Twierdzenie pitagorasa Oblicz pole i wysokość trójkąta równobocznego o boku : a) 4 b) 6 c) 8 d) 10 e) 12

a) [latex] h^{2}[/latex]+[latex] 2^{2}[/latex]=[latex] 4^{2}[/latex] [latex] h^{2}[/latex]+4=16 [latex] h^{2}[/latex]=12  /*  [latex] sqrt{} [/latex] h=2 [latex] sqrt{3} [/latex] P=1/2ah= 1/2*4*2 [latex] sqrt{3} [/latex]=4 [latex] sqrt{3} [/latex] b)[latex] h^{2}[/latex]+[latex] 3^{2}[/latex]=[latex] 6^{2}[/latex] [latex] h^{2}[/latex]+9=36 [latex] h^{2}[/latex]=27 h=3[latex] sqrt{3} [/latex] P=1/2ah= 1/2*6*3[latex] sqrt{3} [/latex] = 9[latex] sqrt{3} [/latex] c)[latex] h^{2}[/latex]+[latex] 4^{2}[/latex]=[latex] 8^{2}[/latex] [latex] h^{2}[/latex]+16=64 [latex] h^{2}[/latex]=48 h=4[latex] sqrt{3} [/latex] P=1/2ah= 1/2*8*4[latex] sqrt{3} [/latex] =16[latex] sqrt{3} [/latex] d)[latex] h^{2}[/latex]+[latex] 5^{2}[/latex]=[latex] 10^{2}[/latex] [latex] h^{2}[/latex]+25=100 [latex] h^{2}[/latex]=75 h=5[latex] sqrt{3} [/latex] P=1/2ah=1/2*10*5[latex] sqrt{3} [/latex] =25[latex] sqrt{3} [/latex] e)[latex] h^{2}[/latex]+[latex] 6^{2}[/latex]=[latex] 12^{2}[/latex] [latex] h^{2}[/latex]+36=144 [latex] h^{2}[/latex]=108 h=6[latex] sqrt{3} [/latex] P=1/2ah= 1/2*12*6[latex] sqrt{3} [/latex] =36[latex] sqrt{3} [/latex]