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By, Michael Thiel and Lauren Larar. First, you must identify what family the quadrilateral falls under. Parallelogram, Trapezoid , quadrilateral. 1. If a parallelogram contains at least one right angle, then it is a rectangle. 2. I the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. *You can also prove that a quadrilateral is rectangle without first showing that it is a parallelogram.* 1. If all four angles of a quadrilateral are right angles, then it is a rectangle. 1. If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it is a kite. 2. If one of the diagonals of a quadrilateral is the perpendicular bisector of the other diagonal, then the quadrilateral is a kite. 1. If a parallelogram contains a pair of consecutive sides that are congruent, then it is a rhombus. 2. If either diagonal of a parallelogram bisects two angles of the parallelogram, then it is a rhombus. *You can also prove that a quadrilateral is a rhombus without first showing that it is a parallelogram.* 1.If the diagonals of a quadrilateral are perpendicular bisectors of each other, then the quadrilateral is a rhombus. 1. If a quadrilateral is both a rectangle and a rhombus, then it is a square. 1.If the nonparallel sides of a trapezoid are congruent, then it is isosceles. 2. If the lower or the upper base angles of a trapezoid are congruent, then it is isosceles. 3. If the diagonals of a trapezoid are congruent, then it is isosceles. First, the quadrilateral must be identified as either :parallelogram, trapezoid, or quadrilateral. Then, if it is a parallelogram: determine if it is a rectangle or rhombus; if either, determine if it is a square. If it is a trapezoid, determine if it is isosceles. J (-2,4) A (6,5) Sample Problem! What is the most descriptive name for quadrilateral JACK with vertices J=(2,4), A=(6,5), C=(5,-7), K=(-3,-8)? K (-3,-8) C (5,-7) A. m JA= 4-5/-2-6= -1/-8= 1/8 m KC= -8-(-7)/-3-5= -1/-8= 1/8 Therefore, JA is parallel to KC (same slope implies parallel lines) m JK= 4-(-8)/-2-(-3)= 12/1= 12 m AC= 5-(-7)/ 6-5= 12/1= 12 Therefore, JK is parallel to AC (same slope implies parallel lines) Therefore, JACK is a parallelogram (opp sides parallelquad. is a parallelogram) B. m JC= 4-(-7)/-2-5= 11/-7 m KA= -8-5/-3-6= -13/-9= 13/9 JC is not perpendicular to KA (No opp reciprocal slopes) Therefore, JACK is not a rhombus (Diag. are not perpendicular) C. m JA= 1/8 m AC= 12….Therefore JACK is not a rectangle (Not at least on rt angle) D. Therefore JACK is a parallelogram. 2 column proof sample problem! S 1.) ME R 1.) Given MI 2.) Given 2.) KM bisects <EMI 3.) <EMT 4.) MT 5.) EMT 6.) <ETM 3.) If a seg bisecvts the <, it divides the < into 2 congruent segs. < IMT MT 4.) Reflexive IMT 5.) SAS (1,3,4) <ITM 6.) CPCTC 7.) CPCTC 7.) ET TI 8.) If two adjacent angles form a straight angle, they are suppl. 9.) If 2 angles are both suppl and cong., they are rt angles 10.) If 2 pts are =dist from the segs endpts, it det. The perp bis of a seg. 8.)<ETM suppl. <TIM 9.) <ETM, <ITN rt <s 10.) KM perp. bis. EI 11.)MIKE is a kite 11.) If one of the diags of a quad is the perp bis of the other diag, then the quad is a kite. M E T K I 1. ---> Rhombus! 2. -----> Isosceles Trapezoid Practice Problems! 1. 1. Given 2 . 2. Perpendicular lines meet to form right angles 3. 4. parallel. 4. In a plane, lines perpendicular to the same line are parallel. 5. 5. A quadrilateral with two sets of opposite 6. 3. The opposite sides of a parallelogram are parallel sides is a parallelogram. 6. A parallelogram with one right angle is a rectangle What is the most descriptive name for quadrilateral ABCD with A=(1,2), B=(2,5), C=(5,7), D=(4,4)? m AB m DC m BC m AD Therefore, AB is parallel to DC (same slope implies parallel lines). BC is parallel to AD (same slope implies parallel lines) Therefore, ABCD is a parallelogram (opp sides parallel implies parallelogram) m AB= 3 m BC= 2/3 AB is not perpendicular to BC (no opp recip. slopes) m BD= 5-4/2-4= 1/-2 m CA= 7-2/5-1=5/4 BD is not perpendicular to CA (no opp recip. slopes) Therefore, ABCD is not a rectangle (not at least one right angle) Therefore, ABCD is not a rhombus (diags not perpendicular) Therefore, ABCD is a parallelogram because it is not a rhombus or a rectangle In order for MIKE to be a rectangle, what must the value of X be? M E I 2x+40+30=90 2x+70=90 2x=20 X=10 2(10)+40+30=90 K In order to prove this quadrilateral to be a rectangle, one side must be right. A right angle is 90 degrees, therefore, the measurements of each angle must be added together and set equal to 90 degrees. Then the value of X must be found. In order for this quadrilateral to be a rectangle, the value of X must be 20. Find the area of the parallelogram (10,8) (-2,6) (-2,0) (10,2) In order to find the area of the parallelogram, first, the length of the sides must be determined. In addition, the opposite sides are congruent due to the fact that it is a parallelogram. Therefore, after finding one side to be 6 and the other side to be 8, multiply the length by width. (6x8= 48) The area is 48. A Greek mathematician became interested in the Golden Ratio, a ratio of about 1:1.618. A rectangle whose side lengths are in the golden ratio is called a Golden Rectangle. These rectangles are believed to be especially pleasing to look at. The Golden Ratio has been found in many works of art and architecture such as the Parthenon at the left. ---The outline of this figure to the left is a quadrilateral. What is a more specific name for it? This Shape would be classified as a square because all sides are congruent and all of the angles are right. "Coordinate Geometry Proofs." Oswego City School District Regents Exam Prep Center. Web. 19 Jan. 2011. <http://www.regentsprep.org/regents/math/geometry/GCG4/CoordinatepRACTICE.htm>. McDougal Littell Geometry Applying, Reasoning, Measuring. Evanston, IL: McDougal Littell, 2005. Print. "Practice with Quadrilaterals." Oswego City School District Regents Exam Prep Center. Web. 19 Jan.2011. <http://www.regentsprep.org/Regents/math/geometry/GP9/PracQuadPf.htm>. Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry for Enjoyment and Challenge. Evanston, IL: McDougal, Littell, 1991. Print