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)= -½Both pairs of opposite sides are parallel,So LMNP is a parallelogram. No sides arePerpendicular 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= √ 5All 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= 3Since 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:Given: Rhombus ABCDProve: 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 measuresof 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 + b²TW=WV=VU=UT, so parallelogram TWVU is a rhombus.