Conditions for Special Parallelograms Entry Task List the 6 ways to prove a quadrilateral is a parallelogram, show a picture of each.

Conditions for Special Parallelograms Rectangle 4 rt. Angles Congruent diagonals Rhombus 4 congruent sides Diagonals that bisect opp. Angles Perpendicular diagonals Square (must have 1 property from rectangle and rhombus) 4 rt. Angles Congruent diagonals 4 congruent sides Diagonals that bisect opp. Angles Perpendicular diagonals Is it a parallelogram??

Holt Geometry 6-6 Properties of Kites and Trapezoids I know and understand the properties of trapezoids and kites Learning targets Success Criteria I can use properties of kites and trapezoids to solve problems.

Holt Geometry 6-6 Properties of Kites and Trapezoids kite trapezoid base of a trapezoid leg of a trapezoid base angle of a trapezoid isosceles trapezoid midsegment of a trapezoid Vocabulary

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

Holt Geometry 6-6 Properties of Kites and Trapezoids *Note – These are NOT parallelograms!

Holt Geometry 6-6 Properties of Kites and Trapezoids 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. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

Holt Geometry 6-6 Properties of Kites and Trapezoids

Holt Geometry 6-6 Properties of Kites and Trapezoids 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.

Holt Geometry 6-6 Properties of Kites and Trapezoids Check It Out! Example 2c Polygon  Sum Thm. Def. of  s Substitute. Simplify. In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR. mSPT + mTRS + mRSP = 180° mSPT = mTRS mTRS + mTRS + mRSP = 180° 59° + 59° + mRSP = 180° mRSP = 62°

Holt Geometry 6-6 Properties of Kites and Trapezoids 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LHJ3. mPKL about in. 81°18°

Holt Geometry 6-6 Properties of Kites and Trapezoids Isos.  trap. s base  Example 3A: Using Properties of Isosceles Trapezoids Find mA. Same-Side Int. s Thm. Substitute 100 for mC. Subtract 100 from both sides. Def. of  s Substitute 80 for mB mC + mB = 180° mB = 180 mB = 80° A  B mA = mB mA = 80°

Holt Geometry 6-6 Properties of Kites and Trapezoids Example 5: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

Holt Geometry 6-6 Properties of Kites and Trapezoids Check It Out! Example 5 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by = 25 + EH Subtract 25 from both sides. 13 = EH = ( 25 + EH ) 2

Holt Geometry 6-6 Properties of Kites and Trapezoids

Holt Geometry 6-6 Properties of Kites and Trapezoids Homework p. 394 #7-23 odds, 26 and odds,47-52 Challenge - 63

Holt Geometry 6-6 Properties of Kites and Trapezoids 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?

Holt Geometry 6-6 Properties of Kites and Trapezoids Example 1 Continued 1 Understand the Problem The answer will be the amount of wood Lucy has left after cutting the dowel. 2 Make a Plan The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find, and. Add these lengths to find the length of.

Holt Geometry 6-6 Properties of Kites and Trapezoids Solve 3 N bisects JM. Pythagorean Thm. Example 1 Continued

Holt Geometry 6-6 Properties of Kites and Trapezoids 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. Example 1 Continued

Holt Geometry 6-6 Properties of Kites and Trapezoids Look Back 4 Example 1 Continued To estimate the length of the diagonal, change the side length into decimals and round., and. The length of the diagonal is approximately = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer.

Holt Geometry 6-6 Properties of Kites and Trapezoids 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. Substitute 12x – 11 for AD and 9x – 2 for BC. Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. AD = BC 12x – 11 = 9x – 2 3x = 9 x = 3

Holt Geometry 6-6 Properties of Kites and Trapezoids Check It Out! Example 4 Find the value of x so that PQST is isosceles. Subtract 2x 2 and add 13 to both sides. x = 4 or x = –4 Divide by 2 and simplify. Trap. with pair base s   isosc. trap. Q  SQ  S Def. of  s Substitute 2x for mQ and 4x 2 – 13 for mS. mQ = mS 2x = 4x 2 – = 2x 2

Holt Geometry 6-6 Properties of Kites and Trapezoids Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = Find FB. Isos.  trap. s base  Def. of  segs. Substitute 32.7 for FM. Seg. Add. Post. Substitute 21.9 for KB and 32.7 for KJ. Subtract 21.9 from both sides. KJ = FM KJ = 32.7 KB + BJ = KJ BJ = 32.7 BJ = 10.8

Holt Geometry 6-6 Properties of Kites and Trapezoids Example 3B Continued Same line. Isos. trap.  s base  Isos. trap.  legs  SAS CPCTC Vert.  s  KFJ  MJF BKF  BMJ FBK  JBM ∆FKJ  ∆JMF

Holt Geometry 6-6 Properties of Kites and Trapezoids Isos. trap.  legs  AAS CPCTC Def. of  segs. Substitute 10.8 for JB. Example 3B Continued ∆FBK  ∆JBM FB = JB FB = 10.8

Holt Geometry 6-6 Properties of Kites and Trapezoids Lesson Quiz: Part II Use the diagram for Items 4 and mWZY = 61°. Find mWXY. 5. XV = 4.6, and WY = Find VZ. 6. Find LP. 119°