Quadrilaterals

[vampirewitchreine]

I'm not quite sure which is the correct answer.... because I'm dealing with a square root squared, I have 2 different answers 2 different ways.



Square WXYZ



a. WY
b. XY


a.:
I found WY as \(\displaystyle 5\sqrt{2} + 5\sqrt{2}\)
Then\(\displaystyle WY~\approx 7.07 + 7.07\)
Ending with \(\displaystyle WY~\approx 14.14\)

b.:
I found XY in 2 ways, and came up with 2 very different answers. What I'd like to know, is which answer would be correct?
\(\displaystyle XY^2 = XV^2 = YV^2\)
\(\displaystyle XY^2 = 5\sqrt {2}^2 + 5\sqrt {2}^2\) OR \(\displaystyle XY^2 = (5\sqrt {2})^2 + (5\sqrt {2})^2\)
\(\displaystyle XY^2 = 10 + 10\) OR \(\displaystyle XY^2 = 50 + 50\)
\(\displaystyle \sqrt {XY^2} = \sqrt{20}\) OR \(\displaystyle \sqrt {XY^2} = \sqrt {100}\)
\(\displaystyle XY ~\approx 4.47\) OR \(\displaystyle XY = 10 \)



(My feeling is the second one, but I'd like to get your input as well)

 

[Mrspi]

I'm not quite sure which is the correct answer.... because I'm dealing with a square root squared, I have 2 different answers 2 different ways.



Square WXYZ
View attachment 1571


a. WY
b. XY


a.:
I found WY as \(\displaystyle 5\sqrt{2} + 5\sqrt{2}\)
Then\(\displaystyle WY~\approx 7.07 + 7.07\)
Ending with \(\displaystyle WY~\approx 14.14\)

b.:
I found XY in 2 ways, and came up with 2 very different answers. What I'd like to know, is which answer would be correct?
\(\displaystyle XY^2 = XV^2 = YV^2\)
\(\displaystyle XY^2 = 5\sqrt {2}^2 + 5\sqrt {2}^2\) OR \(\displaystyle XY^2 = (5\sqrt {2})^2 + (5\sqrt {2})^2\)
\(\displaystyle XY^2 = 10 + 10\) OR \(\displaystyle XY^2 = 50 + 50\)
\(\displaystyle \sqrt {XY^2} = \sqrt{20}\) OR \(\displaystyle \sqrt {XY^2} = \sqrt {100}\)
\(\displaystyle XY ~\approx 4.47\) OR \(\displaystyle XY = 10 \)



(My feeling is the second one, but I'd like to get your input as well)

I had hoped by now you had learned to use the special relationships between the legs and the hypotenuse of an isosceles right triangle to save yourself a bunch of work.

In an isosceles right triangle with each of the two legs equal to "a", the hypotenuse will ALWAYS be

\(\displaystyle a \sqrt {2}\)

XY is the hypotenuse of an isosceles right triangle with legs of \(\displaystyle 5 \sqrt {2}\)

So XY =

\(\displaystyle 5 \sqrt {2} * \sqrt {2}\)

or, 5*2, or 10

 

[vampirewitchreine]

I had hoped by now you had learned to use the special relationships between the legs and the hypotenuse of an isosceles right triangle to save yourself a bunch of work.

In an isosceles right triangle with each of the two legs equal to "a", the hypotenuse will ALWAYS be

\(\displaystyle a \sqrt {2}\)

XY is the hypotenuse of an isosceles right triangle with legs of \(\displaystyle 5 \sqrt {2}\)

So XY =

\(\displaystyle 5 \sqrt {2} * \sqrt {2}\)

or, 5*2, or 10

Okay, so that means that my first answer (4.47) is correct, yes?

 

[Mrspi]

Okay, so that means that my first answer (4.47) is correct, yes?

HUH? Please re-read my response. XY = 10!

 

[vampirewitchreine]

HUH? Please re-read my response. XY = 10!

My book tells me that I need to solve by Pythagorean theorem (this is in a section before special right triangles, something that I should have clarified first). They showed an example with a square, but it was using whole numbers and not a number times a square root. I was only trying to make sure that I didn't have to put the \(\displaystyle 5\sqrt {2}\) in parentheses before squaring it to solve.

 

[vampirewitchreine]

I'm not quite sure which is the correct answer.... because I'm dealing with a square root squared, I have 2 different answers 2 different ways.



Square WXYZ
Attachment 1571


a. WY
b. XY


a.:
I found WY as

This is fine
Then

Does the problem say to approximate? The exact answer is

Ending with

b.:
I found XY in 2 ways, and came up with 2 very different answers. What I'd like to know, is which answer would be correct?

This is wrong.(I forgot to shift on the +/= button and left the problem as \(\displaystyle XY^2 = XV^2 = YV^2\) not \(\displaystyle XY^2 = XV^2 + YV^2\)... it was a typing error. I do have this written correctly on my paper at least)

This is the Pythogorean part of the problem
In the line below, which does not follow from the incorrect line above, you are now getting confused by notation. XY is NOT X times Y. It is the length of the line from point X to point Y. At least in the old days you were supposed to put a bar over the XY to indicate that it was a single number representing a length. (I normally would, but i don't know how to do this with the LaTex)

This is wrong because you are treating Y as a number when it is XY that is the number.
OR

This is correct

This is wrong as explained above OR

This is correct

wrong OR

correct

wrong OR

correct



(My feeling is the second one, but I'd like to get your input as well)

What is important is whether you now understand why the one line of thought is incorrect and the other correct

My problem only says to find the length (though another answer that is in the back of the book has a square root and the approximate to the first decimal place).

I defiantly understand now.

Thank you for taking the time to explain this to me.... It's always some of the most simple things that stump me, but I rarely have difficulties with more complex problems.

 

[vampirewitchreine]

Like mother-in-law ? :razz:

Haha, I'm still in high school, so I'll let you know what happens once I cross that bridge :lol:

 

[vampirewitchreine]

anagram of mother in law: Hitler woman

Haha, I remember seeing that in the Math Odds & Ends.

Sheesh Denis

The poor girl will be scared away from growing up.

Nah, it'd take more than that to scare me..... I watch enough horror movies that I've got a "Take life by the horns" attitude. Besides, growing up is a part of life and defiantly not something that you can prevent from creeping up on you. (Besides, according to my dad, once I become old enough, I miss my childhood and just start acting like a kid again.... so no worries there xD)

Well, my impression of Hazel: she ain't scared of nothing!
She's already showing no mercy for Pythagoras!

Aww, that's not true. Everyone's afraid of something. Otherwise we're either an empty shell or we're robots.
(And Pythagorean Theorem is one of the easy things that I really get.)