Mathematics, 28.01.2020 13:38 georgesk872

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

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Answer from: hrcatena

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!
_________________________________________________

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