Question 948721
x, short side
y, long side
h, hypotenuse
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{{{y=5.0}}}; {{{h=3x}}}.


{{{x^2+y^2=h^2}}}, Pythagorean Theorem.
{{{x^2=(3x)^2-y^2}}}
{{{x^2=9x^2-y^2}}}
{{{x^2-9x^2=-y^2}}}
{{{9x^2-x^2=y^2}}}
{{{8x^2=y^2}}}
{{{x^2=(1/8)y^2}}}
{{{x=sqrt(y^2/8)}}}
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Substitute the value for y from given.
{{{x=sqrt((5^2)/8)}}}


{{{x=5sqrt(1/8)}}}


{{{x=(5/4)sqrt(1/2)}}}  or  {{{(5/4)/sqrt(2)}}}


{{{((5/4)/sqrt(2))(sqrt(2)/sqrt(2))}}}


{{{(5*sqrt(2))/(4*2)}}}


{{{highlight(x=(5/8)sqrt(2))}}}