By: Manuela Velez, Daniel Machado, Andrea Guerra, and Kenneth Rhombuses By: Manuela Velez, Daniel Machado, Andrea Guerra, and Kenneth

Rhombus: Is a parallelogram with four congruent sides.

Classifying Rhombus by Coordinate methods: Step 1: Find the slope of each side Slope of LM= (3-2) / ( 3-1)= ½ Slope of NP= (2-1) / (5-3)= ½ Slope of MN= (3-2) / (3-5)= -½ Slope of LP= (2-1) / (1-3)= -½ Both pairs of opposite sides are parallel, So LMNP is a parallelogram. No sides are Perpendicular so LMNP is not a rectangle. Step 2: Use the Distance Formula to see if any pairs of the sides are congruent. LM= √(3-1)² + (3-2)²= √ 5 NP= √(5-3)² + (2-1)²= √ 5 MN= √(3-5)² + (3-2)²= √ 5 LP= √ (1-3)² + (2-1)²= √ 5 All sides are congruent, so LMNP is a rhombus.

Using properties of Rhombus: Algebra Find the values of the variables. Then find the lengths of the sides of the rhombus. Since all 4 sides are congruent (def of rhombus) 15=3y=5x=4x+3. 4x+3= 15 4x= 12 X= 3 Since x=3 3y= 5(3) 3y= 15 Y= 5 The variables are x=3 and y=5. The lengths of the sides are 15, 15, 15, 15.

Parallelogram/Rhombus properties. Since a rhombus is a parallelogram its consecutive angles, or angles of a polygon that share a side, are supplementary. Also, its opposite angles are congruent. Furthermore, The diagonals bisect each other.

More Rhombus Properties: Theorem 6-9: Each diagonal of a rhombus bisects two angles of the rhombus. Theorem 6-10: The diagonals of a rhombus are perpendicular.

Proving Theorem 6-9: In the next slide is a proof of theorem 6-10 Given: Rhombus ABCD Prove: AC bisects <BAD and < BCD. Proof: ABCD is a Rhombus, so its sides are all congruent. AC is congruent to AC by Reflexive Property of Congruence. Therefore, Δ ABC is congruent to ΔADC by the SSS postulate. <1is congruent to <2 and <3 is congruent to <4 by CPCTC. Therefore, AC bisects <BAD and <BCD by the definition of bisect. In the next slide is a proof of theorem 6-10

Finding Angle measures: MNPQ is a rhombus and m<N= 120. Find measures of the numbered angles. m <1= m <3- ITT m <1 + m <3 + 120= 180-Triangle Angle-Sum Theorem. 2(m <1) + 120= 180 2(m<1)= 60 m<1= 30 Therefore, m<1= m<3= 30. By Theorem 6-9, m<1= m<2 and m<3= m<4. Therefore, m<1= m<2= m<3= m<4= 30.

Converses to Theorems 6-9 and 6-10 Theorem 6-12: If one diagonal of a parallelogram bisects two angles of the parallelogram, then the parallelogram is a rhombus. Theorem 6-13: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

Naming coordinates: Give coordinates for points W and Z without using any new variables. Since the coordinates of two of the vertices are (0,t) and (r,0) then the coordinates of W (on x axis) is (-r,0) and Z (on y axis) is (0,-t).

Rectangle and Rhombus The midpoints of the sides of a rectangle connected form a rhombus.

Algebra: Use coordinate geometry to prove that the quadrilateral formed by connecting the midpoints of the sides of a rectangle is a rhombus. Given: MNPO is a rectangle. T, W, V, U are the midpoints of its sides. Prove: TWVU is a rhombus. From lesson 6-6 example 2, you know that TWVU is a parallelogram. In this example, it was shown that when you draw a quadrilateral by connecting the midpoints of the sides of another quadrilateral, the inner quadrilateral is a parallelogram. Now to prove that all sides are congruent (Definition of Rhombus) you need to use the distance formula. Coordinate proof: By the midpoint formula, the coordinates of the midpoints are T(0,b), W(a,2b), V(2a,b), and U(a,0). By the distance formula: TW= √(a-0)²+ (2b-b)²= √a²+ b² WV= √(2a-a)² + (b-2b)²= √a² + b² VU= √(a-2a)² + (0-b)²= √a² + b² UT= √(0-a)² + (b-0)²= √a² + b² TW=WV=VU=UT, so parallelogram TWVU is a rhombus.