Find the lengths of the diagonals of this trapezoid.
Do you know the distance formula?
d=(x2-x1)^2 + (y2-y1)^2
Yes. If you take the square root of the right hand side expression that is it.
Using the points (-b,c) and (a,0) will give you the length of one diagonal.
Using the points (b,c) and (-a,0) will give you the length of the other diagonal.
Is that all I have to do?
yes. I would simplify each expression enough to show that the diagonals are congruent.
How would I do that exactly?
Plug in the first point and post the result.
to the distance formula?
Yes. Plug the points (-b,c) and (a,0) into the distance formula.
I got (a + b , c) for the first equation and the same for the second time (a + b , c). Is that correct?
Here is the distance formula: \[d=\sqrt{(x _{2}-x _{1})^2-(y _{2}-y _{1})^2}\]
Plug the points into that.
I did that and I got ( a+b,c)
\[d=\sqrt{(a+b)^2+(0-c)^2}\]
Now, how in the world did you get (a+b,c) out of that?
I simplified from where you left off.
\[\sqrt{9+16}= ???\]
What is the answer to that example?
5
By your logic in the previous example, it would be 3+4
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
So if that answer is 5, (which it is) then how can this: \[\sqrt{(x _{2}-x _{1})^2+(y _{2}-y _{1})^2}=(x _{2}-x _{1}+(y _{2}-y _{1})\]
What you just did was take apart the (x2-x1)^2 +(y2-y1)^2 because it was squared
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.
ok. So would (a+b)^2 + c^2 have a 2 in front since its twice of itself?
\[\sqrt{(a+b)^2+c^2}\]
That's what it would be.
That's all?
yes
Then what about the other equation?
Do the same thing but use the points ((b,c) and (-a,0)
Won't it be the same since it's supposed to be congruent to the other diagonal?
It's up to you to show that. The problem said to find the lengths of the diagonals. That's plural.
Thanks for your help
yw
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