[tex]a) Aplicamos\ Pitagoras\\hip^2=base^2+(h)altura^2\\7^2=2^2+h^2\ (usamos\ 2\ na\ base\ que\ eh\ o\ termo\ medio\\ do\ lado\ IA\ para\ que\ seja\ valido\ o\ calculo\ da\ altura)\\\\49=4+h^2\\\sqrt{h^2}=\sqrt{49-4}\\h=\sqrt{45}=\sqrt{3^2.5}\\\\h=3\sqrt{5}[/tex]
[tex]b) Pela\ formula\ de\ Heron\ nao\ precisa\ da\\ altura\ para\ achar\ a\ area,\ veja:\\A=\sqrt{S.(S-B).(S-A).(S-I)}\ \ \ (S:\ semi-perimetro)\\perimetro=7+7+4=18\ semiperimetro=\frac{18}{2}=9\\\\A=\sqrt{9.(9-7).(9-7).(7-4)}=>\sqrt{9.2.2.3}=\sqrt{180}\\A=\sqrt{2^2.3^2.5}=>A=6\sqrt{5}\\\\Pela\ formula\ geral\ da\ A=\frac{base.altura}{2}\\\\A=\frac{4.3\sqrt{5}}{2}=>A=6\sqrt{5}[/tex]
Nesse caso necessitaria do exercício a
