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Mathematics 20 Online
OpenStudy (vampirediaries):

Find the lengths of the diagonals of this trapezoid.

OpenStudy (vampirediaries):

OpenStudy (mertsj):

Do you know the distance formula?

OpenStudy (vampirediaries):

d=(x2-x1)^2 + (y2-y1)^2

OpenStudy (mertsj):

Yes. If you take the square root of the right hand side expression that is it.

OpenStudy (mertsj):

Using the points (-b,c) and (a,0) will give you the length of one diagonal.

OpenStudy (mertsj):

Using the points (b,c) and (-a,0) will give you the length of the other diagonal.

OpenStudy (vampirediaries):

Is that all I have to do?

OpenStudy (mertsj):

yes. I would simplify each expression enough to show that the diagonals are congruent.

OpenStudy (vampirediaries):

How would I do that exactly?

OpenStudy (mertsj):

Plug in the first point and post the result.

OpenStudy (vampirediaries):

to the distance formula?

OpenStudy (mertsj):

Yes. Plug the points (-b,c) and (a,0) into the distance formula.

OpenStudy (vampirediaries):

I got (a + b , c) for the first equation and the same for the second time (a + b , c). Is that correct?

OpenStudy (mertsj):

Here is the distance formula: \[d=\sqrt{(x _{2}-x _{1})^2-(y _{2}-y _{1})^2}\]

OpenStudy (mertsj):

Plug the points into that.

OpenStudy (vampirediaries):

I did that and I got ( a+b,c)

OpenStudy (mertsj):

\[d=\sqrt{(a+b)^2+(0-c)^2}\]

OpenStudy (mertsj):

Now, how in the world did you get (a+b,c) out of that?

OpenStudy (vampirediaries):

I simplified from where you left off.

OpenStudy (mertsj):

\[\sqrt{9+16}= ???\]

OpenStudy (mertsj):

What is the answer to that example?

OpenStudy (vampirediaries):

5

OpenStudy (mertsj):

By your logic in the previous example, it would be 3+4

OpenStudy (vampirediaries):

Why would it be that instead of 5 if you were to add 9+16=25. square 25 to get 5. What you did was like extending or whatever is the correct term you just did, to get 3+4

OpenStudy (mertsj):

So if that answer is 5, (which it is) then how can this: \[\sqrt{(x _{2}-x _{1})^2+(y _{2}-y _{1})^2}=(x _{2}-x _{1}+(y _{2}-y _{1})\]

OpenStudy (vampirediaries):

What you just did was take apart the (x2-x1)^2 +(y2-y1)^2 because it was squared

OpenStudy (mertsj):

You are the one who said that: \[\sqrt{(a+b)^2+c^2}=(a+b)+c\] I am just trying to convince you that it is false.

OpenStudy (vampirediaries):

ok. So would (a+b)^2 + c^2 have a 2 in front since its twice of itself?

OpenStudy (mertsj):

\[\sqrt{(a+b)^2+c^2}\]

OpenStudy (mertsj):

That's what it would be.

OpenStudy (vampirediaries):

That's all?

OpenStudy (mertsj):

yes

OpenStudy (vampirediaries):

Then what about the other equation?

OpenStudy (mertsj):

Do the same thing but use the points ((b,c) and (-a,0)

OpenStudy (vampirediaries):

Won't it be the same since it's supposed to be congruent to the other diagonal?

OpenStudy (mertsj):

It's up to you to show that. The problem said to find the lengths of the diagonals. That's plural.

OpenStudy (vampirediaries):

Thanks for your help

OpenStudy (mertsj):

yw

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