Perimeter of the Trapeziod

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Apr 14, 2020
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Given ΔVXY and ΔVWZ, what is the perimeter of the trapezoid WXYZ? Round your answer to the nearest hundredth.
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Hello, I am having so much trouble with this problem. I have tried looking at videos but can not find one to help me so far.
So far, I split up the two triangles and found the scale factor of VWZ to VXY is 4/1 (I divided 20 by 80 to get 0.25)
I know VZ = 30, but that is it. Am I on the right path? Is VY ≅ YX? Please help!

Thank you.
 
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Where did you get 20 from? I don't see any such thing, and your scale factor is wrong.

The hardest part of the problem, perhaps, is that there are no explicit vertical dimensions. (No, you can't assume that VY ≅ YX .)

To help find WZ and XY, I would draw in a line through W parallel to VY, which will give you another right triangle to work with.
 
Given ΔVXY and ΔVWZ, what is the perimeter of the trapezoid WXYZ? Round your answer to the nearest hundredth.
View attachment 17925

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Hello, I am having so much trouble with this problem. I have tried looking at videos but can not find one to help me so far.
So far, I split up the two triangles and found the scale factor of VWZ to VXY is 4/1 (I divided 20 by 80 to get 0.25)
I know VZ = 30, but that is it. Am I on the right path? Is VY ≅ YX? Please help!

Thank you.
1586901283176.png

Consider this:

In triangle XYV, we have WZ || XY and VZ = 30 and WV = m, then

(m+62.5)/m = 80/30 \(\displaystyle \to \ \ \ \) m = ?
 
OP, look at a proportion. Let the length of the hypotenuse VX be called c.

c is to 62.5 as 80 is to 50.

\(\displaystyle \dfrac{c}{62.5} \ = \ \dfrac{80}{50}\)
 
Okay, I see my mistake. I accidentally thought 80 - 50 was 20 (even though I acknowledged it was 30 the sentence before that), so when I divided it it became 20 ÷ 80 = 0.25, and not 30 ÷ 80 = 0.375

Also you said to draw in a line through W parallel to VY, but wouldn't that just be WZ?
 
Okay, I see my mistake. I accidentally thought 80 - 50 was 20 (even though I acknowledged it was 30 the sentence before that), so when I divided it it became 20 ÷ 80 = 0.25, and not 30 ÷ 80 = 0.375
Good. I could tell you knew better than you wrote.

Also you said to draw in a line through W parallel to VY, but wouldn't that just be WZ?
No, WZ is parallel to XY, not to VY.
 
OP, look at a proportion. Let the length of the hypotenuse VX be called c.

c is to 62.5 as 80 is to 50.

\(\displaystyle \dfrac{c}{62.5} \ = \ \dfrac{80}{50}\)

So if I take:
c/62.5 = 80/50

and then simplify the numbers and get rid of the decimal I would end up with the complex fraction:
c/125/2 = 8/5

After simplifying the complex fraction I would get:
2c/125 = 8/5

And after cross multiplication:
10c = 1000, which would simplify down to:

c = 100


What would the next step be? Would I have to use the Pythagorean Theorem to find XY and WZ?

If so, I would start with solving YX.
80^2 + b^2 = 100^2 (^2 meaning squared)

b^2 would become 3600, simplified to b = 60

Now I would solve for WZ.
30^2 + b^2 = 37.5^2

b^ would become 506.25 simplified to b = 22.5

Now that I have all of the sides I just add them up to find the perimeter?

22.5 + 60 + 50 + 62.5 = 195

I am pretty sure this is the correct answer, please correct me if I am wrong or did something I wasn't supposed to.

Thank you.



 
So if I take:
c/62.5 = 80/50

and then simplify the numbers and get rid of the decimal I would end up with the complex fraction:
c/125/2 = 8/5

After simplifying the complex fraction I would get:
2c/125 = 8/5

And after cross multiplication:
10c = 1000, which would simplify down to:

c = 100

What would the next step be? Would I have to use the Pythagorean Theorem to find XY and WZ?

If so, I would start with solving YX.
80^2 + b^2 = 100^2 (^2 meaning squared)

b^2 would become 3600, simplified to b = 60

Now I would solve for WZ.
30^2 + b^2 = 37.5^2

b^ would become 506.25 simplified to b = 22.5

Now that I have all of the sides I just add them up to find the perimeter?

22.5 + 60 + 50 + 62.5 = 195

I am pretty sure this is the correct answer, please correct me if I am wrong or did something I wasn't supposed to.

Thank you.
So you really are good at math after all !!
 
So you really good at math after all !!
Wow thank you for the help! And thank you all for your help pointing me in the right direction. My main problem was that I didn't know where to start the problem, but once I saw the formula, it all kind of fell into place. I definitely couldn't have done this without you all, and I appreciate the pointers.

Thanks again!
 
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