Question

utilizando a fórmula de herão, calcule a área de um triangulo cujos lados medem: A) 2km,3km e 4km resp:3raiz de 15/2 b) 12 dm,15 dm e 9dm respo, 54dm

Answer 1

Sendo p = $\frac{a+b+c}{2}$, a fórmula de Heron dá-se por: A = $\sqrt{p(p-a)(p-b)(p-c)}$

A) 2km, 3km e 4km 

p = $\frac{a+b+c}{2}$
p = $\frac{2+3+4}{2}$
p = $\frac{9}{2}$

A = $\sqrt{p(p-a)(p-b)(p-c)}$
A = $\sqrt{\frac{9}{2}(\frac{9}{2}-2)(\frac{9}{2}-3)(\frac{9}{2}-4)}$
A = $\sqrt{\frac{9}{2}(\frac{-4+9}{2})(\frac{-6+9}{2})(\frac{-8+9}{2})}$
A = $\sqrt{\frac{9}{2}.(\frac{5}{2}).(\frac{3}{2}).(\frac{1}{2})}$
A = $\sqrt{ \frac{9.5.3.1}{2.2.2.2} }$
A = $\sqrt{9}$ × $\sqrt{ \frac{15}{16} }$
A = 3$\frac{ \sqrt{15} }{4}$ km


b) 12 dm,15 dm e 9dm

p = $\frac{a+b+c}{2}$
p = $\frac{12+15+9}{2}$
p = $\frac{36}{2}$
p = 18

A = $\sqrt{p(p-a)(p-b)(p-c)}$
A = $\sqrt{18(18-12)(18-15)(18-9)}$
A = $\sqrt{18.6.3.9}$
A = $\sqrt{2916}$
A = 54 dm