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

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

7. A trapezoid is a quadrilateral with exactly one pair of parallel sides
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.

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

9. 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.

10. 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.

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

12. 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.

13. 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)

14. Example 3A Continued
Step 1 Graph PQRS.

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

16. Example 3A 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.

17. 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. ∆
CBF  CDF
isos. ∆ base s 
mCBF = mCDF
Def. of   s
mBCD + mCBF + mCDF = 180°
Polygon  Sum Thm.

18. 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°

19. 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 
mADC = mABC
Def. of  s
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°

20. 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.

21. 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.

22. 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 a 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.

23. 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.

24. 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 2x for mQ and 4x2 – 13 for mS.
2x = 4x2 – 13
Subtract 2x2 and add 13 to both sides.
32 = 2x2
Divide by 2 and simplify.
x = 4 or x = –4

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

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