Question

Num prisma hexagonal regular de altura 10√3, a área lateral é o dobro da área da base. Determine a área total e o volume desse prisma.

Answer 1

Olá.

$A_{lateral} = L.H\\\\ A_{lateral} = 2.A_{base}\\\\ 2.A_{base} = L.(10\sqrt3)\\\\\\ A_{base} = 6\frac{L^2\sqrt3}{4}\\\\ A_{base} = 3\frac{L^2\sqrt3}{2}\\\\\\ 2.(3\frac{L^2\sqrt3}{2}) = L.(10\sqrt3)\\\\ 3L^2\sqrt3 = L.(10\sqrt3)$

$3L^2-10L= 0\\\\ \Delta = 10^2-4.3.0\\ \Delta = 100\\\\ L = \frac{10+-10}{2.3}\\\\ L^1 = \frac{10+10}{6} = \frac{20}{6} = \frac{10}{3}\\\\ L^2 = \frac{10-10}{6} = \frac{0}{6} = 0\\\\\\ L = \frac{10}{3}$

$A_{total} = 2.A_{B} + 2.A_{B}\\ A_{total} = 4.A_{B}\\\\ A_{total} = 4.(\frac{L^2\sqrt3}{4})\\\\ A_{total} = (\frac{10}{3})^2.\sqrt3\\\\ A_{total} = \frac{100\sqrt3}{9}\\\\ A_{total} = \frac{100.3^\frac{1}{2}}{3^2}\\\\ A_{total} = 100.3^{\frac{1}{2}-2}\\\\ A_{total} = 100.3^{-\frac{3}{2}}$

$V = A_{B}.H\\\\ V = \frac{L^2\sqrt3}{4}.10\sqrt3\\\\ V = (\frac{10}{3})^2.3.\frac{10}{3}\\\\ V = \frac{300}{9}.\frac{10}{3}\\\\ V = \frac{1000}{9}cm^3$