Question 175457
Let {{{s}}}= a sides of the square
Let {{{A[s]}}} = area of square
{{{A[s] = s^2}}}
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For triangle:
{{{A[t] = (1/2)*b*h}}}
It is given that {{{h = s}}}
and {{{b = 9}}}in
{{{A[t] = (1/2)*s *9}}}
{{{A[t] = (9/2)*s}}}
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Also given is:
{{{A[s] = (1/3)*A[t]}}}
therefore:
{{{s^2 = (1/3)*(9/2)*s}}}
Multiply both sides by {{{6}}}
{{{6s^2 = 9s}}}
{{{6s^2 - 9s = 0}}}
{{{s(6s - 9) = 0}}}
The solutions are:
{{{s = 0}}} This one doesn't work here
{{{s = 3/2}}}
The side of the square is 1.5 in
check:
{{{A[s] = s^2}}}
{{{A[s] = 2.25}}}
{{{A[t] = (9/2)*s}}}
{{{A[t] = 6.75}}}
Also given is:
{{{A[s] = (1/3)*A[t]}}}
{{{2.25 = (1/3)*6.75}}}
{{{6.75 = 6.75}}}
OK