Respostas

2014-05-07T16:00:15-03:00
A) Medida do lado :  \sqrt{60}  ≈7.75

b) diagonal: Teorema de pitagoras.
D= Diagonal
 D^{2}   \sqrt{60} ^{2}   \sqrt{60} ^{2}  ⇔ 
⇔  D^{2} = 60 + 60 ⇔
⇔  D^{2} = 120 ⇔
⇔D= \sqrt{120}  ≈ 11