Question Number 167152 by cortano1 last updated on 08/Mar/22 | ||
Commented by som(math1967) last updated on 08/Mar/22 | ||
$$\:{x}^{\mathrm{2}} =\mathrm{28}\:? \\ $$ | ||
Commented by cortano1 last updated on 08/Mar/22 | ||
$$\mathrm{yes} \\ $$ | ||
Answered by som(math1967) last updated on 08/Mar/22 | ||
Commented by som(math1967) last updated on 08/Mar/22 | ||
$${let}\:{SB}={l}\:{side}\:{of}\:{square}={a} \\ $$$${A}=\frac{{al}}{\mathrm{2}}\:\: \\ $$$$\mathrm{2}{A}=\frac{{a}×{AQ}}{\mathrm{2}} \\ $$$$\therefore{a}×{AQ}=\mathrm{2}{al}\Rightarrow{AQ}=\mathrm{2}{l} \\ $$$$\therefore{BR}={a}−{l}\:\:\:{AR}={a}−\mathrm{2}{l} \\ $$$$\:\mathrm{3}{A}=\frac{\left({a}−{l}\right)\left({a}−\mathrm{2}{l}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{al}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −\mathrm{3}{al}+\mathrm{2}{l}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\Rightarrow{a}^{\mathrm{2}} −\mathrm{6}{al}+\mathrm{2}{l}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} }{{l}^{\mathrm{2}} }\:−\frac{\mathrm{6}{a}}{{l}}\:+\mathrm{2}=\mathrm{0} \\ $$$$\therefore\:\frac{{a}}{{l}}=\frac{\mathrm{6}+\sqrt{\mathrm{28}}}{\mathrm{2}}=\mathrm{3}+\sqrt{\mathrm{7}}\:\left[\:\because\frac{{a}}{{l}}>\mathrm{1}\:\therefore\:−{ve}\:{rejected}\right] \\ $$$${A}=\frac{{al}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{7}}\right)} \\ $$$${ar}.\:{of}\:{yellow}\:\bigtriangleup={Y}\left({let}\right) \\ $$$${Y}={a}^{\mathrm{2}} −\mathrm{6}{A} \\ $$$$={a}^{\mathrm{2}} −\frac{\mathrm{3}{a}^{\mathrm{2}} }{\left(\mathrm{3}+\sqrt{\mathrm{7}}\right)}=\frac{\sqrt{\mathrm{7}}{a}^{\mathrm{2}} }{\left(\mathrm{3}+\sqrt{\mathrm{7}}\right)} \\ $$$$\:\:\frac{{Y}}{{A}}=\mathrm{2}\sqrt{\mathrm{7}}\Rightarrow{Y}=\mathrm{2}\sqrt{\mathrm{7}}{A} \\ $$$$\therefore{x}=\mathrm{2}\sqrt{\mathrm{7}}\:\Rightarrow{x}^{\mathrm{2}} =\mathrm{28} \\ $$ | ||
Commented by Tawa11 last updated on 10/Mar/22 | ||
$$\mathrm{Great}\:\mathrm{sir} \\ $$ | ||