Question Number 210157 by Spillover last updated on 01/Aug/24
Answered by A5T last updated on 01/Aug/24
$${Let}\:{distance}\:{of}\:{vertex},{V},\:{to}\:{centroid},{G},\:{be}\:{GV} \\ $$$$\Rightarrow\frac{{sin}\mathrm{30}°}{{GV}}=\frac{{sin}\mathrm{120}^{°} }{{x}}\Rightarrow{GV}=\frac{\frac{{x}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${H}=\sqrt{{x}^{\mathrm{2}} −{GV}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}=\frac{{x}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$${Area}=\frac{\mathrm{1}}{\mathrm{3}}×{Area}\:{of}\:{base}×{H} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{\mathrm{1}}{\mathrm{2}}×{x}^{\mathrm{2}} ×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)×\frac{{x}\sqrt{\mathrm{6}}}{\mathrm{3}}=\frac{\mathrm{3}{x}^{\mathrm{3}} \sqrt{\mathrm{2}}}{\mathrm{4}×\mathrm{3}×\mathrm{3}}=\frac{{x}^{\mathrm{3}} \sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$
Commented by Spillover last updated on 01/Aug/24
$${nice}\:{solution}. \\ $$
Commented by A5T last updated on 01/Aug/24
$${This}\:{is}\:{for}\:{a}\:\boldsymbol{{triangular}}\:{pyramid},\:{when}\:\boldsymbol{{all}}\:{sides} \\ $$$${are}\:{equilateral}\:{triangles}.\:{I}\:{ignored}\:{the}\:\boldsymbol{{square}} \\ $$$${while}\:{reading}\:{the}\:{question}. \\ $$
Answered by mr W last updated on 01/Aug/24
Commented by mr W last updated on 01/Aug/24
$${base}\:{area}\:{x}^{\mathrm{2}} \\ $$$${height}\:{h}=\sqrt{{x}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}{x}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}{x}}{\:\mathrm{2}} \\ $$$${volume}\:{V}=\frac{{x}^{\mathrm{2}} {h}}{\mathrm{3}}=\frac{\sqrt{\mathrm{2}}{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$
Commented by Spillover last updated on 01/Aug/24
$${nice}\:{solution} \\ $$
Answered by Spillover last updated on 01/Aug/24
