Question Number 156671 by cortano last updated on 14/Oct/21
Answered by som(math1967) last updated on 14/Oct/21
Commented by som(math1967) last updated on 14/Oct/21
$${cos}\alpha=\frac{\mathrm{10}}{\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${sin}\alpha=\frac{\sqrt{\mathrm{11}}}{\mathrm{6}} \\ $$$${sin}\mathrm{2}\alpha=\mathrm{2}×\frac{\sqrt{\mathrm{11}}}{\mathrm{6}}×\frac{\mathrm{5}}{\mathrm{6}}=\frac{\mathrm{5}\sqrt{\mathrm{11}}}{\mathrm{18}} \\ $$$${again}\:{sin}\mathrm{2}\alpha=\frac{{LM}}{{AL}} \\ $$$$\:\:\frac{\mathrm{10}}{{AL}}=\frac{\mathrm{5}\sqrt{\mathrm{11}}}{\mathrm{18}}\Rightarrow{AL}=\frac{\mathrm{36}}{\:\sqrt{\mathrm{11}}}=\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$${DL}^{\mathrm{2}} =\left(\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\right)^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \\ $$$${DL}=\frac{\mathrm{14}}{\:\sqrt{\mathrm{11}}}=\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$${LC}=\left(\mathrm{10}−\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}}\right) \\ $$$${BN}=\sqrt{\mathrm{12}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{11}} \\ $$$${CN}=\mathrm{10}−\mathrm{2}\sqrt{\mathrm{11}} \\ $$$${perimeter}={AL}+{LC}+{CN}+{AN} \\ $$$$=\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\:+\left(\mathrm{10}−\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}}\right)+\mathrm{10}−\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{12} \\ $$$$=\mathrm{32}+\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\:−\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}=\mathrm{32}{unit} \\ $$
Commented by cortano last updated on 14/Oct/21
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{gave}\:\mathrm{kudos} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/Oct/21
$$\mathrm{wow} \\ $$
Commented by Tawa11 last updated on 14/Oct/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by john_santu last updated on 15/Oct/21
Commented by cortano last updated on 15/Oct/21
$$\mathrm{nice} \\ $$