W trójkącie prostok. równoram. mamy kąty 45-45-90, a zatem przeciwprostokątna d ma długość [latex]d=asqrt{2} \ 10 = a sqrt{2} \ a= frac{10}{sqrt{2}} =5 sqrt{2} [/latex] W takim razie pole: [latex]P= frac{1}{2} a^2= frac{1}{2}cdot 5^2cdot 2=25 cm^2[/latex] i obwód [latex]Obw=a+a+d=5 sqrt{2} +5 sqrt{2} +10=10+10 sqrt{2} [/latex]
[latex]a = b[/latex] [latex]c^{2} = a^{2} + b^{2} = a^{2} + a^{2}[/latex] [latex]10^{2} = a^{2} + a^{2} [/latex] [latex]2a^{2} = 10^{2}[/latex] [latex]2a^{2} = 100 /:2[/latex] [latex]a^{2} = 50 /:sqrt{} [/latex] [latex]a = sqrt{50} [/latex] [latex]a = sqrt{25 * 2} [/latex] [latex]a = 5sqrt{2} [/latex] [latex]a = 5sqrt{2}cm[/latex] [latex]a=b[/latex] [latex]P = frac{a*b}{2} [/latex] [latex]P = frac{5 sqrt{2}cm * 5sqrt{2}cm }{2} = frac{50cm^{2} }{2} = 25cm^{2}[/latex] [latex]Obw=a + b + c[/latex] [latex]5 sqrt{2}cm + 5 sqrt{2}cm + 10 cm = 10 sqrt{2}cm + 10 cm = 10(sqrt{2}+1)cm[/latex]
