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Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE .

Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5. 43. 156. 6-6. Properties of Kites and Trapezoids. Holt Geometry. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Example 2A: Using Properties of Kites.

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Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE .

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  1. Warm Up Solve for x. 1.x2 + 38 = 3x2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156

  2. 6-6 Properties of Kites and Trapezoids Holt Geometry

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

  4. 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. 2  sides isos. ∆ isos. ∆base s  CBF  CDF mCBF = mCDF Def. of  s Polygon  Sum Thm. mBCD + mCBF + mCDF = 180°

  5. Example 2A Continued mBCD + mCBF + mCDF = 180° Substitute mCDF for mCBF. mBCD + mCBF+ mCDF= 180° Substitute 52 for mCBF. mBCD + 52°+ 52° = 180° Subtract 104 from both sides. mBCD = 76°

  6. 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.

  7. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid.

  8. Example 3A: Using Properties of Isosceles Trapezoids Find mA. mC + mB = 180° Same-Side Int. s Thm. 100 + mB = 180 Substitute 100 for mC. mB = 80° Subtract 100 from both sides. A  B Isos. trap. s base  mA = mB Def. of  s mA = 80° Substitute 80 for mB

  9. Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos.  trap. s base  KJ = FM Def. of segs. KJ = 32.7 Substitute 32.7 for FM. Seg. Add. Post. KB + BJ = KJ 21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ. BJ = 10.8 Subtract 21.9 from both sides.

  10. Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Isos. trap. s base  Def. of segs. KM = JL JL = JN + NL Segment Add Postulate KM = JN + NL Substitute. KM = 10.6 + 14.8 = 25.4 Substitute and simplify.

  11. Example 4A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. Trap. with pair base s  isosc. trap. S  P mS = mP Def. of s Substitute 2a2 – 54 for mS and a2 + 27 for mP. 2a2 – 54 = a2 + 27 Subtract a2 from both sides and add 54 to both sides. a2 = 81 a = 9 or a = –9 Find the square root of both sides.

  12. Example 4B: 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. Def. of segs. AD = BC Substitute 12x – 11 for AD and 9x – 2 for BC. 12x – 11 = 9x – 2 Subtract 9x from both sides and add 11 to both sides. 3x = 9 x = 3 Divide both sides by 3.

  13. The midsegment of a trapezoidis the segment whose endpoints are the midpoints of the legs.

  14. Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

  15. 1 16.5 = (25 + EH) 2 Check It Out! Example 5 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by 2. 33= 25 + EH Subtract 25 from both sides. 13= EH

  16. Lesson Quiz: Part I 1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite? In kite HJKL, mKLP = 72°, and mHJP = 49.5°. Find each measure. 2. mLHJ 3. mPKL about 191.2 in. 81° 18°

  17. Lesson Quiz: Part II Use the diagram for Items 4 and 5. 4. mWZY = 61°. Find mWXY. 5.XV = 4.6, and WY = 14.2. Find VZ. 6. Find LP. 119° 9.6 18

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