Since the gry triangle is the folded part of the paper which has dimensions
[tex]\begin{gathered} L=x+y \\ W=x+\frac{1}{2}d(\text{small square)} \end{gathered}[/tex]L is the length
x is the side of the small square
y is the side of the big square
d is the diagonal of the small square
The relation between the side of a square and its diagonal is
[tex]d=s\sqrt[]{2}[/tex]s is the side of the square
d is its diagonal
Since p Is the center of the small square, then
The side PQ = 1/2 the diagonal
Since the side of the small square is x, then its diagonal is
[tex]d=x\sqrt[]{2}[/tex]From here we know that
The base of the gry triangle is PQ which equal to
[tex]\begin{gathered} PQ=\frac{1}{2}x\sqrt[]{2} \\ PQ=\frac{\sqrt[]{2}}{2}x \end{gathered}[/tex]The height of the triangle RP is equal to the length of the unfolded figure which is
[tex]RP=x+y[/tex]The formula of the area of a triangle is
[tex]A=\frac{1}{2}bh[/tex]Substitute the base by PQ and the height by PR
[tex]\begin{gathered} A=\frac{1}{2}(\frac{\sqrt[]{2}}{2}x)(x+y) \\ A=\frac{\sqrt[]{2}}{4}x(x+y) \end{gathered}[/tex]
