a)
[tex]4 \times \frac{(4.5 {)}^{2} \sqrt{3} }{4} = 20.25 \sqrt{3} {dm}^{2} \\ \\ 20.25 \sqrt{3} = \\ = 20.25 \times 1.7 = \\ = \underline{34.425 {dm}^{2} }[/tex]
b)
Wysokość ściany bocznej
[tex](1.4 {)}^{2} + {x}^{2} = {5}^{2} \\ 1.96 + {x}^{2} = 25 \\ {x}^{2} = 23.04 \\ x = \sqrt{23.04} dm[/tex]
Pole podstawy
[tex]2.8 \times 2.8 = 7.84 {dm}^{2} [/tex]
Pole boczne
[tex]4 \times \frac{2.8 \times \sqrt{23.04} }{2} = 2 \times 2.8 \sqrt{23.04} = 5.6 \sqrt{23.04} {dm}^{2} [/tex]
Pole całkowite
[tex](5.6 \sqrt{23.04} + 7.84 {)dm}^{2} = \\ = 5.6 \times 4.8 + 7.84 = \\ = 26.88 + 7.84 = \\ = \underline{ 34.72 {dm}^{2} }[/tex]
Większe pole powierzchni całkowitej ma ostrosłup drugi.