Oblicz obwód Trójkąta ACD, jeśli pole trapezu BCDE jest równe 7.   http://img198.imageshack.us/i/zadaniee3.jpg/

P-pole trapezu   P=[latex]frac{|BE|+|CD|}{2}*|BC|[/latex] [latex]ctg45^o=frac{|AB|}{|BE|}\ 1=frac{2,5}{|BE|}\ |BE|=2,5\ [/latex]   [latex]cos45^{o}=frac{|AB|}{AE}\ frac{sqrt{2}}{2}=frac{2,5}{|AE|}\ |AE|=frac{5sqrt{2}}{2}[/latex]   [latex]frac{|AB|}{|BE|}=frac{|AC|}{|CD|}\ |DC|=|BC|+2,5[/latex]   [latex]7=frac{|BE|+|DC|}{2}*|BC|\ 7=frac{2,5+2,5+|BC|}{2}*|BC|\ 14=5|BC|+|BC|^{2}\ |BC|^{2}+5|BC|-14=0\ Delta=25+45=81\ (sqrt{Delta})=9\ |BC|_1=(-5+9)/2=2 \ |BC|_2=(-5-9)/2=-7 nie nalezy do dziedziny [/latex]   |AC|=|DC|=|AB|+|BC|=2,5+2=4,5   [latex]frac{|AE|}{|AB|}=frac{|AD|}{|AC|}\ frac{frac{5sqrt{2}}{2}}{2,5}=frac{|AD|}{|4,5|}\ |AD|=4,5sqrt{2}[/latex]   obwod ACD=|AC|+|CD|+|AD|=4,5+4,5+[latex]4,5sqrt{2}[/latex]=9+[latex]4,5sqrt{2}[/latex]