Geometry: Squares

[vampirewitchreine]

In square WXYZ, find the measure of angle WVX, WX and ZV.




I've got WX as 12 (since it's a square and all sides are congruent)
The measure of angle WVX is 90 [Since the diagonals make the corners all 45 when divided (180-45-45=90).... or the simplified version of the diagonals for perpendicular bisectors]

What I need help with is how to get ZV since ▲WVZ is a right triangle, i've got 2 variables instead of one to work with (Since I would need to know one of the other halves of the diagonals if I use the pre-given information)


Could some one help me to figure out how to solve?

 

[wjm11]

▲WVZ is an *isosceles* right triangle, so both legs are the same length. a^2 + b^2 = c^2 becomes 2a^2 = c^2.

 

[vampirewitchreine]

Okay, so should it work out like this:

2a²=12²
\(\displaystyle \sqrt{2a^​​{​^2}}\)\(\displaystyle =\sqrt{144}\)
\(\displaystyle 2a=12\)
\(\displaystyle a=6\)





(The text coding is having some issues for some reason. But the one should be the square root of 2a²)

 

[Mrspi]

Okay, so should it work out like this:

2a²=12²
\(\displaystyle \sqrt{2a^​​{​^2}}\)\(\displaystyle =\sqrt{144}\)
\(\displaystyle 2a=12\)
\(\displaystyle a=6\)





(The text coding is having some issues for some reason. But the one should be the square root of 2a²)

Well, let's see....

You have 2a² = 12²

Then apparently you took the square root of both sides...unfortunately, sqrt(2a²) is NOT 2a.

I would have done this:

2a² = 144
Divide both sides by 2, and you have
a² = 72

NOW take the square root of both sides:

sqrt(a²) = sqrt(72)

a = sqrt(72)

Simplify sqrt(72).

You are, I believe, expected to know that there is a very special relationship between the lengths of a leg and the hypotenuse in an isosceles right triangle....MEMORIZING that relationship will save you a TON of "grunt work" in your future if you continue to take math classes.

 

[vampirewitchreine]

C'mon Hazel: use the pythagorean theorem as Wjm told you; diagonal = SQRT(12^2 + 12^2).
That's VERY basic stuff.

Okay, so I made it a little more complicated than it should have been.

Well, let's see....

You have 2a² = 12²

Then apparently you took the square root of both sides...unfortunately, sqrt(2a²) is NOT 2a.

I would have done this:

2a² = 144
Divide both sides by 2, and you have
a² = 72

NOW take the square root of both sides:

sqrt(a²) = sqrt(72)

a = sqrt(72)

Simplify sqrt(72).

You are, I believe, expected to know that there is a very special relationship between the lengths of a leg and the hypotenuse in an isosceles right triangle....MEMORIZING that relationship will save you a TON of "grunt work" in your future if you continue to take math classes.

And thanks for the correction of my answer.


I was having the complication of thinking that I needed to use the isosceles right triangle of WVZ instead of using the big triangle of XYZ and then dividing whatever answer that I get for that hypotenuse by half, giving me the same answer that I would get by using the smaller triangle. I only thought of doing this after the post by Mrspi because it didn't click when Denis posted it previously. Both ways did give me the same final product at least.

\(\displaystyle \sqrt{72}~\approx 8.49\) or \(\displaystyle \sqrt{288}~\approx 16.97/2 ~\approx 8.49\)