Question Number 188155 by normans last updated on 26/Feb/23
Answered by HeferH last updated on 26/Feb/23
Commented by HeferH last updated on 26/Feb/23
$$\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{180}}{{k}\:+\:\mathrm{64}} \\ $$$$\:\frac{{c}}{{d}}\:=\:\frac{{k}}{\mathrm{64}} \\ $$$$\frac{{a}}{{c}}\:=\:\frac{{b}}{{d}};\:\frac{{a}}{{b}}\:=\:\frac{{c}}{{d}} \\ $$$$\:\frac{{k}}{\mathrm{64}}\:=\:\frac{\mathrm{180}}{{k}\:+\:\mathrm{64}} \\ $$$$\:{k}\:=\:\mathrm{80} \\ $$$$\:{x}\:=\:\frac{\mathrm{36}\centerdot\mathrm{9}}{\mathrm{4}}\centerdot\mathrm{5}=\:\mathrm{81}\centerdot\mathrm{5}\:=\:\mathrm{405}\: \\ $$$$\:{Blue}\:=\:\mathrm{405}\:+\:\mathrm{80}\:=\:\mathrm{485}\:{u}^{\mathrm{2}} \\ $$
Answered by mr W last updated on 26/Feb/23
Commented by mr W last updated on 26/Feb/23
$$\frac{\mathrm{64}+\mathrm{180}+{B}_{\mathrm{2}} }{\mathrm{64}}=\frac{{B}_{\mathrm{1}} }{{B}_{\mathrm{2}} }=\left(\frac{{b}+{a}}{{a}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{180}}{\mathrm{64}+{B}_{\mathrm{2}} } \\ $$$$\frac{\mathrm{244}+{B}_{\mathrm{2}} }{\mathrm{64}}=\left(\mathrm{1}+\frac{\mathrm{180}}{\mathrm{64}+{B}_{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${B}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{64}{B}_{\mathrm{2}} −\mathrm{64}×\mathrm{180}=\mathrm{0} \\ $$$$\left({B}_{\mathrm{2}} +\mathrm{144}\right)\left({B}_{\mathrm{2}} −\mathrm{80}\right)=\mathrm{0} \\ $$$$\Rightarrow{B}_{\mathrm{2}} =\mathrm{80} \\ $$$$\Rightarrow{B}_{\mathrm{1}} =\frac{\left(\mathrm{244}+{B}_{\mathrm{2}} \right){B}_{\mathrm{2}} }{\mathrm{64}}=\frac{\left(\mathrm{244}+\mathrm{80}\right)×\mathrm{80}}{\mathrm{64}}=\mathrm{405} \\ $$$${B}=\mathrm{405}+\mathrm{80}=\mathrm{485} \\ $$