Pearson Unit 1 Topic 6: Polygons and Quadrilaterals 6-2: Properties of Parallelograms Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (6)(E) Prove a quadrilateral is a parallelogram, rectangle, square, or rhombus using opposite sides, opposite angles, or diagonals and apply these relationships to solve problems. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication.

Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special quadrilaterals are given their own names. Opposite sides of a quadrilateral do not share a vertex. Opposite angles do not share a side. Helpful Hint

A quadrilateral with two pairs of parallel sides is a parallelogram A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol .

Example: 1 Answer: C Consecutive angles in a parallelogram are supplementary.

Example: 2 In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find CF , DF, and mEFC.  diags. bisect each other.  opp. sides  DF = 2DG CF = DE Def. of  segs. DF = 2(31) Substitute 31 for DG. CF = 74 mm Substitute 74 for DE. DF = 62 Simplify. mEFC + mFCD = 180°  consecutive s supp. mEFC + 42 = 180 Substitute 42 for mFCD. mEFC = 138° Subtract 42 from both sides.

Example: 3 In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find KN, LO, and mNML.  opp. sides  LM = KN Def. of  segs. LM = 28 in. Substitute 28 for DE.  diags. bisect each other. LN = 2LO 26 = 2LO Substitute 26 for LN. LO = 13 in. Simplify. NML  LKN  opp. s  mNML = mLKN Def. of  s. mNML = 74° Substitute 74° for mLKN.

Example: 4 WXYZ is a parallelogram. Find YZ and mZ .  opp. s  YZ = XW Def. of  segs. 8a – 4 = 6a + 10 Substitute the given values. Subtract 6a from both sides and add 4 to both sides. 2a = 14 a = 7 Divide both sides by 2. YZ = 8a – 4 = 8(7) – 4 = 52

Example: 4 continued  consecutive s supp. mZ + mW = 180° (9b + 2) + (18b – 11) = 180 Substitute the given values. 27b – 9 = 180 Combine like terms. 27b = 189 Add 9 to both sides. b = 7 Divide by 27. mZ = (9b + 2)° = [9(7) + 2]° = 65°

Example: 5 EFGH is a parallelogram. Find JG and FH.  diags. bisect each other. EJ = JG Def. of  segs. 3w = w + 8 Substitute. 2w = 8 Simplify. w = 4 Divide both sides by 2. JG = w + 8 = 4 + 8 = 12

Example: 5 continued  diags. bisect each other. FJ = JH Def. of  segs. 4z – 9 = 2z Substitute. 2z = 9 Simplify. z = 4.5 Divide both sides by 2. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18

Example: 6 STATEMENT REASON 1. ABCD, AK  MK 1. Given 2. BCD  A 2. Opposite ’s in a are  3.  A  CMD 3. If two sides of ∆ are  then ’s opposite those sides are  4. BCD  CMD 4. Transitive Property

Example: 7 Find the values of x and y in Parallelogram PQRS. Find PR and SQ. y = x + 1 2x = 3y – 7 2x = 3 (x + 1) – 7 2x = 3x + 3 – 7 2x = 3x – 4 -1x = -4 x = 4 y = 4 + 1 = 5 PR = 16 SQ = 10