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6-6 Kites and Trapezoids

6-6 Kites and Trapezoids. Properties and conditions. Kites. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Properties of Kites. Trapezoids.

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6-6 Kites and Trapezoids

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  1. 6-6 Kites and Trapezoids Properties and conditions

  2. Kites A kiteis a quadrilateral with exactly two pairs of congruent consecutive sides.

  3. Properties of Kites

  4. Trapezoids A trapezoidis 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 anglesof a trapezoid are two consecutive angles whose common side is a base.

  5. 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.

  6. Properties of Isosceles Trapezoids

  7. Midsegments of Trapezoids The midsegment of a trapezoidis 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. Trapezoid Midsegment Theorem

  9. Lets apply! Example 1 In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. Kite cons. sides  2  sides isos. ∆ ∆BCD is isos. isos. ∆base s  CBF  CDF mCBF = mCDF Def. of  s mBCD + mCBF+ mCDF= 180° Polygon  Sum Thm.

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

  11. 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.

  12. 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 a2 + 27 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.

  13. 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.

  14. 1 16.5 = (25 + EH) 2 Lets apply! Example 5 Conditions of Midsegments Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. 33= 25 + EH Multiply both sides by 2. 13= EH Subtract 25 from both sides.

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