Question

I have a trapezoid and I need to find bases of it (AD ir BC). Look at the picture below

Answer 1

it is all that I have. Can anyone help me/

Answer 2

If you're looking for a numerical solution, you need a specified dimension or 2 of something...

Answer 3

the answer is 6 and 10, by the way

Answer 4

Are you looking to see how to get to those numbers?

Answer 5

Where did you get those numbers from?

Answer 6

Well, I have the answers. But I dont know how to get them.

Answer 7

MN is midline.
AC is diagonal
thats all that I have

Answer 8

@UnkleRhaukus , I am sorry, can you help me?

Answer 9

Exactly.

You can't get 6 or 10 from the data you gave.

You can't conjure any dimensions from 6 undimensioned lines. At best, you can illustrate ratios, but no values.

OTOH, if 6 and 10 ARE the answers, then there HAS to be dimensions laying around somewhere whether you have them or not.

Answer 10

@qweqwe123123123123111 ummm... "You can't get 6 or 10 from the data you gave." are you saying on the face of it that is not evident, or that you see no path to the answer? Because the data she gave does have a path to it....

Answer 11

- "are you saying on the face of it that is not evident, or that you see no path to the answer?"

Either or...

If you see a path, I'd be interested in seeing how you do it.

Answer 12

@e.mccormick can you explain this to me?

Answer 13

Well, they tell you one of the keys. it is from a properlt of reapazoids, that midpoint conecotre is the average of the two basis. So the total length of the basis is 16.

Answer 14

yes, I get it

Answer 15

Wut?

Answer 16

Then they tell you the total length of that line, and a way to find the ratio of each side.

Answer 17

The ratio of X to Y is also the ratio of the top to the bottom.

Answer 18

What, you mean (ad+bc)/2 ?

That just gives you the ratio of the line that cuts through the center to the top and bottom bases.

Answer 19

They are related ratios. Has to do with the triangles they make.

Answer 20

Of course they're related.

But that still doesn't determine the top and bottom bases to be 6 and 10. It also works if they're 3 and 5.

Answer 21

\[\frac{AD+BC}{2}=8\implies AD+BC=16\]\[MN=x+y=8; \quad x-y=2\implies \\
x=2+y\implies 2+y+y=8\implies y=3\implies x=5\]\(3:5\) ratio of the sides implies a 3 ot 5 spilt of the total, or 16. \(3+5=8\); \(16/8=2\); \(2\cdot (3:5)=(6:10)\)

Answer 22

You have (ad+bc)/2 = 8

8 is Kamille's number, not a given.

Answer 23

it is given number

Answer 24

No, she listed all that as given.

Answer 25

"it is all that I have." AKA: Given.

Answer 26

thank you a lot, @e.mccormick

Answer 27

@qweqwe123123123123111
If you were wondering what triangles I meant:

In this trapezoid, EG is the midline or median.

\(\triangle ACD\sim \triangle FCG\) and \(\triangle ACB\sim \triangle AFE\) \(\therefore\) BC has a ratio to EF and CG has a ratio to AD. Also, because it is a trapezoid, EG=(BC+AD)/2. This means those ratios are related. The ratios of the sides has a direct correlation to the ratios of the midline as bisected by a diagonal.

I looked for that to link, but did not see it, so I decided to finally just draw it.

Answer 28

Oops. FG has a ratio to AD. Small typo there. But I think you should get it.

Answer 29

AH: I see where the confusion began. When she restated her conditions: Kamille, "MN is midline. AC is diagonal thats all that I have"

Yah, that would take the 8 out of the equation.

Answer 30

I was working off the original statement and the picture that impled those were all given. Missed her taking the 8 away. I can understand why anyone would think there was only an undefined ratio at that point. Good thing the 8 really was given!

Answer 31

Yes, the ratios are all obvious, and is what I said was all that could be ascertained from the data he gave.

As far as I could tell, the entire problem he was given consisted of 4 lines forming a trapezoid, a line through the center of the trapezoid parallel to the bases, a transverse line corner to corner, and a line below that showing AD+BC text above the line and a "2" below it.

This, and ONLY this, is what I considered "given" data.

I could have considered his penciled-in numbers and references as valid data except that x-y=2 could not be true because the ratio of x:y is identical to the ratio of the top-line:bottom-line. In fact, x-y=-2, although y-x=2 (on your diagram "x" is EF, and "y" is FG)

While this may all sound trivial, it was enough to cause me to think that ALL the penciled data may have been assumed or in error, so I ignored them.; especially since when I told him that he needed a dimension or 2, he didn't tell me that the data he penciled in WERE actually dimensions/data that he was "given".

In any event, my error was in ignoring his data, not in my perception of the problem.