Can someone walk me through these and help me get the right answer .... A quadrilateral has vertices (2, 0), (0, –2), (–2, 4), and (–4, 2). Which special quadrilateral is
formed by connecting the midpoints of the sides?
kite
rectangle
trapezoid
rhombus
2. Which of the following describes TVS? The vertices are T(1, 1), V(4, 0), and S(3, 5)
isosceles
scalene
right
equilateral
4. Verify that parallelogram ABCD with vertices A(–5, –1), B(–9, 6), C(–1, 5), and D(3, –2) is a
rhombus by showing that it is a parallelogram with perpendicular diagonal

- anonymous

- jamiebookeater

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- anonymous

hi

- anonymous

Heyy :)

- DanJS

Plot each point, then, Connect the middle of each side and see what you get

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## More answers

- DanJS

|dw:1432947141416:dw|

- anonymous

I got a rectangle too! okay thanks

- DanJS

The dotted part is the connected midpoints, that is what the first shape answer is

- anonymous

How would I solve the second one ? @DanJS

- DanJS

The first one is not a rectangle, the dotted shape does not have right angles

- anonymous

hmmm so would it be a trapaoid? or a rhombus?

- DanJS

It is a parallelogram, but that isnt an answer, so i believe this is what it is...
http://en.wikipedia.org/wiki/Kite_%28geometry%29

- DanJS

a rhombus has all 4 sides of equal length, this one here does not

- DanJS

a trapazoid has only one pair of parallel sides, this one has 2 pairs

- DanJS

It is not a rectangle, so it must be the Kite

- anonymous

OHH yhea it would be Kite I see it now

- DanJS

The second prob....
You have to figure the lengths of each side of the triangle...
Equilaterla = all 3 sides the same
isosceles = 2 sides the same
scalene = all sides different
right = 90 degree angle

- DanJS

Did you learn the distance formula between 2 points?

- DanJS

|dw:1432947917511:dw|

- anonymous

Yhea , its a squared + b squared = c squared right ?

- DanJS

\[distance = \sqrt{(5-1)^2+(3-1)^2}\]

- DanJS

Do that for all 3 sides

- anonymous

Ohh , so do I just simplify ?

- DanJS

yes, that is the length of that one side. THen you need to compare that to the length of the other sides..

- DanJS

\[distance2=\sqrt{(0-5)^2+(4-3)^2}\]
\[distance3 = \sqrt{(0-1)^2+(4-1)^2}\]

- anonymous

I got 5 for the fist one
11 for the second one and
and 5 for the third one

- DanJS

In general, the distance between 2 points...
\[D = \sqrt{(y _{2}-y _{1})^2+(x _{2}-x _{1})^2}\]

- DanJS

umm let me calculate and see

- DanJS

\[\sqrt{20}\]
\[\sqrt{26}\]
\[\sqrt{10}\]

- DanJS

all 3 sides different length, scalene triangle

- DanJS

Recall, if it is a right triangle, the 3 sides will satisfy the pythagorean theorem
a^2 + b^2 = c^2
Here it does not, 20 + 10 = 26, NO
Not a right triangle

- anonymous

OHH ok so that therom only works for right triangles .

- DanJS

yes

- DanJS

If not a right triangle, you have to use the law of sines and law of cosines to figure side lengths.

- anonymous

okayy

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