Question Number 220393 by Hanuda354 last updated on 12/May/25
Commented by Hanuda354 last updated on 12/May/25
$$\mathrm{ABCD}\:\:\mathrm{is}\:\:\mathrm{a}\:\:\mathrm{square}.\:\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{value}\:\:\mathrm{of}\:\:{x}. \\ $$
Answered by mr W last updated on 13/May/25
Commented by mr W last updated on 13/May/25
$${a}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +{x}^{\mathrm{2}} ={area}\:{of}\:{square} \\ $$$${area}\:{of}\:\Delta{DCE}=\frac{\mathrm{6}×\left({x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)}{\mathrm{2}}=\frac{{area}\:{of}\:{square}}{\mathrm{2}} \\ $$$$\mathrm{6}\left({x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\right)=\mathrm{6}^{\mathrm{2}} +{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3}+\sqrt{\mathrm{3}\left(\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{5}\right)}\:\approx\mathrm{5}.\mathrm{405}\:{or} \\ $$$$\Rightarrow{x}=\mathrm{3}−\sqrt{\mathrm{3}\left(\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{5}\right)}\:\approx\mathrm{0}.\mathrm{595} \\ $$
Commented by mr W last updated on 13/May/25
Commented by mr W last updated on 13/May/25
