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Re: PLEASE Solve
Volume:
$V = A_b h$
$512 = \left[ 6 \times \frac{1}{2}(x^2)\sin 60^\circ \right] h$
$\frac{3}{2}\sqrt{3}x^2 h = 512$
$h = \dfrac{1024}{3\sqrt{3} \, x^2}$
Wet area:
$A = A_b + A_L$
$512 = 6 \times \frac{1}{2}(x^2)\sin 60^\circ + 6xh$
$512 = \frac{3}{2}\sqrt{3}x^2 + 6x \left( \dfrac{1024}{3\sqrt{3} \, x^2} \right)$
$512 = \dfrac{3\sqrt{3}}{2}x^2 + \dfrac{2048}{\sqrt{3} \, x}$
$\dfrac{3\sqrt{3}}{2}x^3 - 512x + \dfrac{2048}{\sqrt{3}} = 0$
$x = -15.075 ~ \text{(absurd)}, ~ 12.697, ~ \text{and} ~ 2.378$
Try x = 12.697"
$V = \left[ 6 \times \frac{1}{2}(12.697^2)\sin 60^\circ \right] (1.222)$
$V = 511.83 \cong 512 ~ \text{okay}$
Try x = 2.378"
$V = \left[ 6 \times \frac{1}{2}(2.378^2)\sin 60^\circ \right] (34.849)$
$V = 511.995 \cong 512 ~ \text{okay}$
Answer: h = 12.697" or h = 2.378"