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Find the midpoint of each side of the trapezoid. Connect the midpoints. What is the most precise classification of the quadrilateral formed by connecting the midpoints of the sides of the trapezoid?

Mathematics
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|dw:1377490333537:dw|
I just made it a bit bigger. |dw:1377490546842:dw|
|dw:1377490654352:dw|

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Other answers:

So I know the answer is, The rhombus is a square but i need help finding midpoints.
We can label all midpoints.
Im guessing|dw:1377490680684:dw|
|dw:1377490733250:dw|
|dw:1377490731812:dw|
Alright to do that you just add and divide right?
Like the Xs and the Ys respectively?
Right. Add each 2 x's and divide by 2. Add each 2 y's and divide by 2.
Okay, easier then it sounds!
Now we need to see what the quadrilateral made by connecting midpoints is.
A rhombus!
|dw:1377491112017:dw|
Then is you them into right triangles you find that it is a square?
|dw:1377491168215:dw|
YEA:)
Notice that every side of the new quadrilateral is the hypotenuse of a triangle with both legs measuring 2. Therefore, all sides of the quadrilateral are congruent, and the quadrilateral is a rhombus.
The next question is if the rhombus is also a square.
All we need to do is to see if any two consecutive sides are perpendicular.
Which they are!
|dw:1377491486110:dw|
One slope is 1. The other slope is -1. The slopes of two consecutive sides are negative reciprocals, so those two sides are perpendicular.
I get it:)
|dw:1377491738087:dw| Opposite angles of any parallelogram are congruent.
Also all angles added up are 360 degrees.
|dw:1377491805994:dw|
x + x + 90 + 90 = 360 x = 90 All angles measure 90 degrees, and all sides a congruent, so it's a square.
Yup:) hey! can you go back to the first question for a second?
ok

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