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

2. Rhombus:
Is a parallelogram with four congruent sides.

3. 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.

4. 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.

5. 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.

6. 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.

7. 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

9. 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 < = 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.

10. 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.

11. 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).

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

13. 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.