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why the diagonal of a square forms a 45 degree angle with each sides?

Guest Jun 9, 2017
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 #1
avatar+1224 
+2

I think I have an explanation for you that may help you. I have a picture that should help you visualize this phenomenon
 

Source: https://vt-s3-files.s3.amazonaws.com/uploads/problem_question_image/image/1471/square_diagonal.jpg

 

A square is defined as an equiangular and equilateral quadrilateral. This is complicated language. Let's break this down, so it is easier to understand. 

 

"Equiangular" means that all angle measures are congruent and have the same measure. Of course, in a square the measure of all the angles is \(90^{\circ}\), so a square satisfies this condition

 

"Equilateral" means that all sides are congruent and have the same length. This is also true in a square.

 

"Quadrilateral" means a four-sided figure. This fits the description of a square.

 

In the diagram above, let's find the length of \(d\), the length of the diagonal of a square when the side lengths have length \(a\). We'll use Pythagorean's Theorem for the length. \(\angle B\) is a right angle, so the hypotenuse is d, and the legs are a. Let's do it: 

 

\(a^2+a^2=d^2\)Both side lengths are equivalent in a square. Let's solve for by combining both a^2.
\(2a^2=d^2\)Take the square root of both sides to isolate d.
\(\sqrt{2a^2}=d\) 
  

 

Ok, the distance of diagonal of a square is \(\sqrt{2a^2}\). We can actually simplify this further. Let's do that:

 

\(\sqrt{2a^2}\)First, I'm going to use a radical rule stating \(\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\hspace{3mm},a\geq0\hspace{1mm},b\geq0\). Note that this radical rule can only be used if and only if both a and b are nonnegative. Therefore, this radical rule is not universal. However, we can assume that a is nonegative because of the context of geometry; it is nonsensical for a to have a side length of -8, for example. 
\(\sqrt{2}\sqrt{a^2}\)I'll use another radical rule stating that \(\sqrt[n]{a^n}\hspace{3mm},a\geq0\). This, too, can only be used for nonnegative values. Note that this rule make the \(\sqrt{a^2}\) turn into something much prettier.
\(a\sqrt{2}\) 
  

 

Therefore, \(d=a\sqrt{2}\). There is something very special about this relationship. Do you know what this is? Let's look at 45-45-90 triangles:

 

Source: http://d2r5da613aq50s.cloudfront.net/wp-content/uploads/372023.image1.jpg

 

Do you notice any similiarities to this diagram and the one above? I do! The side lengths are the same and the length of d, the diagonal of the square and the hypotenuse of the triangle is equal to \(a\sqrt{2}\)! The converse of the Pythagorean theorem says that if \(a^2+b^2=c^2\), then \(\triangle ABC\) is right. The converse of the 45-45-90 triangle says if the side lengths of a triangle are equal and the length of the diagonal is that length times the square root of two, then the remaining angles are 45 degrees, 

 

The proof is done now.

 

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-TheXSquaredFactor

TheXSquaredFactor  Jun 10, 2017
 #2
avatar+4747 
0

Here's one more explanation... smiley

Let's look at half of  X's  picture:

 

 

A  +  B  +  C      =  180º        B = 90º because it is the corner of a square.

                                                          

A  +  90º  +  C   =  180º        AB = BC,   so ▵ABC is isosceles.. this makes  m∠A   =   m∠C

 

A  +  90º  +  A   =  180º        Subtract  90º  from both sides of this equation.

 

A  +  A  =  90º                       Combine like terms.

 

2A = 90º                               Divide both sides of the equation by 2.

 

A = 45º

hectictar  Jun 10, 2017
 #3
avatar+1224 
0

Yeah, I realized when I posted that convoluted explanation that you could use the isosceles triangle theorem, like you did. We all have our unique approaches...

TheXSquaredFactor  Jun 10, 2017

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