Chapter 6 Section 6: Trapezoids & Kites
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Chapter 6 Section 6: Trapezoids & Kites Goal: Recognize the properties of Kites & Trapezoids. Goal: Use the properties of Trapezoids & Kites to find missing measures. A KITE is a Quadrilateral that has TWO (2) pairs of Congruent-Consecutive Sides BUT Opposite Sides are NOT Congruent. The Diagonals of a KITE are Perpendicular—SEE Page 427 Theorem 6-6-1 IF a Quadrilateral is a KITE, EXACTLY one pair of Opposite Angles are Congruent AND the other Pair of Opposite Angles are Bisected by the Diagonal. SEE Page 427 Theorem 6-6-2 A TRAPEZOID is a Quadrilateral with EXACTLY 1 Pair of Parallel Sides. The Parallel sides are called the BASES and the NON-Parallel sides are called LEGS. A Trapezoid has two (2) sets of Base Angles. BASE ANGLES are formed where the Legs intersect the Bases. An ISOSCELES Trapezoid has exactly 1 pair of parallel sides and the legs are Congruent. A RIGHT Trapezoid has 1 pair of parallel sides and one (1) leg forms TWO 90o angles. Recall that in an Isosceles Triangle the BASE Angles are Congruent. When you “cut” the top off of an Isosceles Triangle an ISOSCELES TRAPEZOID is formed. The Two \ / Legs are congruent and the Base Angles have not changes and they remain Congruent. SEE PAGE 429—Theorem 6-6-3 and Theorem 6-6-4 THEOREM 6-6-5 “A Trapezoid is Isosceles IFF its Diagonals are Congruent.” BECAUSE: In an Isosceles Trapezoid, the legs are Congruent by Definition AND the Base Angles are also \ / Congruent Notice that there are TWO Triangles that can be shown to be Congruent by the SAS Postulate. || Once the Triangles are shown to be Congruent then EVERYTHING about them is Congruent INCLUDING the remaining Side which happens to be the Diagonals A MID-SEGMENT of a Trapezoid is very similar to the Mid-Segment of a Triangle in that it connects the Mid-Points of the two sides, SPECIFICALLY IN A TRAPEZOID, it connects the Mid-Points of the two Legs. TWO CHARACTERISTICS of the Mid-Segment: 1. It is Parallel to BOTH Bases 2. The LENGTH of the Mid-Segment is ½ the SUM of the length of BOTH Bases: MidSgmnt = ½ (b1 + b2) (SEE Page 431 Theorem 6-6-6) .