Question

Encontre o perímetro e a área das figuras

Answer 1

a)

$a = \sqrt{72} \times \sqrt{72} = 72 \\ \\ p = 4 \times \sqrt{72 } = 4 \times 6 \sqrt{2} = 24 \sqrt{2}$

b)

$a = 2 \sqrt{3} \times 3 \sqrt{3} = 6 \sqrt{9} = 6 \times 3 = 18 \\ p = 2 \times 2 \sqrt{3} + 2 \times 3 \sqrt{3} = 4 \sqrt{3} + 6 \sqrt{3} = 10 \sqrt{3}$

c)

$a = \frac{ \sqrt{8} \times \sqrt{18} }{2 } = \frac{2 \sqrt{2} \times 3 \sqrt{2} }{2} = \frac{6 \sqrt{4} }{2} = 3 \times 2 = 6 \\ p = \sqrt{8} + \sqrt{18} + \sqrt{32} = 2 \sqrt{2} + 3 \sqrt{2} + 4 \sqrt{2 } = 9 \sqrt{2}$

d)

$a = \sqrt{108} \times \sqrt{48} = \sqrt{2 ^{2} \times {3}^{3} } \times \sqrt{16 \times 3} = 6 \sqrt{3} \times 4 \sqrt{3} = 24 \sqrt{9} = 24 \times 3 = 72 \\ p = 2 \times \sqrt{108} + 2 \times \sqrt{12} = 2 \times 6 \sqrt{3} + 2 \times 2 \sqrt{3} = 12 \sqrt{3} + 4 \sqrt{3} = 16 \sqrt{3}$