Answer :

[tex]a-the\ sides\ of\ the\ square\\\\A_{\fbox{}}=a^2;\ A_{\fbox{}}=8100ft^2\\\\a^2=8100ft^2\\\\a=\sqrt{8100ft^2}\\\\a=90ft\\---------------\\The\ diagonal\ of\ the\ square:d=a\sqrt2\\------------------\\The\ distance\ from\ home\ plate\ to\ second\ base:\\\\d=90\sqrt2\ ft\approx(90\times1.414)ft=\boxed{127.26ft}[/tex]
kate200468
[tex]A_{square}=(the\ side\ of\ the\ square)^2\\\\A_{square}=8100\ ft^2\ \ \ \Leftrightarrow\ \ \ A_{square}=(90\ ft)^2\\\\(the\ side\ of\ the\ square)^2=(90\ ft)^2\ \ and\ \ \ the\ side\ of\ the\ square>0\\\\the\ side\ of\ the\ square=91\ ft\\\\the\ length\ of\ the\ diagonal= \sqrt{2} \cdot the\ side\ length\ (of\ the\ square)\\\\the\ length\ of\ the\ diagonal= \sqrt{2} \cdot 90\ ft\\\\ \sqrt{2} =1.4142135623...\approx1.4142\\\\ [/tex]

[tex]\Rightarrow\ \ \ the\ length\ of\ the\ diagonal\approx1.4142\cdot90\ ft=127.278\ ft[/tex]

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