Menu Close

Question-172195




Question Number 172195 by cherokeesay last updated on 24/Jun/22
Answered by som(math1967) last updated on 24/Jun/22
let side of biggest square=rad. of  quarter circle =r unit  ∴ AB=AL=r  ∴ side of S_2 =(r/( (√2)))   AC=r(√2)  ∴LC=r(√2)−r=r((√2)−1)  ∴ side of S_1 =((r((√2)−1))/( (√2)))    (S_1 /S_2 )=((ar.of S_1 )/(ar. of S_2 ))=(((r^2 ((√2)−1)^2 )/2)/(r^2 /2))        =((3−2(√2))/1)=17.15% (approx)
$${let}\:{side}\:{of}\:{biggest}\:{square}={rad}.\:{of} \\ $$$${quarter}\:{circle}\:={r}\:{unit} \\ $$$$\therefore\:{AB}={AL}={r} \\ $$$$\therefore\:{side}\:{of}\:{S}_{\mathrm{2}} =\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:{AC}={r}\sqrt{\mathrm{2}} \\ $$$$\therefore{LC}={r}\sqrt{\mathrm{2}}−{r}={r}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\therefore\:{side}\:{of}\:{S}_{\mathrm{1}} =\frac{{r}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }=\frac{{ar}.{of}\:{S}_{\mathrm{1}} }{{ar}.\:{of}\:{S}_{\mathrm{2}} }=\frac{\frac{{r}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}}{\frac{{r}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{1}}=\mathrm{17}.\mathrm{15\%}\:\left({approx}\right) \\ $$
Commented by som(math1967) last updated on 24/Jun/22
Commented by cherokeesay last updated on 24/Jun/22
thank sir !
$${thank}\:{sir}\:! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *