Question Number 34361 by mondodotto@gmail.com last updated on 05/May/18
$$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{prism}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{regular}} \\ $$$$\boldsymbol{\mathrm{hexagonal}}\:\boldsymbol{\mathrm{base}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{length}}\:\mathrm{15}\boldsymbol{\mathrm{cm}},\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{height}}\: \\ $$$$\boldsymbol{\mathrm{of}}\:\mathrm{20}\boldsymbol{\mathrm{cm}}.\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{volume}}\:\boldsymbol{\mathrm{and}} \\ $$$$\boldsymbol{\mathrm{total}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}. \\ $$
Answered by MJS last updated on 05/May/18
$${V}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{s}^{\mathrm{2}} {h}=\mathrm{6750}\sqrt{\mathrm{3}} \\ $$$${S}=\mathrm{3}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} +\mathrm{6}{sh}=\mathrm{1800}+\mathrm{675}\sqrt{\mathrm{3}} \\ $$