Question Number 49680 by peter frank last updated on 09/Dec/18
Answered by math1967 last updated on 09/Dec/18
$${Total}\:{surface}\:{area}\:=\mathrm{2}×\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{20}^{\mathrm{2}} +\mathrm{6}×\mathrm{20}×\mathrm{15} \\ $$$$=\mathrm{3878}.\mathrm{46}{cm}^{\mathrm{2}} \left({aprox}\right) \\ $$$${Volume}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{6}×\mathrm{20}^{\mathrm{2}} ×\mathrm{15}=\mathrm{233826}.\mathrm{85}{cm}^{\mathrm{3}} \left({aprox}\right) \\ $$
Commented by peter frank last updated on 09/Dec/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}...\mathrm{where}\:\mathrm{this}\:\mathrm{come}\:\mathrm{from}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:? \\ $$
Commented by math1967 last updated on 09/Dec/18
$${Regular}\:{hexago}\:{splits}\:{to}\:\mathrm{6}{equilateral} \\ $$$${triangles}\:{so}\:{area}\:{of}\:{hexagon}=\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\left({side}\right)^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 09/Dec/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{once}\:\mathrm{again} \\ $$
Commented by math1967 last updated on 09/Dec/18
$${welcome} \\ $$