Question 670822
Apply Pythagorean Theorem
...
{{{a^2 + b^2 = c^2}}}
except in this canse
{{{a^2 = b^2}}}
...
{{{a^2 + a^2 = c^2}}}
or
{{{2a^2 = 40^2}}}
...
{{{sqrt(2a^2) = sqrt(40^2)}}}
{{{(a)sqrt(2) = 40}}}
...
a = {{{40/sqrt(2)}}}
= {{{(40/sqrt(2))(sqrt(2)/sqrt(2))}}}
= {{{40sqrt(2)/2}}}
= {{{(20cross(40)sqrt(2))/1cross(2)}}}
{{{highlight_green(a = 20sqrt(2))}}}cm for each leg.
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