Question Number 218528 by Spillover last updated on 11/Apr/25
Answered by mr W last updated on 11/Apr/25
Commented by mr W last updated on 11/Apr/25
$$\mathrm{3}\theta=\frac{\pi}{\mathrm{2}}\:\Rightarrow\theta=\frac{\pi}{\mathrm{6}}=\mathrm{30}° \\ $$$${a}\:\mathrm{sin}\:\theta+\mathrm{4}\:\mathrm{cos}\:\theta=\mathrm{4} \\ $$$$\Rightarrow{a}=\mathrm{8}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${length}\:{of}\:{big}\:{rectangle}\:{b}: \\ $$$${b}={a}+{a}\:\mathrm{cos}\:\theta+\mathrm{2}{a}\:\mathrm{sin}\:\theta+\frac{{a}}{\mathrm{tan}\:\theta}+\mathrm{4} \\ $$$${b}=\mathrm{8}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{3}}\right)+\mathrm{4}=\mathrm{2}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${area}\:{of}\:{big}\:{rectangle}: \\ $$$$\mathrm{4}×{b}=\mathrm{8}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)\:\checkmark \\ $$$${area}\:{of}\:{small}\:{rectangle}: \\ $$$$\mathrm{4}×{a}=\mathrm{16}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\checkmark \\ $$
Answered by Spillover last updated on 11/Apr/25
Answered by Spillover last updated on 11/Apr/25
