Squares and Rhombi Lesson 6-5.

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Presentation transcript:

Squares and Rhombi Lesson 6-5

Vocabulary: Rhombus: A parallelogram with all four sides congruent. Square: A parallelogram with all four sides congruent and all four right angles.

A rhombus (plural is rhombi) is a parallelogram with all four sides congruent. A rhombus has all the characteristics of a parallelogram, with two additional characteristics.

A) According to Theorem 6 A) According to Theorem 6.16, if the figure is a rhombus, then each diagonal bisects a pair of angles. Since figure FGHJ is defined as a rhombus, this theorem applies. So, if m∠FJH = 82, then m∠KJH + m∠KJF = 82. Since GJ bisects these two angles, m∠KJH = m∠KJF, so m∠KJH is half of 82, which is 41 degrees. So m∠KJH + m∠KHJ + 90 = 180 41+ m∠KHJ = 90 So m∠KHJ = 49 B) The definition of a rhombus is a parallelogram with all four sides congruent, so GH ≅ JH, thus GH = JH, so x+9=5x-2. Then we can solve it using algebra. x+9=5x-2 -x -x 9= 4x-2 +2 +2 11= 4x (divide both sides by 4) So x= 11/4

A square is a parallelogram with four congruent sides and four right angles. It is also a rectangle and a rhombus because it satisfies all of those requirements as well.

All of the properties of parallelograms, rectangles, and rhombi also apply to squares. --the diagonals of a parallelogram bisect each other --the diagonals of a rectangle are congruent. --the diagonals of a rhombus are perpendicular

Since JKLM is a parallelogram, we know that we can use the properties of parallelograms, rectangles, squares, and rhombi to determine its true name. If ΔJKL is isosceles, then we know that side JK ≅ side KL. According to Theorem 6.19, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. So, JKLM is a rhombus.

Step 1: Draw a graph. This will help you visualize the problem, help you make conjectures, and determine reasonableness of your answer. Step 2: Ask yourself if you know what it looks like and how do you know. The figure looks like a parallelogram. Maybe a Rhombus, definitely not a square or rectangle.

Continue the previous example: Step 3: Prove it is a parallelogram. I am going to check using the distance formula of all the sides. (Remember you can use slope, distance, or midpoint formulas). Since all sides are congruent, it must be a Rhombus. Step 4: Justify your answer. Since all four sides are congruent, this is not just a parallelogram, but also a Rhombus according to the definition of a rhombus.

Continue the previous example: Step 5: Remember, just because it doesn’t LOOK like a square or a rectangle, doesn’t mean that we can assume that they are not. We must PROVE it. According to Theorem 6.13, if a parallelogram is a rectangle, then its diagonals are congruent. And Theorem 6.20 states that a quadrilateral has to be BOTH a rhombus AND a rectangle to be a square. So, using the Distance formula, we can determine if the diagonals are congruent. Since 68 ≠256 , the diagonals are NOT congruent, so the figure CANNOT be a rectangle and therefore, cannot be a square. So it is a Rhombus!!!