ofof Special Parallelograms 6 4 Properties Special Parallelograms

ofof Special Parallelograms 6 -4 Properties Special Parallelograms Warm Up Lesson Presentation Lesson Quiz Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Below are some conditions you can use to determine whether a parallelogram is a rhombus. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 2 A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. WV = XT 13 b – 9 = 3 b + 4 10 b = 13 b = 1. 3 Holt Mc. Dougal Geometry Def. of rhombus Substitute given values. Subtract 3 b from both sides and add 9 to both sides. Divide both sides by 10.

6 -4 Properties of Special Parallelograms Example 2 A Continued TV = XT Def. of rhombus TV = 3 b + 4 Substitute 3 b + 4 for XT. TV = 3(1. 3) + 4 = 7. 9 Substitute 1. 3 for b and simplify. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 2 B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find m VTZ. m VZT = 90° 14 a + 20 = 90° a=5 Holt Mc. Dougal Geometry Rhombus diag. Substitute 14 a + 20 for m VTZ. Subtract 20 from both sides and divide both sides by 14.

6 -4 Properties of Special Parallelograms Example 2 B Continued m VTZ = m ZTX Rhombus each diag. bisects opp. s m VTZ = (5 a – 5)° Substitute 5 a – 5 for m VTZ = [5(5) – 5)]° Substitute 5 for a and simplify. = 20° Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 3 A: Identifying Special Parallelograms in the Coordinate Plane Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(– 1, 4), Q(2, 6), R(4, 3), S(1, 1) Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 3 A Continued Step 1 Graph Holt Mc. Dougal Geometry PQRS.

6 -4 Properties of Special Parallelograms Example 3 A Continued Step 2 Find PR and QS to determine if PQRS is a rectangle. Since , the diagonals are congruent. PQRS is a rectangle. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 3 A Continued Step 3 Determine if PQRS is a rhombus. Since , PQRS is a rhombus. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 2 A: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m BCD. Kite cons. sides ∆BCD is isos. 2 sides isos. ∆ CBF CDF isos. ∆ base s m CBF = m CDF Def. of s m BCD + m CBF + m CDF = 180° Polygon Sum Thm. Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 2 A Continued m BCD + m CBF + m CDF = 180° Substitute m CDF m BCD + m CDF = 180° for m CBF. m BCD + 52° = 180° m BCD = 76° Holt Mc. Dougal Geometry Substitute 52 for m CDF. Subtract 104 from both sides.

6 -4 Properties of Special Parallelograms Example 2 B: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m ABC. ADC ABC Kite one pair opp. s Def. of s Polygon Sum Thm. m ABC + m BCD + m ADC + m DAB = 360° m ADC = m ABC Substitute m ABC for m ADC. m ABC + m BCD + m ABC + m DAB = 360° Holt Mc. Dougal Geometry

6 -4 Properties of Special Parallelograms Example 2 B Continued m ABC + m BCD + m ABC + m DAB = 360° m ABC + 76° + m ABC + 54° = 360° 2 m ABC = 230° m ABC = 115° Holt Mc. Dougal Geometry Substitute. Simplify. Solve.

6 -4 Properties of Special Parallelograms Example 2 C: Using Properties of Kites In kite ABCD, m DAB = 54°, and m CDF = 52°. Find m FDA. CDA ABC Kite one pair opp. s m CDA = m ABC Def. of s m CDF + m FDA = m ABC Add. Post. 52° + m FDA = 115° m FDA = 63° Holt Mc. Dougal Geometry Substitute. Solve.

6 -4 Properties of Special Parallelograms Example 4 A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. Trap. with pair base s isosc. trap. S P m S = m P 2 a 2 – 54 = a 2 Def. of s Substitute 2 a 2 – 54 for m S and + 27 2 a + 27 for m P. = 81 a = 9 or a = – 9 Holt Mc. Dougal Geometry Subtract a 2 from both sides and add 54 to both sides. Find the square root of both sides.

6 -4 Properties of Special Parallelograms Example 4 B: Applying Conditions for Isosceles Trapezoids AD = 12 x – 11, and BC = 9 x – 2. Find the value of x so that ABCD is isosceles. Diags. isosc. trap. AD = BC Def. of segs. Substitute 12 x – 11 for AD and 12 x – 11 = 9 x – 2 for BC. 3 x = 9 x=3 Holt Mc. Dougal Geometry Subtract 9 x from both sides and add 11 to both sides. Divide both sides by 3.

6 -4 Properties of Special Parallelograms Check It Out! Example 4 Find the value of x so that PQST is isosceles. Q S m Q = m S Trap. with pair base s isosc. trap. Def. of s 2 + 19 for m Q Substitute 2 x 2 x 2 + 19 = 4 x 2 – 13 and 4 x 2 – 13 for m S. 32 = 2 x 2 x = 4 or x = – 4 Holt Mc. Dougal Geometry Subtract 2 x 2 and add 13 to both sides. Divide by 2 and simplify.

6 -4 Properties of Special Parallelograms Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10. 75 Holt Mc. Dougal Geometry Solve.

6 -4 Properties of Special Parallelograms Check It Out! Example 5 Find EH. Trap. Midsegment Thm. 1 16. 5 = 2 (25 + EH) Substitute the given values. Simplify. 33 = 25 + EH Multiply both sides by 2. 13 = EH Subtract 25 from both sides. Holt Mc. Dougal Geometry