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# Traps and kites updated2014

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### Traps and kites updated2014

1. 1. GT Geometry 2/12/14 Drill Put HW and pen on the corner of your desk Solve for x. 5 or –5 1. x2 + 38 = 3x2 – 12 43 2. 137 + x = 180 156 3. 4. Find FE.
2. 2. Objectives Use properties of kites to solve problems. Use properties of trapezoids to solve problems.
3. 3. Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid
4. 4. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
5. 5. 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.
6. 6. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.
7. 7. 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.
8. 8. Example 1: Problem-Solving Application Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along . She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?
9. 9. Example 1 Continued 3 Solve N bisects JM. Pythagorean Thm. Pythagorean Thm.
10. 10. Example 1 Continued Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is, 36 – 32.4 3.6 cm Lucy will have 3.6 cm of wood left over after the cut.
11. 11. 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. CBF CDF m CBF = m CDF 2 sides isos. ∆ isos. ∆ base Def. of s s m BCD + m CBF + m CDF = 180° Polygon Sum Thm.
12. 12. Example 2A Continued m BCD + m CBF + m CDF = 180° Substitute m CDF m BCD + m CDF + m CDF = 180° for m CBF. m BCD + 52° + 52° = 180° m BCD = 76° Substitute 52 for m CDF. Subtract 104 from both sides.
13. 13. Example 3A: Using Properties of Isosceles Trapezoids Find m A. m C + m B = 180° 100 + m B = 180 Same-Side Int. s Thm. Substitute 100 for m C. m B = 80° A B Subtract 100 from both sides. Isos. trap. s base m A=m B Def. of m A = 80° Substitute 80 for m B s
14. 14. Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9 and MF = 32.7. Find FB. Isos.  trap. KJ = FM Def. of s base segs. KJ = 32.7 Substitute 32.7 for FM. KB + BJ = KJ Seg. Add. Post. 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.
15. 15. Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. EF = 10.75 Solve.