Question Number 101767 by 175mohamed last updated on 04/Jul/20
Commented by mr W last updated on 04/Jul/20
$${a}\:{CD}\:{can}\:{cover}\:{a}\:{wall}\:{area}\:{of} \\ $$$$\mathrm{2}×\frac{{D}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×{D}=\frac{\sqrt{\mathrm{3}}{D}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${number}\:{of}\:{CDs}\:{needed}\:{to}\:{cover} \\ $$$$\mathrm{6}{m}×\mathrm{4}{m}\:{wall}\:{area}: \\ $$$$\frac{\mathrm{6}×\mathrm{4}\:{m}^{\mathrm{2}} }{\frac{\sqrt{\mathrm{3}}×\mathrm{0}.\mathrm{12}^{\mathrm{2}} }{\mathrm{2}}{m}^{\mathrm{2}} }\approx\mathrm{1425} \\ $$
Commented by mr W last updated on 04/Jul/20
