At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
this is the first two
Do you want the answer or how to do it?
so i know how to do it in the future
I'll let the professor do it XD
These are all quadrilaterals. Are they all squares?
Ok good, that cancels out option B, we've determined that all quadrilaterals are NOT necessarily squares.
Are they all rectangles? :)
no but they are all parallelograms?
Yes, all the ones I drew are parallelograms, my mistake :) Lemme draw ANOTHER quadrilateral.
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? :)
yeah so then it would be c ?
Good, yes. Option C: All rectangles are `quadrilaterals` basically is saying this: All rectangles `have 4 sides`.
i see lol
k next one
k sec, reading >.<
|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).
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.
Will any of the other options work >.< hmm thinking
|dw:1450406084898:dw|What would happen if we were to stretch out one of these diagonals so they were the same length?
|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).
I guess option B is what we were looking for :) But hmm that one is tricky.
I shoulda let Cycle help you, Geometry is not really my thing XD lol
yeah thanks tho help me with a few more?
Bahhh I need a Geometry break XD @NeverEndingCycle
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
think of it in terms of venn diagrams |dw:1450408037282:dw|
i already have that one i need the one i just posted
it's a rectangle, so the diagonals are equal KM = LN 6x+16 = 49 x = ??
oh lol 49
i dont know im confused i suck at this
you need to isolate x in `6x+16 = 49`
-16 on both sides?
yep, then divide both sides by 6
cool help me with a few more
I'll help with one more
wait wrong one
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`