This is a special right triangle, what is the missing side length?

Question
Answer:
The answer is:  "x = [tex] \frac{4 \sqrt{3} }{3} [/tex]  ." ; 

          AND:       "y =  [tex] \frac{4 \sqrt{3} }{3} [/tex]  ."
_______________________________________________________Explanation:______________________________________________________The sides of a "45-45-90" (right triangle) are:  "a", "a" ; and "a√2" .

Note that:  "a√2"  is the hypotenuse length— and the other 2 (TWO) sides of the triangle are of equal length— {since:  "a = a" .}._______________________________________________________
As such:  "x = y" ;  and the hypotenuse,  "x√2", equals:
  " [tex] \frac{4 \sqrt{6} }{3} [/tex] " .
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Note: The Pythagorean theorem (for the side lengths of right triangles):

       →  " a²  +  b²  = c²  ; 

in which:  "c = the hypotenuse length" ; 
                 "a = one of the other side lengths"
                 "b = the remaining side length" .
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Note that:  "x = y" ;
 
so:    " x² + x² = 2x " ; 

2x² = x√2 ; 

2x² = c² ; in which "c" is the hypotenuse;  Solve for "x" and "y" ;  Since "x = y" ; solve for "x" ; 

2x² = c² ;

→  Given (from image attached);  " c = [tex] \frac{4 \sqrt{6} }{3} [/tex] " . 

→  c² = ( [tex] \frac{4 \sqrt{6} }{3} [/tex] )² ;  

          =  [tex] \frac{(4 \sqrt{6})^2 }{3 ^{2} } [/tex] ;

          =  [tex] \frac{4 ^{2}( \sqrt{6} ) ^{2} }{3 ^{2} } [/tex] ;
 
          = [tex] \frac{(16*6)}{9} [/tex]  ;  

          = [tex] \frac{32}{3} [/tex] ; 
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         →  2x²  =  [tex] \frac{32}{3} [/tex]

Divide each side of the equation by "2" ; 

2x² / 2 = [tex] \frac{32}{3} [/tex]) ÷ 2 ;


x² = [tex] \frac{32}{3} * \frac{1}{2}[/tex] ;  

Note:  The "32" cancels out to "16"; and the "2" cancels out to "1" ; 

→  {since:  "32 ÷ 2 = 16" ; and since: "2 ÷ 2 = 1 " } l  

And we have;  

x² = [tex] \frac{16}{3} * \frac{1}{1} = \frac{16}{3} * 1  ;

    = \frac{16}{3} ;

→  x² = \frac{16}{3} ; 

Take the positive square root of each side of the equation; 
 to isolate "x" on one side of the equation; & to solve for "x" ; 

→  ⁺√(x²) = ⁺√(\frac{16}{3}) ; 

→  x = ⁺\frac({√16}{√3})  =  [tex] \frac{4}{ \sqrt{3} } [/tex] ; 
→  Multiply by " [tex] \frac{ \sqrt{3} }{ \sqrt{3} } [/tex]" ; to eliminate the "√3" in the "denominator" ; 

→    [tex] \frac{4}{ \sqrt{3} } [/tex]  * [tex] \frac{ \sqrt{3} }{ \sqrt{3} } [/tex]  ;

       = [tex] \frac{4}{ \sqrt{3} } [/tex]  ÷  [tex] \frac{ \sqrt{3} }{ \sqrt{3} } [/tex]  ;

       =    " [tex] \frac{4 \sqrt{3} }{3} [/tex] " .
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The answer is:  " x = [tex] \frac{4 \sqrt{3} }{3} [/tex]  ." ; 

          AND:       " y =  [tex] \frac{4 \sqrt{3} }{3} [/tex]  ."
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Does "x√2" = the hypotenuse length shown?

that is:  Does "x√2" =  "[tex] \frac{4 \sqrt{6} }{3} [/tex]" ?

Note:  " x =  [tex] \frac{4 \sqrt{3} }{3} [/tex] " ;  (from our calculated answer) .
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→  Multiply this value by "√2" ; and see if we get the same values as the given hypotenuse: 

→  [tex] \frac{4 \sqrt{3} }{3} [/tex]  * √2  ;

 =  [tex] \frac{4 \sqrt{3}* \sqrt{2} }{3} [/tex]  ?? ; 

→  Note:  "√3 * √2  =  √(3 * 2) = √6 " ; 
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→   [tex] \frac{4 \sqrt{3}* \sqrt{2} }{3} [/tex]  ; 

 =  [tex] \frac{4 \sqrt{6} }{3} [/tex] ;

→  which is the value of the hypotenuse shown in the figure!  
Yes; the answer does make sense!
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solved
general 10 months ago 3607