1. 6-5: Conditions of Special Parallelograms
Page 412
10) 29
11) 25
12)
13)
14.5
14)
31.5
15)
𝒎∠𝑽𝑾𝑿=𝟏𝟑𝟐°
𝒎∠𝑾𝒀𝑿=𝟔𝟔°
18)
𝒎∠𝟏=𝟐𝟗°
𝒎∠𝟐=𝟔𝟏°
𝒎∠𝟑=90°
𝒎∠𝟒=29°
𝒎∠𝟓=𝟗𝟎°
19)
𝒎∠𝟏=𝟓𝟒°
𝒎∠𝟐=36°
𝒎∠𝟑=54°
𝒎∠𝟒=108°
𝒎∠𝟓=72°
20)
𝒎∠𝟏=𝟗𝟎°
𝒎∠𝟐=45°
𝒎∠𝟑=45°
𝒎∠𝟒=45°
𝒎∠𝟓=45°
21)
𝒎∠𝟏=𝟏𝟐𝟔°
𝒎∠𝟐=27°
𝒎∠𝟑=27°
𝒎∠𝟒=126°
𝒎∠𝟓=27°
22)
𝒎∠𝟏=𝟓𝟓°
𝒎∠𝟐=55°
𝒎∠𝟑=55°
𝒎∠𝟒=70°
𝒎∠𝟓=55°
23)
𝒎∠𝟏=64°
𝒎∠𝟐=64°
𝒎∠𝟑=26°
𝒎∠𝟒=90°
𝒎∠𝟓=64°
24)
Always
25)
Sometimes
26)
27)
28)
29)
30)
31)
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6-5: Conditions of Special Parallelograms

2. 6-5: Conditions of Special Parallelograms
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6-5: Conditions of Special Parallelograms

3. Properties of Kites and Trapezoids
Section 6-5
Geometry PreAP, Revised ©2013
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6-5: Conditions of Special Parallelograms

4. 6-5: Conditions of Special Parallelograms
Definitions
Kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
Trapezoid is a quadrilateral with exactly one pair of parallel sides
Each parallel side is called a base
Base angles of a trapezoid are two consecutive angles whose common side is a base
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6-5: Conditions of Special Parallelograms

5. 6-5: Conditions of Special Parallelograms
Properties of Kites
A kite is a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent.
Just as in an isosceles triangle, the angles between each pair of congruent sides are vertex angles. The other pair of angles are nonvertex angles.
If a quadrilateral is a kite, then its diagonals are perpendicular
If a quadrilateral is a kite, then exactly one pair of opposite angles are congruent
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6-5: Conditions of Special Parallelograms

6. 6-5: Conditions of Special Parallelograms
Kites
If a quadrilateral is a kite, then the nonvertex angles are congruent.
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6-5: Conditions of Special Parallelograms

7. 6-5: Conditions of Special Parallelograms
Kites
If a quadrilateral is a kite, then the diagonal connecting the vertex angles is the perpendicular bisector of the other diagonal.
and CE  AE.
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6-5: Conditions of Special Parallelograms

8. 6-5: Conditions of Special Parallelograms
Kites
If a quadrilateral is a kite, then a diagonal bisects the opposite non-congruent vertex angles.
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6-5: Conditions of Special Parallelograms

9. 6-5: Conditions of Special Parallelograms
Example 1
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD.
mBCD + mCBF + mCDF = 180°
mBCD + 52° + 52° = 180°
mBCD = 76°
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6-5: Conditions of Special Parallelograms

10. 6-5: Conditions of Special Parallelograms
Example 2
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC.
ADC  ABC
mADC = mABC
mABC + mBCD + mADC + mDAB = 360°
mABC + mBCD + mABC + mDAB = 360°
mABC + 76° + mABC + 54° = 360°
2mABC = 230°
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6-5: Conditions of Special Parallelograms

11. 6-5: Conditions of Special Parallelograms
Your Turn
In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT.
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6-5: Conditions of Special Parallelograms

12. Properties of Trapezoids
The parallel sides are called bases
The non-parallel sides are called legs
A trapezoid has two pairs of base angles
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6-5: Conditions of Special Parallelograms

13. 6-5: Conditions of Special Parallelograms
Trapezoids
A. If a quadrilateral is a trapezoid, then the consecutive angles between the bases are supplementary.
If ABCD is a trapezoid, then x + y = 180° and r + t = 180°.
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6-5: Conditions of Special Parallelograms

14. Midsegment Trapezoids
B. A midsegment of a trapezoid is a segment that connects the midpoints of the legs of a trapezoids.
If ABCD is a trapezoid, then x + y = 180° and r + t = 180°.
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6-5: Conditions of Special Parallelograms

15. 6-5: Conditions of Special Parallelograms
Trapezoids
C. If a trapezoid is isosceles, then each pair of base angles is congruent.
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6-5: Conditions of Special Parallelograms

16. 6-5: Conditions of Special Parallelograms
Trapezoids
D. The midsegment of a trapezoid is parallel to each base and its length is one half the sum of the lengths of the bases. If
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6-5: Conditions of Special Parallelograms

17. 6-5: Conditions of Special Parallelograms
Example 3
KP = 21.9m and FM = 32.7m. Solve for FB.
KP = FM
KJ = 32.7
KB + BP = KB
BP = 10.8
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6-5: Conditions of Special Parallelograms

18. 6-5: Conditions of Special Parallelograms
Example 4
Find the value of a so that PQRS is isosceles.
S  P
mS = mP
2a2 – 54 = a2 + 27
a2 = 81
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6-5: Conditions of Special Parallelograms

19. 6-5: Conditions of Special Parallelograms
Your Turn
Find the value of x so that PQST is isosceles.
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6-5: Conditions of Special Parallelograms

20. 6-5: Conditions of Special Parallelograms
Example 5
Find EH. 1
16.5 = (25 + EH) 2
33 = 25 + EH
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6-5: Conditions of Special Parallelograms

21. 6-5: Conditions of Special Parallelograms
Your Turn
Solve for LP
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6-5: Conditions of Special Parallelograms

22. 6-5: Conditions of Special Parallelograms
Assignment
Pg 432: 14-25, all (omit 34)
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6-5: Conditions of Special Parallelograms