Expert Guidance for Your Pythagorean Theorem Questions

Algebra Logical Pythagorean theorem help
A wire is attached to the top of a pole. The pole is 2 feet shorter than the wire, and the distance from the wire on the ground to the bottom of the pole is 9 feet less than the length of the wire. Find the length of the wire and the height of the pole.
Hint: Use the pythagorean theorem then set up a quadratic equation equal to zero and solve. I got c=17 and c=5 but I don't know what the length of the wire is or the height of the pole.

Vector addition and Pythagorean theorem
Finding length or magnitude using vector addition and the Pythagorean theorem.
I am trying to understand why vector addition and the Pythagorean theorem are giving different results?

Vector Addition :
According to diagram (A) : $\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}=\stackrel{\to <!--\; \to \; -->}{c}$now suppose : magnitude of $\stackrel{\to <!--\; \to \; -->}{a}=3$, magnitude of $\stackrel{\to <!--\; \to \; -->}{b}=4$ then
$\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{c}\Vert <!--\; \Vert \; -->=\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{a}\Vert <!--\; \Vert \; -->+\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{b}\Vert <!--\; \Vert \; -->$
$\Vert <!--\; \Vert \; -->\stackrel{\to <!--\; \to \; -->}{c}\Vert <!--\; \Vert \; -->=3+4$
magnitude of c = 7
Pythagorean Theorem
Now when we consider this as a triangle shown in diagram (B)
Similarly, length of a=3, length of b=4
so according to Pythagorean theorem
${\text{hypotenuse}}^2={\text{opposite}}^2+{\text{adjacent}}^2$
$c^2=a^2+b^2$i.e.
$c^2=3^2+4^2$
$c=\sqrt{9+16}=\sqrt{25}$
length of c = 5
Why there is inconsistency in the results? I am doing something wrong?
Thanks!

How does the Pythagorean theorem describe a circle?
The Pythagorean theorem states, for a right triangle with legs a,b and hypotenuse c,
$a^2+b^2=c^2$
By replacing c with r, radius this equation becomes the equation of circle at centre (0,0).
How does Pythagoras' equation end up describing the circle?

Deducing the Pythagorean Theorem from a particular dissection

How can I deduce the pythagorean theorem from the follow image?
I have been draw some parallels and I got the figure but I don't know how to deduct, some hint?

Where does the Pythagorean theorem "fit" within modern mathematics?
I am interested in how today's professional mathematicians view the Pythagorean theorem, in terms of how the theorem fits within the axiomatic framework of mathematics. I often come across textbooks that define length by the Pythagorean theorem, so that the theorem is in essence a definition or axiom. In more modern mathematics such as linear algebra, is the Pythagorean theorem generally just used as the definition of length? Is it more conventional today to treat the Pythagorean theorem as a definition (or axiom) rather than a theorem? Are there any modern proofs of the Pythagorean theorem that don't rely on Euclidean geometry (like a proof that utilizes linear algebra/the dot product, etc.)?

How is the Pythagorean Theorem related to the Equation of a Circle
My teacher wants me to figure out how the Pythagorean Theorem and the Equation of a Circle are related. I can't figure this out because I view them as being two completely different things. I understand what the equation for a circle is, but how does that relate to the Pythagorean Theorem?

Why is the Pythagorean Theorem not the Pythagorean Law?
Why do we call the Pythagorean Theorem a theorem, not a law?
As far as I know, we call a theorem a theorem because though it's reliable in every observable case, its truthfulness cannot be proven for every case. However I've looked and it seems as though we (as the human race) have very extensive proofs of the Pythagorean Theorem, considering every case. Why is it that it hasn't made the transition from theorem to law? Is my terminology confused? Is it not as proven as I think it is? Is there some other condition for being a law that I don't know about?

Euclidean norm vs pythagorean theorem?
I recently stumbled upon an Euclidian norm. First I thought there are the powers and square root to deal with possible negative values (like in Standard deviation formula) but then I realized, the final number (sum of squares and square root of it) is not same as sum of absolute numbers.
So then I noticed it is more related to Pythagorean theorem (Euclidian distance).
But if looked in google for PT in 3d space i found the formula like
$(a^2+b^2+c^2{)}^{\frac{1}{2}}$
But in the Euclidian norm it is
$SQRT(a^2+b^2+c^2)$
How comes, it is different?
Second question, how comes that the theorem works for any count of dimensions?

Pythagorean theorem
We can make a square into four equal squares. Fine, if we want to make into five.. Then there is a problem. Please discuss, How to make five squares from a single square by using a Pythagorean theorem. Is there any other way to make five squares from one square without using Pythagorean theorem? Please discuss. Thanking you, KKRG

Pythagorean Theorem intuition
I actually have a degree in Pure Mathematics, and this has always bugged me. Like a tiny, hardly-noticeable stone in your shoe you just can't get out.
Despite being at the core of geometry, arithmetic, and infinity, despite being able to prove it a dozen different ways, and despite it being the first "formula" anyone learns and one of the oldest known mathematical ideas - I can only see shadows and silhouettes of its truth.
Looked at from the point of view of geometry, I can see the statement plainly as a statement regarding the intrinsic notions of angle and area (using the proof of similar right-angle triangles) - however, the last step requires a correspondence between the notions of length and area to make complete (namely, the area of a shape increases quadratically as its scale increases linearly). This has always bothered me, since, the Theorem is an idea solely between angle and length.
There are, of course, simple proofs that don't depend on the notion of area (ex: Pythagorean Theorem Proof Without Words 6), however, sadly, they don't further my intuition, despite being clever/cute.
Staring at the statement blank in the face, I understand it far less intuitively then I do half of the ideas in algebra, analysis, number theory, etc.

Pythagorean Theorem for imaginary numbers
If we let one leg be real-valued and the other leg equal bi then the Pythagorean Theorem changes to $a^2-<!--\; -\; -->b^2=c^2$ which results in some kooky numbers.
For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary?

Approximation using pythagorean theorem
Im looking at the following diagram,

Here, I am interested in a relation between a,s,c, assuming that $r>>1$ my first thought is to relate then using the pythagorean theorem,
$c^2=a^2+s^2$
Is making sure an approximation a good approach and is there an alternate way to relate my variables? Thank you in advance!

Pythagorean theorem and orthocenter
Point H is the orthocenter of $\mathrm{△<!--\; △\; -->}ABC$. Let AD and CM be the altitudes. Show that $B{M}^2-<!--\; -\; -->B{D}^2=H{D}^2-<!--\; -\; -->H{M}^2$.
By Pythagorean theorem: 1) △$\mathrm{△<!--\; △\; -->}BCM\to <!--\; \to \; -->B{M}^2=B{C}^2-<!--\; -\; -->C{M}^2$ and 2) $\mathrm{△<!--\; △\; -->}ABD\to <!--\; \to \; -->B{D}^2-<!--\; -\; -->A{B}^2-<!--\; -\; -->A{D}^2$ How to approach the problem further?

Pythagorean Theorem - Incomplete proof
The image below is supposed to prove Pythagorean Theorem (without words). However, I see that it is assuming that the square of side c can be inscribed in the square of side a+b and would result in the 4 triangles of equal area. This fact is only assumed, so would the proof be considered incomplete?

Pythagorean theorem
The Pythagorean theorem is typically written as
$a^2+b^2=c^2$
where a and b are the two shorter sides (legs) of a right triangle and c is its longest side (the hypotenuse).
Given the exponent powers $c^{2\times <!--\; \times \; -->1/2}=c$, why cant I express "$a=c-<!--\; -\; -->b$".
The correct answer given was $\sqrt{c^2-<!--\; -\; -->b^2}$

Can the Pythagorean Theorem be extended like this?
What is the significance, if any, of the fact that
$3^3$ + $4^3$ + $5^3$ = $6^3$
?
How curious this is. This would be like the Pythagorean Theorem exploding into a higher dimension, on steroids.

Strange proof of Schwarz Inequality with Pythagorean Theorem
Does anyone know what is going on in this proof of the Schwarz inequality? Most importantly: how can one assume that $c^2\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$, or later on, that $c^2\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$? This would imply that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$, or in the latter case $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$, could also be equal to zero (seeing as it's a smaller-than-or-equal-to sign). But since we're using the Pythagorean theorem to arrive at these relations, $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$ or $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$ could not possibly be zero, as then one of the sides of our 'triangle' would be zero, and we wouldn't be able to use the Pythagorean theorem to justify this conclusion. So why isn't it just a '<'-sign? As I said, I don't see how vector A−cE or A−cB can have length zero if we're using the Pythagorean theorem, that is, if we're presupposing the existence of a triangle. If either A−cE or A−cB equals zero, there wouldn't be a triangle anymore, and we wouldn't be able to apply the Pythagorean theorem.
Edit: fixed link.
Edit2: I must add that the author maintains an unusual definition for the scalar component; instead of the conventional $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$, i.e., $\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}$, the author defines $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\mathbf{B}\cdot <!--\; \cdot \; -->\mathbf{B}}$ to be the scalar component, i.e., $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$. In other words, the author divides the 'conventional' scalar component by the the norm of B, thus making 'his' scalar component a number which is in fact the proportion of the 'regular' scalar component to the entire length of the vector on which this component is projected.
Edit3: Since people don't seem to understand my confusion, let me phrase my question as explicitely as possible. My question is: why is the $\ge <!--\; \ge \; -->$ sign used? Why not just ">"? In what situation could $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$ (the $\ge <!--\; \ge \; -->$ sign means greater than OR EQUAL TO, so in what situation could the left hand side ever equal the right hand side?) The problem is: since we're using the Pythagorean theorem for our proof, we're presupposing the existence of a triangle (or otherwise we wouldn't be able to use the Pythagorean theorem), and as such, $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ can never equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$, because this would mean that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->tE{\Vert <!--\; \Vert \; -->}^2$ is equal to zero, and that one of the sides of our triangle has length zero. Then we wouldn't have a triangle, and we wouldn't be justified in using the Pythagorean theorem in our particular situation. So let me ask it again: why is the $\ge <!--\; \ge \; -->$ sign used, instead of the ">"-sign?

Extended Pythagorean Theorem
Extended Pythagorean Theorem We all well familiar with the basic Pythagorean statement – I will use a different description that will serve the discussion that follows – Given two points in 2D space, (a,0) and (0,b) with distance d between them, than the relation between a,b, and d follows
$a^2+b^2=d^2$
In 3D this is extended to three points (a,0,0), (0,b,0) and (0,0,c) with an area of s for the triangle created by the three points - than the following equation holds:
${}^1{/}_4(a^2+b^2+c^2)=s^2$
I wonder if this is true for higher dimensions, as example 4D - given four points (a,0,0,0), (0,b,0,0), (0,0,c,0), (0,0,0,d) and the volume for the pyramid created by the four point to be v than
${}^1{/}_9(a^2+b^2+c^2+d^2)=v^2$

Deduce the Pythagorean Theorem [duplicate]
Let V be an inner product space,and suppose that x and y are orthogonal vectors in V.We also know ${\Vert x+y\Vert }^2={\Vert x\Vert }^2+{\Vert y\Vert }^2$.My question is how can we deduce the pythagorean theorem in ${\mathbb{R}}^2$ from it.If possible give geometrical views.
Any help will be greatly appreciated.
thanks!! in advance.

Pythagorean theorem in functional analysis
Prove the Pythagorean theorem and its converse in R : f is orthogonal to g if and only if
$\Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2$
LHS -> RHS
$\begin{array}{rl}& \Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2\\ =& (f,f)-<!--\; -\; -->2(f,g)+(g,g)\\ =& \Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2-<!--\; -\; -->2(f,g)+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2\end{array}$
In order for LHS=RHS −2(f,g) has to be 0 which means that f and g are othogonal. If −2(f,g)=0 Then
$=\Vert <!--\; \Vert \; -->f\Vert <!--\; \Vert \; -->+\Vert <!--\; \Vert \; -->g\Vert <!--\; \Vert \; -->$
RHS-> LHS
$\begin{array}{rl}& \Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2\\ =& (f,f)+(g,g)\\ =& \Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2\text{ iff }-<!--\; -\; -->2(f,g)=0,\text{ iff }f\text{ and }g\text{ are orthogonal.}\end{array}$