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Published byPaul Chase Modified over 7 years ago

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Work as a team to solve the following problem: In rectangle ABCD, X and Y are mid- points of AB and CD and PD QC. Compare the area of quadrilateral XQYP with the area of ABCD. Prove your conjecture.

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A kite = A △ ABD + A △ DBC B C D Two formulas A DB = 10m BC = 13m <BAD is a right angle Find the area of the kite. A = ½(10)(5) + ½ (10)(12) = 25 + 60 = 85 m 2

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Theorem 105: The area of a kite equals half the product of its diagonals. A kite = d 1 d 2 A kite = ½ (10)(17) = 85m 2 B C D A DB = 10m AC = 17m Find the area of the kite.

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Find the area of a rhombus whose perimeter is 20 and whose longer diagonal is 8. A rhombus is a parallelogram, so its diagonals bisect each other. It is also a kite, so its diagonals are perpendicular to each other. XZ = 8 & XP = 4 The perimeter is 20 so XB = 5. Δ BPX is a right triangle so BP = 3 & BY = 6. A = ½ d 1 d 2 A = ½ (6)(8) A = 24

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