i need help with a lot of questions i literally have no clue what to do

- edbf123

i need help with a lot of questions i literally have no clue what to do

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- edbf123

##### 1 Attachment

- edbf123

this is the first two

- NeverEndingCycle

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## More answers

- edbf123

both

- edbf123

so i know how to do it in the future

- NeverEndingCycle

I'll let the professor do it XD

- zepdrix

|dw:1450405325789:dw|

- zepdrix

These are all quadrilaterals.
Are they all squares?

- edbf123

no

- zepdrix

Ok good, that cancels out option B,
we've determined that all quadrilaterals are NOT necessarily squares.

- zepdrix

Are they all rectangles? :)

- edbf123

no but they are all parallelograms?

- zepdrix

Yes, all the ones I drew are parallelograms, my mistake :)
Lemme draw ANOTHER quadrilateral.

- zepdrix

|dw:1450405485531:dw|

- zepdrix

That new shape is still a quadrilateral (4-sided figure).
But it's not a parallelogram.
So if it's a quadrilateral, it's not necessarily a parallelogram, ya? :)

- edbf123

yeah so then it would be c ?

- zepdrix

Good, yes.
Option C:
All rectangles are `quadrilaterals`
basically is saying this:
All rectangles `have 4 sides`.

- edbf123

i see lol

- edbf123

k next one

- zepdrix

k sec, reading >.<

- zepdrix

|dw:1450405799077:dw|Here is an example of a specific quadrilateral, a Rhombus.
All the side lengths are equal,
and the diagonals bisect one another (intersect at the middle of their length).

- zepdrix

So this shows us that option A is not sufficient.
We have the side lengths equal, we have diagonals bisected,
but we didn't end up with a square.

- zepdrix

Will any of the other options work >.< hmm thinking

- edbf123

c maybe

- zepdrix

|dw:1450406084898:dw|What would happen if we were to stretch out one of these diagonals so they were the same length?

- zepdrix

|dw:1450406156699:dw|I drew it a little sloppy lol
But we should end up with a square, ya?
The diagonals being equal length,
and all other sides being equal length forces the diagonals to bisect perpendicularly (at right angles).

- zepdrix

I guess option B is what we were looking for :)
But hmm that one is tricky.

- zepdrix

I shoulda let Cycle help you,
Geometry is not really my thing XD lol

- edbf123

yeah thanks tho help me with a few more?

- zepdrix

Bahhh I need a Geometry break XD
@NeverEndingCycle

- edbf123

k thanks

- edbf123

lol

- edbf123

##### 1 Attachment

- jim_thompson5910

If a quadrilateral has 4 equal sides, then it is a rhombus
If a quadrilateral is a parallelogram with equal diagonals, then it is a rectangle
if a figure is a rhombus and a rectangle, then it must be a square

- jim_thompson5910

think of it in terms of venn diagrams
|dw:1450408037282:dw|

- jim_thompson5910

|dw:1450408060867:dw|

- edbf123

i already have that one i need the one i just posted

- jim_thompson5910

it's a rectangle, so the diagonals are equal
KM = LN
6x+16 = 49
x = ??

- edbf123

oh lol 49

- edbf123

right?

- jim_thompson5910

no

- edbf123

i dont know im confused i suck at this

- jim_thompson5910

you need to isolate x in `6x+16 = 49`

- edbf123

-16 on both sides?

- jim_thompson5910

yep, then divide both sides by 6

- edbf123

so 5.5?

- jim_thompson5910

correct

- edbf123

cool help me with a few more

- jim_thompson5910

I'll help with one more

- edbf123

##### 1 Attachment

- edbf123

wait wrong one

- edbf123

##### 1 Attachment

- jim_thompson5910

since the two horizontal sides are parallel, we know that the same side interior angles are supplementary
that means
`a+36 = 180`
and
`113+b = 180`

- edbf123

thank you

- jim_thompson5910

no problem

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