Question 1209368: Sketch two noncongruent parallelograms ABCD and EFGH such that AC=EG and BD=FH.
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Get some graph paper or use a tool like GeoGebra
I'll use the 2nd option.
The particular GeoGebra work book I made is shown here
P = intersection of diagonals AC and BD
The diagonals of any parallelogram always bisect each other.
So for parallelogram ABCD we have AP = PC and BP and PD
In the GeoGebra work book shown above, I constructed ABCD by starting at P and radiating outward.
P = (0,0) = the origin
I moved 2 units to the right and left from this center point to form points A(2,0) and C(-2,0)
The choice of moving 2 units is arbitrary.
Since AP = PC = 2, this shows P bisects AC.
Also from point P, I moved up and down 5 units to arrive at B(0,5) and D(0,-5)
BP = PD = 5 which shows P bisects BD.
By construction, diagonals AC and BD bisect each other at P.
This shows ABCD is a parallelogram.
Another way to see it's a parallelogram is to use the slope formula
You need to show that: slopeAB = slopeCD and slopeBC = slopeAD.
I'll let the student handle this portion.
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Now onto the other parallelogram.
Q = intersection of diagonals EG and FH.
Parallelogram EFGH will have points E and G 2 units away from the center point Q.
The placement of E and G are the same as A and C earlier.
This is to have AC = EG.
I placed Q at (10,0) as an arbitrary decision to give some buffer room between the parallelograms.
Points F and H are on the dashed circle.
The circle is centered at Q and has radius 5.
The 5 refers to the distance from P to B; which ensures that we keep the same diagonal lengths as from ABCD.
To go from Q to F, we move 3 units left and 4 units up. Note that I'm using a 3-4-5 right triangle to have point F be a lattice point.
But point F doesn't have to be a lattice point.
The red point F can be moved around to show various other parallelograms possible that have the same diagonal configuration.
According to the diagram, EFGH is not congruent to ABCD if F is not directly above Q (or directly below Q)
To prove ABCD isn't congruent to EFGH, you can use the distance formula to find the side lengths.
AB = sqrt(29) = 5.38516 approximately
EF = sqrt(41) = 6.40312 approximately
This is when point F is at (7,4)
If you move F to other locations on the circle, then EF will change in length.
Since AB and EF aren't the same length, this proves ABCD and EFGH are not congruent parallelograms.
Congruent parallelograms must have the same corresponding lengths.
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