Question 1113575
{{{ drawing(300,300, -10, 10, -10, 10, grid(0),
                    line(-4.24, -4.24, -4.24, 4.24),
                    line(-4.24,4.24, 4.24,4.24),
                    line(4.24,4.24,4.24,-4.24),
                    line(4.24,-4.24, -4.24,-4.24),
                   green(line(-2,-4.2,4.2,2)),
                   green(line(4.2,2, 2,4.2)),
                green(line(2,4.2, -4.2,-2)),
                green(line(-4.2,-2, -2,-4.2))
)}}}
                    
Given information:
(i)  The diagonal of the square is given as 12m —> sides are {{{6sqrt(2)}}} m
(ii)   L = 2W for the inscribed rectangle.

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I used areas to find the dimensions of the rectangle:

 ( area of the rectangle ) + ( area of the 4 triangles ) =  ( area of the square )

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    LW +  Area(4 triangles) = {{{ ( 6sqrt(2) )^2 }}} = 72             (eq 1)
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The short legs of the small triangle are of length k, where {{{k^2 + k^2 = W^2 }}} —> {{{ k^2 = W^2/2 }}}
so  the area of one small triangle = {{{ (1/2)(k^2) = (1/2)(W^2/2) = (1/4)W^2 }}}
Area of both small triangles = {{{ (1/2)W^2 }}}
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By a similar approach, the area of both large triangles is:
 {{{ (1/2)L^2 }}} = {{{ (1/2)(2W)^2 =  2W^2 }}} 
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Substituting the triangle areas (and L=2W), (1) becomes:

    {{{ (2W)(W) + (1/2)W^2 + 2W^2 = 72 }}}
    {{{ (9/2)W^2  = 72 }}}
    {{{  W^2 = 144/9 = 16 }}}  —>  W=4 —> L=8
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Ans:  The length and width of the inscribed rectangle are {{{ highlight( L=8m )}}} and {{{ highlight( W=4m) }}}  
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Check by summing areas:   
Small triangles:   {{{ (1/2)(4^2) = 8 }}}
Large triangles:  {{{ 2(4^2) = 32 }}}
Rectangle:          {{{  4*8 = 32 }}}

8+32+32 = 72 = area of square,  looks ok.