Question Number 77026 by TawaTawa last updated on 02/Jan/20
Commented by $@ty@m123 last updated on 03/Jan/20
$${Which}\:{one}? \\ $$$$\mathrm{20}\:{or}\:\mathrm{21} \\ $$
Commented by TawaTawa last updated on 03/Jan/20
$$\mathrm{Both}\:\mathrm{sir} \\ $$
Answered by $@ty@m123 last updated on 03/Jan/20
$$\mathrm{21}.\:{Let}\:{x}^{\mathrm{2}} ={y} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\mathrm{34}{y}+\mathrm{225}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{34}\pm\sqrt{\mathrm{1036}−\mathrm{1000}}}{\mathrm{2}}=\frac{\mathrm{34}\pm\mathrm{6}}{\mathrm{2}}=\mathrm{20},\mathrm{14} \\ $$$${x}=\pm\sqrt{\mathrm{20}},\:\pm\sqrt{\mathrm{14}} \\ $$$$\therefore\:{smallest}\:{solution}\:=−\sqrt{\mathrm{20}}\:=−\mathrm{2}\sqrt{\mathrm{5}} \\ $$
Commented by TawaTawa last updated on 03/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by $@ty@m123 last updated on 03/Jan/20
$${Outer}\:{perimeter}=\mathrm{24}×\mathrm{4}''+\mathrm{4}×\mathrm{4}''+\mathrm{23}×\mathrm{5}' \\ $$$$=\mathrm{96}''+\mathrm{16}''+\mathrm{115}' \\ $$$$=\mathrm{8}'+\mathrm{1}\frac{\mathrm{4}}{\mathrm{12}}'+\mathrm{115}^{'} \\ $$$$=\mathrm{124}\frac{\mathrm{1}}{\mathrm{3}}' \\ $$
Commented by TawaTawa last updated on 03/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$