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
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