Question Number 207482 by cherokeesay last updated on 17/May/24
Answered by mr W last updated on 17/May/24
$${x}\left({a}−{x}\right)={yb}={zc} \\ $$$$\Rightarrow{y}=\frac{{x}\left({a}−{x}\right)}{{b}} \\ $$$$\Rightarrow{z}=\frac{{x}\left({a}−{x}\right)}{{c}} \\ $$$${d}^{\mathrm{2}} ={y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{yz}×\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$${d}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$\left(\frac{\sqrt{\mathrm{11}}{a}}{\mathrm{12}}\right)^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} {a}^{\mathrm{2}} }{{b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }=\frac{\mathrm{11}}{\mathrm{144}} \\ $$$$\frac{{yellow}+{blue}}{{yellow}}=\frac{{bc}}{{yz}}=\frac{{b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({a}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{144}}{\mathrm{11}} \\ $$$$\Rightarrow\frac{{blue}}{{yellow}}=\frac{\mathrm{144}}{\mathrm{11}}−\mathrm{1}=\frac{\mathrm{133}}{\mathrm{11}}\:\checkmark \\ $$
Commented by cherokeesay last updated on 17/May/24
$${Great}\:!\:{thank}\:{you}\:{master}\:! \\ $$