Question 375801
Use the Pythagorean theorem,
a){{{s^2+s^2=d^2}}}
{{{highlight( d=sqrt(2)*s)}}}
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b){{{b^2+h^2=d^2}}}
{{{highlight( d=sqrt(b^2+h^2))}}}
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c)Use Heron's formula,{{{A=sqrt(s(s-a)(s-b)(s-c))}}}
In this case, {{{b=c=a}}}
{{{s=(1/2)(a+a+a)=(3/2)a}}}
{{{s-a=(1/2)a}}}
{{{A=sqrt((3/2)a(1/2)a(1/2)a(1/2)a)}}}
{{{A=sqrt((3/16))a^4)}}}
{{{highlight( A=(sqrt(3)/4)a^2)}}}