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Mastering Properties of Special Parallelograms

Explore the conditions for identifying rhombuses, trapezoids, and kites, solving geometric problems, and utilizing theorems in Holt McDougal Geometry. Practice finding measures, identifying shapes in the coordinate plane, and applying rules in warm-up exercises.

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Mastering Properties of Special Parallelograms

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  1. 6-4 Properties of Special Parallelograms Warm Up Lesson Presentation Lesson Quiz Holt McDougal Geometry

  2. Below are some conditions you can use to determine whether a parallelogram is a rhombus.

  3. A trapezoidis 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 anglesof a trapezoid are two consecutive angles whose common side is a base.

  4. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

  5. Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. WV = XT Def. of rhombus 13b – 9=3b + 4 Substitute given values. 10b =13 Subtract 3b from both sides and add 9 to both sides. b =1.3 Divide both sides by 10.

  6. Example 2A Continued TV = XT Def. of rhombus Substitute 3b + 4 for XT. TV =3b + 4 TV =3(1.3)+ 4 = 7.9 Substitute 1.3 for b and simplify.

  7. Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find mVTZ. mVZT =90° Rhombus  diag.  Substitute 14a + 20 for mVTZ. 14a + 20=90° Subtract 20 from both sides and divide both sides by 14. a=5

  8. Example 2B Continued Rhombus  each diag. bisects opp. s mVTZ =mZTX mVTZ =(5a – 5)° Substitute 5a – 5 for mVTZ. mVTZ =[5(5) – 5)]° = 20° Substitute 5 for a and simplify.

  9. Example 3A: 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)

  10. Example 3A Continued Step 1 Graph PQRS.

  11. Since , the diagonals are congruent. PQRS is a rectangle. Example 3A Continued Step 2 Find PR and QS to determine if PQRS is a rectangle.

  12. Since , PQRS is a rhombus. Example 3A Continued Step 3 Determine if 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.

  13. Example 2A: 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. ∆ isos. ∆base s  CBF  CDF mCBF = mCDF Def. of  s Polygon  Sum Thm. mBCD + mCBF + mCDF = 180°

  14. Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCDF + mCDF = 180° Substitute 52 for mCDF. mBCD + 52° + 52° = 180° Subtract 104 from both sides. mBCD = 76°

  15. Example 2B: 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 mADC = mABC Polygon  Sum Thm. mABC + mBCD + mADC + mDAB = 360° Substitute mABC for mADC. mABC + mBCD + mABC + mDAB = 360°

  16. Example 2B Continued mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

  17. Example 2C: 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° Substitute. mFDA = 63° Solve.

  18. Example 4A: 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 Def. of s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 2a2 – 54 = a2 + 27 Subtract a2 from both sides and add 54 to both sides. a2 = 81 a = 9 or a = –9 Find the square root of both sides.

  19. Example 4B: Applying Conditions for Isosceles Trapezoids AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles. Diags.  isosc. trap. Def. of segs. AD = BC Substitute 12x – 11 for AD and 9x – 2 for BC. 12x – 11 = 9x – 2 Subtract 9x from both sides and add 11 to both sides. 3x = 9 x = 3 Divide both sides by 3.

  20. Check It Out! Example 4 Find the value of x so that PQST is isosceles. Trap. with pair base s  isosc. trap. Q  S mQ = mS Def. of s Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS. 2x2 + 19 = 4x2 – 13 Subtract 2x2 and add 13 to both sides. 32 = 2x2 Divide by 2 and simplify. x = 4 or x = –4

  21. Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

  22. 1 16.5 = (25 + EH) 2 Check It Out! Example 5 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by 2. 33 = 25 + EH Subtract 25 from both sides. 13 = EH

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