Question

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

Answer 1

I would say 90 but I could be wrong.

Good Luck!!

Answer 2

The answer is:  "x = $\frac{4 \sqrt{3} }{3}$  ." ; 

          AND:       "y =  $\frac{4 \sqrt{3} }{3}$  ."
_______________________________________________________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:
  " $\frac{4 \sqrt{6} }{3}$ " .
__________________________________________________
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" .
____________________________________________________
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 = $\frac{4 \sqrt{6} }{3}$ " . 

→  c² = ( $\frac{4 \sqrt{6} }{3}$ )² ;  

          =  $\frac{(4 \sqrt{6})^2 }{3 ^{2} }$ ;

          =  $\frac{4 ^{2}( \sqrt{6} ) ^{2} }{3 ^{2} }$ ;
 
          = $\frac{(16*6)}{9}$  ;  

          = $\frac{32}{3}$ ; 
____________________________________________________ 
         →  2x²  =  $\frac{32}{3}$

Divide each side of the equation by "2" ; 

2x² / 2 = $\frac{32}{3}$) ÷ 2 ;


x² = $\frac{32}{3} * \frac{1}{2}$ ;  

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 = $\frac{16}{3} * \frac{1}{1} = \frac{16}{3} * 1 ;\\ = \frac{16}{3} ;\\→ x² = \frac{16}{3} ; \\Take the positive square root of each side of the equation; to isolate "x" on one side of the equation; & to solve for "x" ; \\→ ⁺√(x²) = ⁺√(\frac{16}{3}) ; \\→ x = ⁺\frac({√16}{√3}) = [tex] \frac{4}{ \sqrt{3} }$ ; 
→  Multiply by " $\frac{ \sqrt{3} }{ \sqrt{3} }$" ; to eliminate the "√3" in the "denominator" ; 

→    $\frac{4}{ \sqrt{3} }$  * $\frac{ \sqrt{3} }{ \sqrt{3} }$  ;

       = $\frac{4}{ \sqrt{3} }$  ÷  $\frac{ \sqrt{3} }{ \sqrt{3} }$  ;

       =    " $\frac{4 \sqrt{3} }{3}$ " .
_____________________________________________________
The answer is:  " x = $\frac{4 \sqrt{3} }{3}$  ." ; 

          AND:       " y =  $\frac{4 \sqrt{3} }{3}$  ."
_____________________________________________________

Does "x√2" = the hypotenuse length shown?

that is:  Does "x√2" =  "$\frac{4 \sqrt{6} }{3}$" ?

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

→  $\frac{4 \sqrt{3} }{3}$  * √2  ;

 =  $\frac{4 \sqrt{3}* \sqrt{2} }{3}$  ?? ; 

→  Note:  "√3 * √2  =  √(3 * 2) = √6 " ; 
_________________________________________

→   $\frac{4 \sqrt{3}* \sqrt{2} }{3}$  ; 

 =  $\frac{4 \sqrt{6} }{3}$ ;

→  which is the value of the hypotenuse shown in the figure!  
Yes; the answer does make sense!
_________________________________________________