1. 1 PROBLEM 1A PROBLEM 2A PROBLEM 3A PROBLEM 4A PROBLEM 1B PROBLEM 4B PROBLEM 2B PROBLEM 3B PARALLELOGRAM PROPERTIES PROBLEM 5APROBLEM 5B Standard 7 PROBLEM 6APROBLEM 6B RECTANGLE PROPERTIES END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 STANDARD 7: Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and properties of circles. ESTÁNDAR 7: Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 1. Two pairs of parallel sides. 2. Two pairs of congruent sides. 3. Diagonals bisect each other 4. Opposite angles are congruent. A D B C PARALLELOGRAM 5. Consecutive angles are supplementary. mBmC+=180° mAmB+= mB A B A B B C B C mCmD+= D C D C mDmA+= A D A D Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 Find angles and sides in the parallelogram above. Consecutive angles are supplementary: -60° Opposite angles are congruent: Angles form a Linear Pair: x = 120° m x + 60° = 180° m y= 60° m z = 120° m z + w = 180° mm 120° + w = 180° m -120 w = 60° m Opposite sides are congruent: u = 20 Standard 7 w 60° x y z u 20 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 Find angles and sides in the parallelogram above. Consecutive angles are supplementary: -70° Opposite angles are congruent: Angles form a Linear Pair: f = 110° m f + 70° = 180° m g= 70° m h = 110° m h + j = 180° mm 110° + j = 180° m -110 j = 70° m Opposite sides are congruent: k = 40 Standard 7 j 70° f g h k 40 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 a. This is a parallelogram, so opposite angles are congruent: 2X + 20 = 3Y + 9 -20 b. Segments are parallel c. Alternate interior angles congruent: Standard 7 2X+20 4Y + 12 3Y + 9 3X + 25 4Y + 12 = 3X + 25 2X = 3Y - 11 -3Y 2X – 3Y = -11 -12 4Y = 3X + 13 -3X -3X +4Y = 13 First equation Second equation Solving both equations together: 2X – 3Y = -11 -3X + 4Y = 13 3 3 2 2 6X – 9Y = -33 -6X + 8Y = 26 -Y = -7 Y = 7 Substituting in the First equation: 2X – 3Y = -11 2X – 3( ) = -11 7 2X -21 = -11 +21 2X = 10 2 X = 5 Find the value for X and Y in the figure at the right. Suppose is a parallelogram. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 a. This is a parallelogram, so opposite angles are congruent: 4X + 12 = 3Y + 8 -12 b. Segments are parallel c. Alternate interior angles congruent: Standard 7 4X+12 4Y + 14 3Y + 8 3X + 24 4Y + 14 = 3X + 24 4X = 3Y - 4 -3Y 4X – 3Y = -4 -14 4Y = 3X + 10 -3X -3X + 4Y = 10 First equation Second equation Solving both equations together: 4X – 3Y = -4 -3X + 4Y = 10 4 4 3 3 16X – 12Y = -16 -9X + 12Y = 30 7X = 14 X = 2 Substituting in the First equation: 4X – 3Y = -4 4( ) – 3Y = -4 2 8 – 3Y = -4 -8 -3Y = -12 -3 Y = 4 Find the value for X and Y in the figure at the right. Suppose is a parallelogram. 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 a. This is a parallelogram, so opposite sides are parallel. b. Alterante interior angles congruent: r = 20° m 20° c. Interior Angle Sum in a triangle is 180°: 20° + 120° + s = 180° m 140° + s = 180° m -140° s = 40° m 40° d. Angles form a Linear Pair and are supplementary: 120° + t = 180° m -120° t = 60° m Standard 7 20° t r s 120° Find the value for r, s, and t in the figure below: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 a. This is a parallelogram, so opposite sides are parallel. b. Alterante interior angles congruent: a = 22° m 22° c. Interior Angle Sum in a triangle is 180°: 22° + 130° + b = 180° m 152° + b = 180° m -152° b = 28° m 28° d. Angles form a Linear Pair and are supplementary: 130° + c = 180° m -130° c = 50° m Standard 7 22° c a b 130° Find the value for a, b, and c in the figure below: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 a. Let’s assume it is a parallelogram b. Diagonals bisect each other, so: j + 8 = 3k 15j = 5k -8 j = 3k – 8 First equation Second equation c. Solving both equations together: 15( ) = 5k 3k – 8 45k – 120 = 5k -45k - 120 = -40k k = 3 d. Using first equation: j = 3k – 8 = 3( ) – 8 3 = 9 – 8 j = 1 Standard 7 3k 5k 15j j+8 Find the values for k and j in the figure below: -40 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 a. Let’s assume it is a parallelogram b. Diagonals bisect each other, so: y + 15 = 4z 28y = 7z -15 y = 4z – 15 First equation Second equation c. Solving both equations together: 28( ) = 7z 4z – 15 112z – 420 = 7z -112z - 420 = -105z z = 4 d. Using first equation: y = 4z – 15 = 4( ) – 15 4 = 16 – 15 y = 1 Standard 7 4z 7z 28y y+15 Find the values for z and y in the figure below: -105 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 K L M N 50 4Z+32 4X-2 6Z i. In a parallelogram opposite sides are congruent: 50 = 4X- 2 +2 52=4X 4 4 X=13 4Z+32=6Z -4Z -4Z 32 = 2Z 2 2 Z=16 ii) Finding LM: LM= 4Z+32 LM= 4( )+32 16 =64 + 32 LM=96 Standard 7 KLMN is a parallelogram with KL=50, LM=4Z+32, MN=4X-2, NK=6Z. What is the value of X, Z, and the measure of LM. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 K H I J 86 3U+25 6T-4 8U i. In a parallelogram opposite sides are congruent: 86 = 6T- 4 +4 90=6T 6 6 T=15 3U+25=8U -3U -3U 25 = 5U 5 5 U=5 ii) Finding HI: HI= 3U+25 HI= 3( )+25 5 =15 + 25 HI=40 Standard 7 KHIJ is a parallelogram with KH=86, HI=3U+25, IJ=6T-4, JK=8U. Find the value of T, U, and the measure of HI. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 1. Two pairs of parallel sides. 2. Two pairs of congruent sides. 5. Diagonals bisect each other 6. Opposite angles are congruent. 7. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 4. Diagonals are congruent 3. All angles are right. RECTANGLE Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 7X+3 9X – 13 In a rectangle diagonals bisect each other, then: 7X + 3 = 9X – 13 -7X -7X 3 = 2X – 13 +13 16 = 2X 2 2 X=8 Standard 7 R S T U RSTU is a Rectangle RQ =7X + 3 QT = 9X – 13 Find the value for X and QS. Since all four segments formed when the diagonals bisect are congruent, finding one we’ll know the value for all. RQ = 7X + 3 = 7( ) + 3 8 = 56 + 3 = 59 Q The length of QS is 59. RQ TQ QS UQ RQ = TQ = QS = UQ PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. 16 3X+5 6X – 10 In a rectangle diagonals bisect each other, then: 3X + 5 = 6X – 10 -3X -3X 5 = 3X – 10 +10 15 = 3X 3 3 X=5 Standard 7 A B C D ABCD is a Rectangle BQ =3X + 5 CQ = 6X – 10 Find the value for X and DQ. Since all four segments formed when the diagonals bisect are congruent, finding one we’ll know the value for all. BQ = 3X + 5 =3( ) + 5 5 = 15 + 5 = 20 Q The length of DQ is 20. BQ DQ AQ CQ BQ = DQ = AQ = CQ PRESENTATION CREATED BY SIMON PEREZ. All rights reserved