Pole deltoidu (latawca) wyznaczamy ze wzoru: [latex]P=frac{d_{1}*d_{2}}{2}[/latex] gdzie [latex]d_{1} i d_{2}[/latex] to przekątne. ------------------------------------------------------ [latex]a) d_{1}=sqrt{27}; d_{2}=sqrt{3}\ \ P=frac{sqrt{27}*sqrt{3}}{2}\ \ P=frac{sqrt{27*3}}{2}\ \ P=frac{sqrt{81}}{2}\ \ P=frac{sqrt{9^{2}}}{2}\ \ P=frac{9}{2}\ \ P=4frac{1}{2} [j^{2}][/latex] ------------------------------------------------------ [latex]b) d_{1}=sqrt[3]{24}; d_{2}=sqrt[3]{9}\ \ P=frac{sqrt[3]{24}*sqrt[3]{9}}{2}\ \ P=frac{sqrt[3]{24*9}}{2}\ \ P=frac{sqrt[3]{216}}{2}\ \ P=frac{sqrt[3]{6^{3}}}{2}\ \ P=frac{6}{2}\ \ P=3 [j^{2}][/latex] ------------------------------------------------------ [latex]7*sqrt{216}*sqrt{36}=7*sqrt{36*6}*sqrt{6^{2}}=7*sqrt{6^{2}*6}*6=\ \ =7*6sqrt{6}*6=252sqrt{6}[/latex]
[latex]Pole deltoidu o przekatnych e i f:\\P=frac{ef}{2}[/latex] [latex]1.\a) e=sqrt{27}; f=sqrt{3}\\P=frac{sqrt{27}cdotsqrt3}{2}=frac{sqrt{27cdot3}}{2}=frac{sqrt{81}}{2}=frac{9}{2}=4,5\\b) e=sqrt[3]{24}; f=sqrt[3]{9}\\P=frac{sqrt[3]{24}cdotsqrt[3]9}{2}=frac{sqrt[3]{24cdot9}}{2}=frac{sqrt[3]{216}}{2}=frac{6}{2}=3\\\2.\7cdotsqrt{216}cdotsqrt{36}=7cdotsqrt{36cdot6}cdot6=42cdotsqrt{36}cdotsqrt6=42cdot6sqrt6=252sqrt6[/latex] ©DRK
