Question Number 222624 by Lekhraj last updated on 02/Jul/25
Answered by mr W last updated on 02/Jul/25
Commented by mr W last updated on 02/Jul/25
$${s}={side}\:{length}\:{of}\:{square} \\ $$$${PB}=\sqrt{{x}^{\mathrm{2}} −{s}^{\mathrm{2}} } \\ $$$$\frac{{CP}}{{y}}=\frac{{s}}{{x}}\:\Rightarrow{CP}=\frac{{ys}}{{x}} \\ $$$${CP}+{PB}={s} \\ $$$$\frac{{ys}}{{x}}+\sqrt{{x}^{\mathrm{2}} −{s}^{\mathrm{2}} }={s} \\ $$$${x}^{\mathrm{2}} −{s}^{\mathrm{2}} =\left(\mathrm{1}−\frac{{y}}{{x}}\right)^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{1}−\frac{{y}}{{x}}\right)^{\mathrm{2}} }={area}\:{of}\:{square} \\ $$
Commented by Lekhraj last updated on 02/Jul/25
$$\mathrm{Thank}\:\mathrm{you}\:.\: \\ $$