1. Properties of Special Parallelograms
Math 132
Day 14
Properties of Special Parallelograms
Holt McDougal Geometry

2. 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.

3. Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

4. Example 1: Craft Application
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.

5. Check It Out! Example 1a
Carpentry The rectangular gate has diagonal braces.
Find HJ.
Rect.  diags. 
HJ = GK = 48
Def. of  segs.

6. Check It Out! Example 1b
Carpentry The rectangular gate has diagonal braces.
Find HK.
Rect.  diags. 
Rect.  diagonals bisect each other
JL = LG
Def. of  segs.
JG = 2JL = 2(30.8) = 61.6
Substitute and simplify.

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

9. Like a rectangle, a rhombus is a parallelogram
Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

10. 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.

11. 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.

12. 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

13. 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.

14. Check It Out! Example 2a
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

15. Check It Out! Example 2b
CDFG is a rhombus.
Find the measure.
mGCH if mGCD = (b + 3)°
and mCDF = (6b – 40)°
mGCD + mCDF = 180°
Def. of rhombus
b b – 40 = 180°
Substitute.
7b = 217°
Simplify.
b = 31°
Divide both sides by 7.

16. Check It Out! Example 2b 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.

17. 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.

18. Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms.
Helpful Hint

19. Check It Out! Example 4
Given: PQTS is a rhombus with diagonal
Prove:

20. Check It Out! Example 4 Continued
Statements
Reasons
1. PQTS is a rhombus.
1. Given.
2. Rhombus → each
diag. bisects opp. s 2.
3. QPR  SPR
3. Def. of  bisector. 4.
4. Def. of rhombus. 5.
5. Reflex. Prop. of  6.
6. SAS 7.
7. CPCTC

21. Lesson Quiz: Part I
A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length.
1. TR CE
35 ft
29 ft

22. Lesson Quiz: Part II
PQRS is a rhombus. Find each measure.
3. QP mQRP 42
51°

23. Lesson Quiz: Part IV
6. Given: ABCD is a rhombus.
Prove: 

24. Properties and conditions
Kites and Trapezoids
Properties and conditions

25. Kites
A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

26. Properties of Kites

27. Trapezoids
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.

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

29. Properties of Isosceles Trapezoids

30. Midsegments of Trapezoids
The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.

31. Trapezoid Midsegment Theorem

32. Lets apply!
Example 1
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.

33. Lets apply! Example 1 Continued mBCD + mCBF + mCDF = 180°
mBCD + mCDF + mCDF = 180°
Substitute mCDF for mCBF.
mBCD + 52° + 52° = 180°
Substitute 52 for mCDF.
mBCD = 76°
Subtract 104 from both sides.

34. Lets apply!
Example 2
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.

35. Example 3: Applying Conditions for Isosceles Trapezoids
Lets apply!
Example 3: Applying Conditions for Isosceles Trapezoids
Find the value of a so that PQRS is isosceles.
S  P
Trap. with pair base s   isosc. trap.
mS = mP
Def. of  s
2a2 – 54 = a2 + 27
Substitute 2a2 – 54 for mS and a for mP.
a2 = 81
Subtract a2 from both sides and add 54 to both sides.
a = 9 or a = –9
Find the square root of both sides.

36. Example 4: Applying Conditions for Isosceles Trapezoids
Lets apply!
Example 4: 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.
AD = BC
Def. of  segs.
12x – 11 = 9x – 2
Substitute 12x – 11 for AD and 9x – 2 for BC.
3x = 9
Subtract 9x from both sides and add 11 to both sides.
x = 3
Divide both sides by 3.

37. Example 5 Conditions of Midsegments
Lets apply!
Example 5 Conditions of Midsegments
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.