Pearson Unit 1 Topic 6: Polygons and Quadrilaterals 6-4: Properties of Rhombuses, Rectangles, and Squares Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (5)(A) Investigate patterns to make conjectures about geometric relationships, including angles formed by parallel lines cut by a transversal, criteria required for triangle congruence, special segments of triangles, diagonals of quadrilaterals, interior and exterior angles of polygons, and special segments and angles of circles choosing from a variety of tools. (1)(C) Select tools, including real objects, manipulatives paper and pencil, and technology as appropriate, and techniques, including mental math, estimations, and number sense as appropriate, to solve problems. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (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.

A second type of special quadrilateral is a rectangle A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

A rhombus is another special quadrilateral A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Example: 1 Is ABCD a rhombus, rectangle or square? Explain.

Example: 2 A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect.  diags.  KM = JL = 86 Def. of  segs.  diags. bisect each other Substitute and simplify.

Example: 3 Carpentry: The rectangular gate has diagonal braces. Find HJ and HK. Rect.  diags.  HJ = GK = 48 Def. of  segs. Rect.  diags.  Rect.  diagonals bisect each other JL = LG Def. of  segs. JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

Example: 4 TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9 = 3b + 4 Substitute given values. Subtract 3b from both sides and add 9 to both sides. 10b = 13 b = 1.3 Divide both sides by 10. TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV = 3b + 4 Substitute 1.3 for b and simplify. TV = 3(1.3) + 4 = 7.9

Example: 5 TVWX is a rhombus. Find mVTZ. mVZT = 90° Rhombus  diag.  14a + 20 = 90° Substitute 14a + 20 for mVTZ. a = 5 Subtract 20 from both sides and divide both sides by 14. mVTZ = mZTX Rhombus  each diag. bisects opp. s mVTZ = (5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ = [5(5) – 5)]° = 20° Substitute 5 for a and simplify.

Example: 6 CDFG is a rhombus. Find CD. CG = GF Def. of rhombus 5a = 3a + 17 Substitute a = 8.5 Simplify GF = 3a + 17 = 42.5 Substitute CD = GF Def. of rhombus CD = 42.5 Substitute

Example: 7 CDFG is a rhombus. Find mGCH if mGCD = (b + 3)° and mCDF = (6b – 40)° mGCD + mCDF = 180° Def. of rhombus b + 3 + 6b – 40 = 180° Substitute. Simplify. 7b = 217° Divide both sides by 7. b = 31°

Example: 7 continued mGCH + mHCD = mGCD Rhombus  each diag. bisects opp. s 2mGCH = mGCD 2mGCH = (b + 3) Substitute. 2mGCH = (31 + 3) Substitute. mGCH = 17° Simplify and divide both sides by 2.

Example: 8 12 feet A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle? 8 ft 8 ft 12 feet Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one  of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.