Trapezoids and Kites LESSON 6–6. Lesson Menu Five-Minute Check (over Lesson 6–5) TEKS Then/Now New Vocabulary Theorems: Isosceles Trapezoids Proof: Part.

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Presentation transcript:

Trapezoids and Kites LESSON 6–6

Lesson Menu Five-Minute Check (over Lesson 6–5) TEKS Then/Now New Vocabulary Theorems: Isosceles Trapezoids Proof: Part of Theorem 6.23 Example 1:Real-World Example: Use Properties of Isosceles Trapezoids Example 2:Isosceles Trapezoids and Coordinate Geometry Theorem 6.24: Trapezoid Midsegment Theorem Example 3:Midsegment of a Trapezoid Theorems: Kites Example 4:Use Properties of Kites

Over Lesson 6–5 5-Minute Check 1 A.5 B.7 C.10 D.12 LMNO is a rhombus. Find x.

Over Lesson 6–5 5-Minute Check 2 A.6.75 B C.10.5 D.12 LMNO is a rhombus. Find y.

Over Lesson 6–5 5-Minute Check 3 A B.9 C D.6.5 QRST is a square. Find n if m  TQR = 8n + 8.

Over Lesson 6–5 5-Minute Check 4 QRST is a square. Find w if QR = 5w + 4 and RS = 2(4w – 7). A.6 B.5 C.4 D.3.3 _

Over Lesson 6–5 5-Minute Check 5 A.9 B.10 C.54 D.65 QRST is a square. Find QU if QS = 16t – 14 and QU = 6t + 11.

Over Lesson 6–5 5-Minute Check 6 Which statement is true about the figure shown, whether it is a square or a rhombus? A. B. C.JM║LM D.

TEKS Targeted TEKS G.2(B) Derive and use the distance, slope, and midpoint formulas to verify geometric relationships, including congruence of segments and parallelism or perpendicularity of pairs of lines. Mathematical Processes G.1(F), G.1(G)

Then/Now You used properties of special parallelograms. Apply properties of trapezoids. Apply properties of kites.

Vocabulary trapezoid bases legs of a trapezoid base angles isosceles trapezoid midsegment of a trapezoid kite

Concept 1

Example 1A Use Properties of Isosceles Trapezoids A. BASKET Each side of the basket shown is an isosceles trapezoid. If m  JML = 130, KN = 6.7 feet, and MN= 3.6 feet, find m  MJK.

Example 1A Use Properties of Isosceles Trapezoids Since JKLM is a trapezoid, JK║LM. m  JML + m  MJK=180Consecutive Interior Angles Theorem m  MJK =180Substitution m  MJK=50Subtract 130 from each side. Answer: m  MJK = 50

Example 1B Use Properties of Isosceles Trapezoids B. BASKET Each side of the basket shown is an isosceles trapezoid. If m  JML = 130, KN = 6.7 feet, and MN is 10.3 feet, find JL.

Example 1B Use Properties of Isosceles Trapezoids JL=KMDefinition of congruent JL=KN + MNSegment Addition JL= Substitution JL=10.3Add. Answer: JL= 10.3 Since JKLM is an isosceles trapezoid, diagonals JL and KM are congruent.

Example 1A A.124 B.62 C.56 D.112 A. Each side of the basket shown is an isosceles trapezoid. If m  FGH = 124, FI = 9.8 feet, and IG = 4.3 feet, find m  EFG.

Example 1B A.4.3 ft B.8.6 ft C.9.8 ft D.14.1 ft B. Each side of the basket shown is an isosceles trapezoid. If m  FGH = 124, FI = 9.8 feet, and EG = 14.1 feet, find IH.

Concept 3

Example 3 In the figure, MN is the midsegment of trapezoid FGJK. What is the value of x? Midsegment of a Trapezoid

Example 3 Read the Item You are given the measure of the midsegment of a trapezoid and the measures of one of its bases. You are asked to find the measure of the other base. Solve the Item Trapezoid Midsegment Theorem Substitution Midsegment of a Trapezoid

Example 3 Multiply each side by 2. Subtract 20 from each side. Answer: x = 40 Midsegment of a Trapezoid

Example 3 A.XY = 32 B.XY = 25 C.XY = 21.5 D.XY = 11 WXYZ is an isosceles trapezoid with median Find XY if JK = 18 and WZ = 25.

Concept 4

Example 4A Use Properties of Kites A. If WXYZ is a kite, find m  XYZ.

Example 4A Use Properties of Kites Since a kite only has one pair of congruent angles, which are between the two non-congruent sides,  WXY   WZY. So,  WZY = 121 . m  W + m  X + m  Y + m  Z=360Polygon Interior Angles Sum Theorem m  Y + 121=360Substitution m  Y=45Simplify. Answer: m  XYZ = 45

Example 4B Use Properties of Kites B. If MNPQ is a kite, find NP.

Example 4B Use Properties of Kites Since the diagonals of a kite are perpendicular, they divide MNPQ into four right triangles. Use the Pythagorean Theorem to find MN, the length of the hypotenuse of right ΔMNR. NR 2 + MR 2 =MN 2 Pythagorean Theorem (6) 2 + (8) 2 =MN 2 Substitution =MN 2 Simplify. 100=MN 2 Add. 10=MNTake the square root of each side.

Example 4B Use Properties of Kites Answer: NP = 10 Since MN  NP, MN = NP. By substitution, NP = 10.

Example 4A A.28° B.36° C.42° D.55° A. If BCDE is a kite, find m  CDE.

Example 4B A.5 B.6 C.7 D.8 B. If JKLM is a kite, find KL.

Trapezoids and Kites LESSON 6–6