A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

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Presentation transcript:

A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

Example 1: 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?

Example 1 Continued ? ? 15cm Use Pythagorean Theorem to find missing lengths 9.9cm 22.4 cm KL = =32.3 Lucy needs to cut the dowel to be 32.3 cm long. The amount of wood that will remain after the cut is, 36 – 32.3  3.7 cm Lucy will have 3.6 cm of wood left over after the cut.

Kite  cons. sides  Example 2: In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. ∆BCD is isos. 2  sides  isos. ∆ isos. ∆  base s  Polygon  Sum Thm. CBF  CDF mBCD + mCBF + mCDF = 180° mBCD + 52° + 52° = 180°Substitute 52 for mCBF. mBCD = 76°

Example 3: Def. of  s Polygon  Sum Thm. In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC mADC = mABC mABC + mBCD + mADC + mDAB = 360° Kite  one pair opp. s  mABC + 76° + mABC + 54° = 360° Substitute. 2mABC = 230° Simplify. mABC = 115° Solve.

Example 4: Def. of  s  Add. Post. Substitute. Solve. In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA. CDA  ABC mCDA = mABC mCDF + mFDA = mABC 52° + mFDA = 115° mFDA = 63° Kite  one pair opp. s 

Example 5: In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. Kite  cons. sides  ∆PQR is isos. 2  sides  isos. ∆ isos. ∆  base s RPQ  PRQ mPQR + mQRP + mQPR = 180° Polygon  Sum Thm. 78° + mQRT + mQPT = 180° Substitute 78 for mPQR. 78° + 2mQRT = 180° Simplify. 2mQRT = 102° mQRT = 51°

Assignment #48 Pg. 432 #2-6 all

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.

Isos.  trap. s base  Example 1 Find mA. Same-Side Int. s Thm. (Remember Parallel Lines) 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°

Example 2 KB = 21m and MF = 32 Find FB. Isos.  trap. s base  Def. of  segs. Substitute 32 for FM. Seg. Add. Post. Substitute 21 for KB and 32 for KJ. Subtract 21 from both sides. KJ = FM KJ = 32 KB + JB = KJ 21 + JB = 32 JB = 11 FB = 11 Diagonals of Isosceles Trapezoids are 

Example 3 JN = 10, and NL = 14 Find KM. Segment Add Postulate Substitute. Substitute and simplify. Isos.  trap. s base  JL = JN + NL KM = JN + NL KM = = 24

Example 4: 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

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.

Example 5 Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

Example 6 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

Assignment #50 Pg. 432 #7-16 all *Assignment #49 – Section 6-6 Guided Notes *Assignment #48 – Pg. 432 #2-6