linear pairs gives ou 180n [(n ) 180 ] =. 5. In order to have 50 = 3 more triangles formed b the 180 diagonals, the new polgon will need 3 more sides. Maintaining Mathematical Proficienc 53. x + 79 = 180 x = x = x = 180 x = x = (8x 16) + (3x + 0) = x + = x = 176 x = (6x 19) + (3x + 10) = 180 9x 9 = 180 9x = 189 x = 1 7. xplorations (p. 367) 1. a. heck students work; onstruct and a line parallel to through point. onstruct and a line parallel to through point. onstruct a point at the intersection of the line drawn parallel to and the line drawn parallel to. Finall, construct,,, and b removing the rest of the parallel lines drawn. b. heck students work. (For sample in text, m = m = 63.3 and m = m = ); Opposite angles are congruent, and consecutive angles are supplementar. c. heck students work. (For sample in text, = =. and = =.); Opposite sides are congruent. d. heck students work; Opposite angles of a parallelogram are congruent. onsecutive angles of a parallelogram are supplementar. Opposite sides of a parallelogram are congruent.. a. heck students work. b. heck students work. c. heck students work. (For sample in text, = = 1.58 and = =.55.) Point bisects and. d. The diagonals of a parallelogram bisect each other. 3. parallelogram is a quadrilateral where both pairs of opposite sides are congruent and parallel, opposite angles are congruent, consecutive angles are supplementar, and the diagonals bisect each other. 7. Monitoring Progress (pp ) 1. m G = m m = 60 m G = 60 FG = H H = 8 FG = 8 In parallelogram GHF, FG = 8 and m G = 60.. m J = m L x = 50 x = 5 JK = ML 18 = = In parallelogram JKLM, x = 5 and = 15. opright ig Ideas Learning, LL Geometr 37

8 3. m + m = 180 m + m = 180 3m = 180 m = 60. Given and GF are parallelograms. Prove and F are supplementar angles. G STTMNTS 1. and GF are parallelograms.. and are supplementar angles. F RSONS 1. Given. Parallelogram onsecutive ngles Theorem (Thm. 7.5) 3. m + m = efinition of supplementar angles. F. Parallelogram Opposite ngles Theorem (Thm. 7.) 5. m = m F 5. efinition of congruent angles 6. m + m F = Substitution Propert of qualit 7. and F are supplementar angles. 7. efinition of supplementar angles 5. the Parallelogram iagonals Theorem (Thm. 7.6), the diagonals of a parallelogram bisect each other. Midpoint of TV : ( 1 + 3, ) = (, 6 = (, 3) ) Midpoint of SU : ( , ) = (, 6 = (, 3) ) The coordinates of the intersection of the diagonals are (, 3). 6. Slope of = 5 = 3 Starting at, go up units and left 3 units. So, the coordinates of are (0, 1). x 7. xercises (pp ) Vocabular and ore oncept heck 1. In order to be a quadrilateral, a polgon must have sides, and parallelograms alwas have sides. In order to be a parallelogram, a polgon must have sides with opposite sides parallel. Quadrilaterals alwas have sides, but do not alwas have opposite sides parallel.. The two angles that are consecutive to the given angle are supplementar to it. So, ou can find each of their measures b subtracting the measure of the given angle from 180. The angle opposite the given angle is congruent and therefore has the same measure. Monitoring Progress and Modeling with Mathematics 3. The Parallelogram Opposite Sides Theorem (Thm. 7.3) applies here. Therefore, x = 9 and = 15.. the Parallelogram Opposite Sides Theorem (Thm. 7.3), n = 1 and m + 1 = 6. Therefore, m = Parallelogram Opposite Sides Theorem (Thm. 7.3): z 8 = 0 z = 8 Parallelogram Opposite ngles Theorem (Thm. 7.): (d 1) = 105 d = 16 Therefore, z = 8 and d = Parallelogram Opposite Sides Theorem (Thm. 7.3): 16 h = 7 h = 9 h = 9 Parallelogram Opposite ngles Theorem (Thm. 7.): (g + ) = 65 g = 61 Therefore, h = 9 and g = m + m = m = 180 m = m M + m N = m N = 180 m N = LM = 13; the Parallelogram Opposite Sides Theorem (Thm. 7.3), LM = QN. 10. LP = 7; the Parallelogram iagonals Theorem (Thm. 7.6), LP = PN. 11. LQ = 8; the Parallelogram Opposite Sides Theorem (Thm. 7.3), LQ = MN. 38 Geometr opright ig Ideas Learning, LL

9 1. the Parallelogram iagonals Theorem (Thm. 7.6), MP = PQ. MQ = MP = 8. = Parallelogram onsecutive ngles Theorem (Thm. 7.5) m LMN + m MLQ = 180 m LMN = 180 m LMN = Parallelogram Opposite ngles Theorem (Thm. 7.) m NQL = m NML m NQL = Parallelogram Opposite ngles Theorem (Thm. 7.) m MNQ = m MLQ m MNQ = lternate Interior ngles Theorem (Thm. 3.) m LMQ = m NQM m LMQ = 9 1. In a parallelogram, consecutive angles are supplementar; ecause quadrilateral STUV is a parallelogram, S and V are supplementar. So, m V = = In a parallelogram, the diagonals bisect each other. So the two parts of GJ are congruent to each other; ecause quadrilateral GHJK is a parallelogram, GF FJ. 3. Given and F are parallelograms. Prove F STTMNTS 1. and F are parallelograms.., F 3. F RSONS 1. Given. Parallelogram Opposite Sides Theorem (Thm. 7.3) 3. Transitive Propert of ongruence (Thm..1) F 17. n + 70 = 180 n = 110 m = 70 m = 35 So, n = 110 and m = (b 10) + (b + 10) = 180 b = 180 d = (b + 10) d = b = 90 d = 100 c = (b 10) c = c = 80 So, b = 90, c = 80, and d = k + = 11 k = 7 m = 8 So, k = 7 and m = u + = 5u 10 u = 5u 1 3u = 1 3u 3 = 1 3 u = v 3 = 6 v = 18 So, u = and v = 18.. Given, GF, and HJK are parallelograms. Prove 3 H STTMNTS 1., GF, and HJK are parallelograms.. 1, 3, 1 RSONS 1. Given opright ig Ideas Learning, LL Geometr 39 F J 3 K 1 G. Parallelogram Opposite ngles Theorem (Thm. 7.) Transitive Propert of ongruence (Thm..). 3. Transitive Propert of ongruence (Thm..) 5. the Parallelogram iagonals Theorem (Thm. 7.6), the diagonals of a parallelogram bisect each other. Midpoint of WY : ( , ) = (, 5 = (1,.5) ) Midpoint of ZX : ( + 0, ) = (, 5 = (1,.5) ) The coordinates of the intersection of the diagonals are (1,.5). 6. the Parallelogram iagonals Theorem (Thm. 7.6), the diagonals of a parallelogram bisect each other. Midpoint of QS : ( ( ), ) = ( 0, 1 = (0, 0.5) ) Midpoint of TR : ( 5 + ( 5) + ( 1), ) = ( 0, 1 = (0, 0.5) ) The coordinates of the intersection of the diagonals are (0, 0.5).

10 F G x Slope of = = 3 1 = 3 Starting at G, go up 3 units and left 1 unit. So, the coordinates of F are (3, 3). F G x Slope of FG = = 7 1 = 7 Starting at, go up 7 units and left 1 unit. So, the coordinates of are ( 3, 3). G F x Slope of = 1 ( ) 3 () = = 3 1 = 3 Starting at F, go down 3 units and left 1 unit. So, the coordinates of G are (, 0). 31. x + 0.5x + x + 0.5x =.5x = 360 x = 1 0.5x = 0.5(1 ) = 36 The angles are 36 and x + (x + 50) + x + (x + 50) = 10x = x = 60 x = 6 (x + 50) = ( ) = 15 The angles are 15 and The quadrilateral could not be a parallelogram because and are opposite angles, but m m. 3. m J + m K = 180 (3x + 7) + (5x 11) = 180 8x = 180 8x = 18 x = 3 m J = ( ) = 76 m K = (5 3 11) = Sample answer: When ou fold the parallelogram so that vertex is on vertex, the fold will pass through the point where the diagonals intersect, which demonstrates that this point of intersection is also the midpoint of. Similarl, when ou fold the parallelogram so that vertex is on vertex, the fold will pass through the point where the diagonals intersect, which demonstrates that this point of intersection is also the midpoint of F G 8 x Slope of FG = = 6 3 = Starting at, go down 6 units and right 3 units. So, the coordinates of are (, ). 0 Geometr opright ig Ideas Learning, LL

11 36. Sample answer: F 1 m 1 = m G because corresponding parts of congruent figures are congruent. Note that FG. So, m 1 = m and m G = m G because the are pairs of alternate interior angles, and m 1 = m G b the Transitive Propert of qualit. lso, b the ngle ddition Postulate (Post. 1.), m = m + m G. substituting, ou get m = m 1 + m 1 = m Given is a parallelogram. Prove, STTMNTS RSONS 1. is a 1. Given parallelogram..,. efinition of parallelogram 3., 3. lternate Interior ngles Theorem (Thm. 3.).. Reflexive Propert of ongruence (Thm..1) S ongruence Theorem (Thm. 5.10) orresponding parts of congruent triangles are congruent. 7. m = m, m = m 8. m = m + m m = m + m 9. m = m + m 7. efinition of congruent angles G 8. ngle ddition Postulate (Post. 1.) 9. Substitution Propert of qualit 10. m = m 10. Transitive Propert of qualit efinition of congruent angles 38. Given PQRS is a parallelogram. Prove x + = 180 P Q R x x S STTMNTS RSONS 1. PQRS is a parallelogram. 1. Given. QR PS. efinition of parallelogram 3. Q and P are supplementar. 3. onsecutive Interior ngles Theorem (Thm. 3.). x + = 180. efinition of supplementar angles 39. Q + 1 x + 37 x 5 M MQ = NP x + 37 = x 5 3x = x = MQ = = = 9 NP = 1 5 = 9 QP = MN + 1 = = 9 = 3 QP = = 17 MN = = = 17 The perimeter of MNPQ is = 5 units. 0. LM MN = x M 3x N 3x x + 3x + x + 3x = 8 1x = 8 x = LM = = 8 units x N x P L 3x P 1. no; Two parallelograms with congruent corresponding sides ma or ma not have congruent corresponding angles. opright ig Ideas Learning, LL Geometr 1

12 . a. decreases; ecause P and Q are supplementar, as m Q increases, m P must decrease so that their total is still 180. b. increases; s m Q decreases, the parallelogram gets skinnier, which means that Q and S get farther apart and QS increases. c. The mirror gets closer to the wall; s m Q decreases, the parallelograms get skinnier, which means that P, R, and the other corresponding vertices all get closer together. So, the distance between the mirror and the wall gets smaller. 3. m USV + m TSU = m TUV (x ) + 3 = 1x x 1x + 3 = 0 (x 8)(x ) = 0 (x 8) = 0 x = 8 m USV = (x ) = (8 ) = 6 Starting at W, go down units and right 3 units. So, the coordinates of are (, 0). 6 W X Y 6 x Let F be the fourth vertex. Slope of XW = = = Starting at Y, go up units and right units. So, the coordinates of F are (8, 8). 8 6 X F (x ) = 0 x = m USV = (x ) = ( ) = 16. es; F n triangle, such as, can be partitioned into four congruent triangles b drawing the midsegment triangle, such as F. Then, one triangle, such as, can be rotated 180 about a vertex, such as, to create a parallelogram as shown. 5. Three parallelograms can be created. Let be the fourth vertex. Slope of XY = = 3 = 3 Starting at W, go up units and left 3 units. So, the coordinates of are (, ). 6 W X Y 6 x Let be the fourth vertex. Slope of YX = = 3 = 3 W Y 6 8 x 6. Given K bisects FH, FJ bisects FG, and FGH is a parallelogram. Prove K FJ STTMNTS 1. K bisects FH and FJ bisects FG. FGH is a parallelogram.. m PH = m PF, m PF = m PFG 3. m HF = m PH + m PF, m FG = m PF + m PFG. m HF = m PF + m PF, m FG = m PF + m PF 5. m HF = (m PF), m FG = (m PF) 6. m HF + m FG = (m PF) + (m PF) = 180 P H J K RSONS 1. Given G. efinition of angle bisector 3. ngle ddition Postulate (Post. 1.). Substitution Propert of qualit 5. istributive Propert 6. Parallelogram onsecutive ngles Theorem (Thm. 7.5) 7. Substitution Propert of qualit F Geometr opright ig Ideas Learning, LL

13 8. (m PF + m PF) = m PF + m PF = m PF + m PF + m PF = istributive Propert 9. ivision Propert of qualit 10. Triangle Sum Theorem (Thm. 5.1) m PF = Substitution Propert of qualit 1. m PF = Subtraction Propert of qualit 13. PF is a right angle. 13. efinition of right angle 1. K FJ 1. efinition of perpendicular lines 7. Given GH JK LM, GJ JL Prove HK KM onstruct KP and MQ, such that KP GJ and MQ JL, thus KPGJ is a parallelogram and MQJL is a parallelogram. G J L P H Q K M STTMNTS 1. GH JK LM, GJ JL. onstruct PK and QM such that PK GL QM. 3. GPKJ and JQML are parallelograms.. GHK JKM, PKQ QML 5. GJ PK, JL QM 6. PK QM 7. HPK PKQ, KQM QML RSONS 1. Given. onstruction 3. efinition of parallelogram. orresponding ngles Theorem (Thm. 3.1) 5. Parallelogram Opposite Sides Theorem (Thm. 7.3) 6. Transitive Propert of ongruence (Thm..1) 7. lternate Interior ngles Theorem (Thm. 3.) 8. HPK QML 8. Transitive Propert of ongruence (Thm..) 9. HPK KQM 9. Transitive Propert of ongruence (Thm..) 10. PHK QKM 10. S ongruence Theorem (Thm. 5.11) 11. HK KM 11. orresponding sides of congruent triangles are congruent. Maintaining Mathematical Proficienc 8. es; m b the lternate Interior ngles onverse (Thm. 3.6). 9. es; m b the lternate xterior ngles onverse (Thm. 3.7). 50. no; the onsecutive Interior ngles onverse (Thm. 3.8), consecutive interior angles need to be supplementar for the lines to be parallel, and the consecutive interior angles are not supplementar. 7.3 xplorations (p. 375) 1. a. heck students work. b. es; Slope of 7 3 = 8 ( 1) = = = 1 Slope of = 3 ( 1) 5 ( 9) = = = 1 Slope of 3 ( 1) = 1 ( 9) = = 3 = 3 Slope of 7 3 = 8 ( 5) = = 3 = 3 ecause the slope of equals the slope of,. ecause the slope of equals the slope of,. So, quadrilateral is a parallelogram. c. heck students work. If the opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. d. If a quadrilateral is a parallelogram, then its opposite sides are congruent. The converse is true. This is the Parallelogram Opposite Sides Theorem (Thm. 7.3).. a. heck students work. b. Yes, the quadrilateral is a parallelogram. The opposite sides have the same slope, so the are parallel. c. heck students work. If the opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram. d. If a quadrilateral is a parallelogram, then its opposite angles are congruent. The converse is true. This is the Parallelogram Opposite ngles Theorem (Thm. 7.). 3. To prove a quadrilateral is a parallelogram, show that the opposite sides are congruent or that the opposite angles are congruent.. In the figure, m = m and. ecause the opposite angles are congruent, ou can conclude that is a parallelogram. opright ig Ideas Learning, LL Geometr 3

14 7.3 Monitoring Progress (pp ) 1. WXYZ is a parallelogram because opposite angles are congruent; W Y and X Z. So, m Z = the Parallelogram Opposite ngles onverse (Thm. 7.8), if both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram. So, solve 3x 3 = x for x and = + 87 for. 3x 3 = x 3 = x x = = 87 = 3 9 = So, x = 3 and = Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). Parallelogram Opposite Sides onverse (Thm. 7.7) 5. Parallelogram Opposite ngles onverse (Thm. 7.8) 6. the Parallelogram iagonals onverse (Thm. 7.10), if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. So, solve x = 10 3x for x. x = 10 3x 5x = 10 x = So, x =. 7. Slope of LM = = 6 1 = 6 Slope of JK 5 1 = 3 () = = 6 1 = 6 ecause the slope of LM equals the slope of JK, LM JK. LM = (3 ) + ( 3 3) = (1) + ( 6) = 37 JK = ( 3 ()) + ( 5 1) = (1) + ( 6) = 37 ecause LM = JK = 37, LM JK. So, JK and LM are congruent and parallel, which means that JKLM is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 8. Sample answer: Find the slopes of all four sides and show that opposite sides are parallel. nother wa is to find the point of intersection of the diagonals and show that the diagonals bisect each other. 7.3 xercises (pp ) Vocabular and ore oncept heck 1. es; If all four sides are congruent, then both pairs of opposite sides are congruent. So, the quadrilateral is a parallelogram b the Parallelogram Opposite Sides onverse (Thm. 7.7).. The statement that is different is onstruct a quadrilateral with one pair of parallel sides. Monitoring Progress and Modeling with Mathematics 3. Parallelogram Opposite ngles onverse (Thm. 7.8). Parallelogram Opposite Sides onverse (Thm. 7.7) 5. Parallelogram iagonals onverse (Thm. 7.10) 6. Parallelogram Opposite ngles onverse (Thm. 7.8) 7. Opposite Sides Parallel and ongruent Theorem (Thm. 7.9) 8. Parallelogram iagonals onverse (Thm. 7.10) 9. x = 11 and = 66 b the Parallelogram Opposite ngles onverse (Thm. 7.8). 10. x = 16 and = 9, b the Parallelogram Opposite Sides onverse (Thm. 7.7). 11. the Parallelogram Opposite Sides onverse (Thm. 7.7): x + 6 = 7x 3 6 = 3x 3 9 = 3x 3 = x 3 = = 1 = So, x = 3 and =. 1. the Parallelogram Opposite ngles onverse (Thm. 7.8): (x + 13) = (5x 1) 13 = x 1 5 = x ( + 7) = (3x 8) + 7 = = 75 8 = 60 = 15 So, x = 5 and = the Parallelogram iagonals onverse (Thm. 7.10): x + = 5x 6 = x 6 8 = x So, x = 8. Geometr opright ig Ideas Learning, LL

15 1. the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9): x + 3 = x + 7 x + 3 = 7 x = So, x =. 15. the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9): 3x + 5 = 5x 9 x + 5 = 9 x = 1 x = 7 So, x = the Parallelogram iagonals onverse (Thm. 7.10): 6x = 3x + 3x = x = 3 So, x = x Slope of = 1 = 0 8 = 0 Slope of = = 0 8 = 0 The slope of equals the slope of, so. = (1 ) + ( ) = (8) + (0) = 6 = 8 = (8 0) + (1 1) = (8) + (0) = 6 = 8 ecause = = 8,. and are opposite sides that are both congruent and parallel. So, is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 18. F x G H Slope of F 0 = 3 ( 3) = 0 = undefined Slope of GH 5 ( 1) = = = undefined The slope of F equals the slope of GH, so F GH. F = ( 3 ( 3)) + ( 0) = 0 + = 16 = GH = (3 3) + ( 5 ( 1)) = (0) + () = 16 = ecause F = GH =, F GH. F and GH are opposite sides that are both congruent and parallel. So, FGH is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 19. K 8 L J M 6 8 x Slope of KL 6 7 = 3 ( 5) = = 1 8 Slope of JM 3 = 6 ( ) = = 1 8 The slope of JK equals the slope of LM, so JK LM. JK = ( 5 ( )) + (7 3) = ( 5 + ) + = = 5 = 5 LM = (6 3) + ( 6) = (3) + () = = 5 = 5 ecause JK = LM = 5, JK LM. JK and LM are opposite sides that are both congruent and parallel. So, JKLM is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). opright ig Ideas Learning, LL Geometr 5

16 0. N 5 P Q 3 x R Slope of NP 0 = 0 ( 5) = 5 Slope of PQ = = 3 Slope of QR = 0 3 = 5 = 5 Slope of NR 0 = ( 5) = + 5 = 3 The slope of NP equals the slope of QR, so NP QR. The slope of PQ equals the slope of NR, so PQ NR. ecause both pairs of opposite sides are parallel, NPQR is a parallelogram b definition. 1. In order to be a parallelogram, the quadrilateral must have two pairs of opposite sides that are congruent, not consecutive sides. FG is not a parallelogram.. In order to determine that JKLM is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9), ou would need to know that JM KL. There is not enough information provided to determine whether JKLM is a parallelogram. 5. quadrilateral is a parallelogram if and onl if both pairs of opposite sides are congruent. 6. quadrilateral is a parallelogram if and onl if both pairs of opposite angles are congruent. 7. quadrilateral is a parallelogram if and onl if the diagonals bisect each other. 8. Sample answer: raw two horizontal segments that are the same length and connect the endpoints x 9. heck students work. ecause the diagonals bisect each other, this quadrilateral is a parallelogram b the Parallelogram iagonals onverse (Thm. 7.10). 30. both; If ou show that QR TS and QT RS, then QRST is a parallelogram b definition. If ou show that QR TS and QT RS, then QRST is a parallelogram b the Parallelogram Opposite Sides onverse (Thm. 7.7). 31. Sample answer: 3. The diagonals must bisect each other, so solve for x using either x + 1 = x + 6 or x = 3x + 3. lso, the opposite sides must be congruent, so solve for x using either 3x + 1 = x or 3x + 10 = 5x. x + 1 = x + 6 x + 1 = 6 x = 5 x = 3x + 3 x = 3 x = 5 3x + 10 = 5x 10 = x x = 5 3x + 1 = x 1 = x 5 = x So, x = Sample answer: a. ecause m F = 63 and F is a right angle (m F = 90 ), m F = = 7. b. From the diagram, F FG. So, m FG = 7. ecause m FG = 90, m FG = 90 7 = 63. c. From the diagram, F GH H. So, m GH = m H = 7. d. es; HF HGF because the both are adjacent to two congruent angles that together add up to 180, and HG GF for the same reason. So, FGH is a parallelogram b the Parallelogram Opposite ngles onverse (Thm. 7.8).. es; the onsecutive Interior ngles onverse (Thm. 3.8), WX ZY. ecause WX and ZY are also congruent, WXYZ is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 6 Geometr opright ig Ideas Learning, LL

17 3. a. ecause JK LM and KL JM, JKLM is a parallelogram b the Parallelogram Opposite Sides onverse (Thm. 7.7). b. ecause m JKL = 60, m JML = 60 b the Parallelogram Opposite ngles onverse (Thm. 7.8). m KJM = = 10 b the Parallelogram onsecutive ngles Theorem (Thm. 7.5). m KLM = 10 b the Parallelogram Opposite ngles onverse (Thm. 7.8). c. Transitive Propert of Parallel Lines (Thm. 3.9) 35. You can use the lternate Interior ngles onverse (Thm. 3.6) to show that. Then, and are both congruent and parallel. So, is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 36. You can use the lternate Interior ngles onverse (Thm. 3.6) to show that and. ecause both pairs of opposite sides are parallel, is a parallelogram b definition. 37. Use the orresponding ngles onverse to show that and the lternate Interior ngles onverse to show that. So, is a parallelogram b definition. 38. the Parallelogram Opposite Sides Theorem (Thm. 7.3), JM LK. lso, ou can use the Linear Pair Postulate (Thm..8) and the ongruent Supplements Theorem (Thm..) to show that GJM HLK. ecause JGM and LHK are congruent right angles, ou can now state that MGJ KHL b the S ongruence Theorem (Thm. 5.11). 39. Given, Prove is a parallelogram. 8. and are supplementar. and are supplementar. 9., 10. is a parallelogram. 0. Given QR PS, QR PS Prove PQRS is a parallelogram. STTMNTS 1. QR PS, QR PS 8. efinition of supplementar angles 9. onsecutive Interior ngles onverse (Thm. 3.8) 10. efinition of parallelogram P RSONS 1. Given. RQS PSQ. lternate Interior ngles Theorem (Thm. 3.) 3. QS QS Q 3. Reflexive Propert of onguence (Thm..1). PQS RSQ. SS ongruence Theorem (Thm. 5.5) 5. QSR SQP 5. orresponding parts of congruent triangles are congruent. 6. QP RS 7. PQRS is a parallelogram. 6. lternate Interior ngles onverse (Thm. 3.6) 7. efinition of parallelogram S R STTMNTS RSONS 1., 1. Given. Let m = m = x and m = m = 3. m + m + m + m = x + + x + =. efinition of congruent angles 3. orollar to the Polgon Interior ngles Theorem (or. 7.1). (x ) + ( ) =. Simplif 5. (x + ) = 5. istributive Propert 6. x + = ivision Propert of qualit 7. m + m = 180, m + m = Substitution Propert of qualit opright ig Ideas Learning, LL Geometr 7

18 1. Given iagonals JL and KM bisect each other. Prove JKLM is a parallelogram. STTMNTS 1. iagonals JL and KM bisect each other.. JP LP, KP MP J K RSONS 1. Given P M. efinition of segment bisector 3. KPL MPJ 3. Vertical ngles Theorem (Thm..6). KPL MPJ. SS ongruence Theorem (Thm. 5.5) 5. MKL KMJ, KL MJ 6. KL MJ 7. JKLM is a parallelogram. 5. orresponding parts of congruent triangles are congruent. 6. lternate Interior ngles onverse (Thm. 3.6) 7. Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). Given F is a parallelogram. = F Prove is a parallelogram. STTMNTS 1. F is a parallelogram, = F. F, F F RSONS 1. Given. Parallelogram Opposite Sides Theorem (Thm. 7.3) 3. F 3. Parallelogram Opposite ngles Theorem (Thm. 7.). and form a linear pair. F and F form a linear pair. 5. and are supplementar. F and F are supplementar.. efinition of linear pair 5. Linear Pair Postulate (Post..8) 6. F 6. ongruent Supplements Theorem (Thm..) L 7. F 7. efinition of congruent segments 8. F 8. SS ongruence Theorem (Thm. 5.5) = +, = F + F 9. orresponding parts of congruent triangles are congruent. 10. Segment ddition Postulate (Post. 1.) 11. F = 11. efinition of congruent segments 1. = F + F 1. Substitution Propert of qualit 13. = 13. Transitive Propert of qualit efinition of congruent segments 15. is a parallelogram. 15. Parallelogram Opposite Sides onverse (Thm. 7.7) 3. no; The fourth angle will be 113 because of the orollar to the Polgon Interior ngles Theorem (or. 7.1), but these could also be the angle measures of an isosceles trapezoid with base angles that are each 67.. The segments that remain parallel as the stand is folded are F, F, and F. 5. the Parallelogram Opposite Sides Theorem (Thm. 7.3),. lso, and F are congruent alternate interior angles of parallel segments and. Then, ou can use the Segment ddition Postulate (Post. 1.), the Substitution Propert of qualit, and the Reflexive Propert of ongruence (Thm..1) to show that F. So, F b the SS ongruence Theorem (Thm. 5.5), which means that = F = 8 because corresponding parts of congruent triangles are congruent. 8 Geometr opright ig Ideas Learning, LL

19 6. raw the first parallelogram: Rotate 10 about : 9. Given quadrilateral with midpoints, F, G, and H that are joined to form a quadrilateral, ou can construct diagonal. Then FG is a midsegment of, and H is a midsegment of. So, b the Triangle Midsegment Theorem (Thm. 6.8), FG 1, FG =, H, and H = 1. So, b the Transitive Propert of Parallel Lines (Thm. 3.9), H FG and b the Transitive Propert of qualit, H = FG. ecause one pair of opposite sides is both congruent and parallel, FGH is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). Then rotate 10 about : F G 7. onverse to the Parallelogram onsecutive ngles Theorem: If ever pair of consecutive angles of a quadrilateral are supplementar, then the quadrilateral is a parallelogram. In, ou are given that and are supplementar, and and are supplementar. So, m = m. lso, and are supplementar, and and are supplementar. So, m = m. So, is a parallelogram b the Parallelogram Opposite ngles onverse (Thm. 7.8). 8. Given is a parallelogram and is a right angle. Prove,, and are right angles. (0, d) (b, d) (0, 0) (b, 0) x the definition of a right angle, m = 90. ecause is a parallelogram, and opposite angles of a parallelogram are congruent, m = m = 90. ecause consecutive angles of a parallelogram are supplementar, and are supplementar, and and are supplementar. So, 90 + m = 180 and 90 + m = 180. This gives ou m = m = 90. So,,, and are right angles. H 50. ased on the given information, GH is a midsegment of, and FJ is a midsegment of. So, b the Triangle Midsegment Theorem (Thm. 6.8), GH, GH = 1, FJ, and FJ = 1. lso, b the Parallelogram Opposite Sides Theorem (Thm. 7.3) and the definition of a parallelogram, and are congruent and parallel. So, b the Transitive Propert of Parallel Lines (Thm. 3.9), FJ GH and b the Transitive Propert of qualit, 1 = GH = FJ = 1. ecause one pair of opposite sides is both congruent and parallel, FGHJ is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). Maintaining Mathematical Proficienc 51. The quadrilateral is a parallelogram b the definition of a parallelogram (a quadrilateral with both pairs of opposite sides parallel). 5. The quadrilateral is a rectangle b the definition of a rectangle (a quadrilateral with four right angles). 53. The quadrilateral is a square b the definition of a square (a quadrilateral with four right angles and four congruent sides). 5. The quadrilateral is a rhombus b the definition of a rhombus (a quadrilateral with four congruent sides) What id You Learn? (p. 385) 1. The relationship between the 50 increase and the answer is that the interior angle value added is 50 or the Parallelogram iagonals Theorem (Thm. 7.6), the diagonals of a parallelogram bisect each other. So, the diagonals will have the same midpoint, and it will also be the point where the diagonals intersect. Therefore, with an parallelogram, ou can find the midpoint of either diagonal, and it will be the coordinates of the intersection of the diagonals; Instead of this method, ou could also find the equations of the lines that define each diagonal, set them equal to each other and solve for the values of the coordinates where the lines intersect. opright ig Ideas Learning, LL Geometr 9

20 3. Sample answer: Solve 5x = 3x + 10; opposite sides of a parallelogram are congruent; es; You could start b setting the two parts of either diagonal equal to each other b the Parallelogram iagonals Theorem (Thm. 7.6) Quiz (p. 386) x = 80 + x = 360 x = 80. (5 ) 180 = = x = x = 50 x = x =. Interior angle = (10 ) = = 1 x + 63 = 360 x = 97 = xterior angle = 10 = 36 In a regular decagon, the measure of each interior angle is 1 and the measure of each exterior angle is Interior angle = (15 ) = = 156 = xterior angle = 15 = In a regular 15-gon, the measure of each interior angle is 156 and the measure of each exterior angle is. 6. Interior angle = ( ) 180 = 3960 = 165 = 180 xterior angle = = 15 In a regular -gon, the measure of each interior angle is 165 and the measure of each exterior angle is Interior angle = (60 ) 180 = = 10,0 = xterior angle = 60 = 6 In a regular 60-gon, the measure of each interior angle is 17 and the measure of each exterior angle is = 16; the Parallelogram Opposite Sides Theorem (Thm. 7.3), =. 9. = 7; the Parallelogram Opposite Sides Theorem (Thm. 7.3), =. 10. = 7; the Parallelogram iagonals Theorem (Thm. 7.6), =. 11. = 10. = 0.; the Parallelogram iagonals Theorem (Thm. 7.6), =. 1. m = 10 ; the Parallelogram Opposite ngles Theorem (Thm. 7.), m = m. 13. the Parallelogram onsecutive ngles Theorem (Thm. 7.5), and are supplementar. So, m = = the Parallelogram onsecutive ngles Theorem (Thm. 7.5), and are supplementar. So, m = = m = 3 ; the lternate Interior ngles Theorem (Thm. 3.), m = m. 16. The quadrilateral is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 17. The quadrilateral is a parallelogram b the Parallelogram iagonals onverse (Thm. 7.10). 18. The quadrilateral is a parallelogram b the Parallelogram Opposite ngles onverse (Thm. 7.8) Q T x R S Slope of QR ( ) = = + 3 ( 5) = 0 8 = 0 Slope of ST 6 ( 6) = = = = 0 The slope of QR equals slope of ST, so QR ST. QR = (3 ( 5)) + ( ( )) = (3 + 5) + ( + ) = 8 = 6 = 8 ecause QR = ST = 8, QR ST. QR and ST are opposite sides that are both congruent and parallel. So, QRST is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). 50 Geometr opright ig Ideas Learning, LL

21 0. Z 6 5 W X 3 x Y Slope of WX 3 7 = 3 ( 3) = = 6 = 3 Slope of XY = = 6 = 3 Slope of YZ = 1 ( 3) 5 1 = = 6 = 3 Slope of WZ 1 7 = 5 ( 3) = = 6 = 3 ecause the slope of WX equals the slope of YZ, WX YZ and because the slope of XY equals the slope of WZ, XY WZ. ecause both pairs of opposite sides are parallel, WXYZ is a parallelogram b definition. 1. a. The stop sign is a regular octagon. b. (8 ) 180 = = 1080 = The measure of each interior angle is = 5 The measure of each exterior angle is 5.. a. JK ML b the Parallelogram Opposite Sides Theorem (Thm. 7.3). JM KL b the Parallelogram Opposite Sides Theorem (Thm. 7.3). J KLM b the Parallelogram Opposite ngles Theorem (Thm. 7.). M JKL b the Parallelogram Opposite ngles Theorem (Thm. 7.). b. ecause QT RS and QT = RS, QRST is a parallelogram b the Opposite Sides Parallel and ongruent Theorem (Thm. 7.9). c. ST = 3 feet, because ST = QR b the Parallelogram Opposite Sides Theorem (Thm. 7.3). m QTS = 13, because m QTS = m QRS b the Parallelogram Opposite ngles Theorem (Thm. 7.). m TQR = 57, because TQR and QTS are consecutive interior angles and the are supplementar. So, m TQR = = 57. ecause TSR and TQR are opposite angles b the Parallelogram Opposite ngles Theorem (Thm. 7.), m TSR = xplorations (p. 387) 1. a. heck students work. b. heck students work. c. d. es; es; no; no; ecause all points on a circle are the same distance from the center,. So, the diagonals of quadrilateral bisect each other, which means it is a parallelogram b the Parallelogram iagonals onverse (Thm. 7.10). ecause all angles of are right angles, it is a rectangle. is neither a rhombus nor a square because and are not necessaril the same length as and. e. heck students work. The quadrilateral formed b the endpoints of two diameters is a rectangle (and a parallelogram). In other words, a quadrilateral is a rectangle if and onl if its diagonals are congruent and bisect each other.. a. heck students work. b. c. es; no; es; no; ecause the diagonals bisect each other, is a parallelogram b the Parallelogram iagonals onverse (Thm. 7.10). ecause = = =, is a rhombus. is neither a rectangle nor a square because its angles are not necessaril right angles. d. heck students work. quadrilateral is a rhombus if and onl if the diagonals are perpendicular bisectors of each other. 3. ecause rectangles, rhombuses, and squares are all parallelograms, their diagonals bisect each other b the Parallelogram iagonals Theorem (Thm. 7.6). The diagonals of a rectangle are congruent. The diagonals of a rhombus are perpendicular. The diagonals of a square are congruent and perpendicular.. es; no; es; no; RSTU is a parallelogram because the diagonals bisect each other. RSTU is not a rectangle because the diagonals are not congruent. RSTU is a rhombus because the diagonals are perpendicular. RSTU is not a square because the diagonals are not congruent. 5. rectangle has congruent diagonals that bisect each other. opright ig Ideas Learning, LL Geometr 51

22 7. Monitoring Progress (pp ) 1. For an square JKLM, it is alwas true that JK KL, because b definition, a square has four right angles.. For an rectangle FGH, it is sometimes true that FG GH, because some rectangles are squares. 3. The quadrilateral is a square.. m = 9 = 58 because each diagonal of a rhombus bisects a pair of opposite angles. m = 61 = 1 because each diagonal of a rhombus bisects a pair of opposite angles. 5. m G = = 6 b the Parallelogram onsecutive ngles Theorem (Thm. 7.5). m 1 = 6 = 31 because each diagonal of a rhombus bisects a pair of opposite angles. m = m 1 = 31 because each diagonal of a rhombus bisects a pair of opposite angles. m FG = m G = 6 because opposite angles of a parallelogram are congruent, and a rhombus is a parallelogram. m 3 = 31 because each diagonal of a rhombus bisects a pair of opposite angles. m = m 3 = 31 because each diagonal of a rhombus bisects a pair of opposite angles. 6. no; The quadrilateral might not be a parallelogram. 7. QS = RT x 15 = 3x + 8 x 15 = 8 x = 3 Lengths of the diagonals: QS = 3 15 = 9 15 = 77 RT = = = P( 5, ), Q(0, ), R(, 1), S( 3, 3) P Q PQ = ( 5 0) + ( ) = 5 + = 9 QR = (0 ) + ( ( 1)) = + 5 = 9 ecause PQ = QR, the adjacent sides PQ and QR are congruent. So, PQRS is a square, a rectangle, and a rhombus. 7. xercises (pp ) Vocabular and ore oncept heck 1. nother name for an equilateral rectangle is a square.. If two consecutive sides of a parallelogram are congruent, then the parallelogram is also a rhombus. Monitoring Progress and Modeling with Mathematics 3. L is sometimes congruent to M. Some rhombuses are squares. J K M L. K is alwas congruent to M. rhombus is a parallelogram and the opposite angles of a parallelogram are congruent. J K M L 5. JM is alwas congruent to KL. definition, a rhombus is a parallelogram, and opposite sides of a parallelogram are congruent. J K M L 6. JK is alwas congruent to KL. definition, a rhombus is a parallelogram with four congruent sides. J K M L 7. JL is sometimes congruent to KM. Some rhombuses are squares. J K S x R M L PR = ( 5 ) + ( ( 1)) = = 58 QS = (0 ( 3)) + ( ( 3)) = = 58 ecause PR = QS, the diagonals are congruent, so the quadrilateral is either a square or rectangle. 5 Geometr opright ig Ideas Learning, LL