Pearson Unit 1 Topic 6 Polygons and Quadrilaterals

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. Helpful Hint Opposite sides of a quadrilateral do not share a vertex. Opposite angles do not share a side.

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 CDEF, DE = 74 mm, DG = 31 mm, and m FCD = 42°. Find CF , DF, and m EFC. In opp. sides diags. DF = 2 DG bisect each other. 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° m EFC + 42 = 180 m EFC = 138° consecutive s supp. Substitute 42 for m FCD. 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. LN = 2 LO diags. bisect each other. 26 = 2 LO 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 8 a – 4 = 6 a + 10 2 a = 14 a=7 Def. of segs. Substitute the given values. Subtract 6 a from both sides and add 4 to both sides. Divide both sides by 2. YZ = 8 a – 4 = 8(7) – 4 = 52

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

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

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

Example: 6 STATEMENT 1. ABCD, AK MK REASON 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 2 x = 3 y – 7 2 x = 3 (x + 1) – 7 2 x = 3 x + 3 – 7 2 x = 3 x – 4 -1 x = -4 x=4 y=4+1=5 PR = 16 SQ = 10