2. Lesson 6.6 Trapezoids and Kites

3. Definition  Kite – a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent.

4. Parts of a Kite Vertex angles: <B & <D Non-Vertex diagonal CA Non-Vertex angles: <A & <C Vertex diagonal BD

5. Properties of Kites  The diagonals of a kite are perpendicular  Non-vertex angles (a)are congruent  Vertex diagonal bisects the vertex angles  Vertex diagonal bisects the non-vertex diagonal

6. Perpendicular Diagonals of a Kite  If a quadrilateral is a kite, then its diagonals are perpendicular.

7. Non-Vertex Angles of a Kite  If a quadrilateral is a kite, then non- vertex angles are congruent  A   C,  B   D

8. Vertex diagonals bisect vertex angles If a quadrilateral is a kite then the vertex diagonal bisects the vertex angles.

9. Vertex diagonal bisects the non-vertex diagonal If a quadrilateral is a kite then the vertex diagonal bisects the non-vertex diagonal

10. Definition-a quadrilateral with exactly one pair of parallel sides. Leg Leg Base Base A B CD › › Trapezoid

11. <A + <C = 180 <A + <C = 180 <B + <D = 180 <B + <D = 180 A B CD › › Leg Angles are Supplementary Property of a Trapezoid

12. Isosceles Trapezoid Definition - A trapezoid with congruent legs.

13. Isosceles Trapezoid - Properties | | 1) Base Angles Are Congruent 2) Diagonals Are Congruent

14. Example PQRS is an isosceles trapezoid. Find m P, m Q and mR. m  R = 50  since base angles are congruent m  P = 130  and m  Q = 130  (consecutive angles of parallel lines cut by a transversal are  )

15. Find the measures of the angles in trapezoid 48 m< A = 132 m< B = 132 m< D = 48

16. Find BE  AC = 17.5, AE = 9.6 E

17. Example  Find the side lengths of the kite.

18. Example Continued We can use the Pythagorean Theorem to find the side lengths. 12 2 + 20 2 = (WX) 2 144 + 400 = (WX) 2 544 = (WX) 2 12 2 + 12 2 = (XY) 2 144 + 144 = (XY) 2 288 = (XY) 2

19. Find the lengths of the sides of the kite W X Y Z 4 55 8

20. Find the lengths of the sides of kite to the nearest tenth 4 2 2 7

21. Example 3  Find mG and mJ. Since GHJK is a kite  G   J So 2(m  G) + 132  + 60  = 360  2(m  G) =168  m  G = 84  and m  J = 84 

22. Try This!  RSTU is a kite. Find mR, mS and mT. x +30 + 125 + 125 + x = 360 2x + 280 = 360 2x = 80 x = 40 So m  R = 70 , m  T = 40  and m  S = 125 