Question Number 201599 by mr W last updated on 09/Dec/23
Commented by AST last updated on 09/Dec/23
$${Are}\:{those}\:{numbers}\:{areas}\:{or}\:{lengths}? \\ $$
Commented by mr W last updated on 09/Dec/23
$${they}\:{are}\:{lengths}. \\ $$
Answered by mr W last updated on 10/Dec/23
Commented by mr W last updated on 10/Dec/23
$$\mathrm{3}{x}={y}+{z} \\ $$$$\mathrm{3}{x}+\frac{\mathrm{20}−{p}}{{p}}{y}=\mathrm{3}×\frac{\mathrm{20}−{p}}{\mathrm{3}{p}}{z} \\ $$$$\Rightarrow{y}+{z}=\frac{\left(\mathrm{20}−{p}\right)\left({z}−{y}\right)}{{p}} \\ $$$${x}+\frac{\mathrm{20}−{p}}{\mathrm{3}{p}}{z}=\frac{\mathrm{20}−{p}}{{p}}{y} \\ $$$$\Rightarrow{y}+{z}=\frac{\left(\mathrm{20}−{p}\right)\left(\mathrm{3}{y}−{z}\right)}{{p}} \\ $$$${z}−{y}=\mathrm{3}{y}−{z}\Rightarrow{z}=\mathrm{2}{y}\:\Rightarrow{x}={y} \\ $$$${y}+\mathrm{2}{y}=\frac{\left(\mathrm{20}−{p}\right)\left(\mathrm{2}{y}−{y}\right)}{{p}} \\ $$$$\Rightarrow\mathrm{3}=\frac{\mathrm{20}−{p}}{{p}}\:\Rightarrow{p}=\mathrm{5} \\ $$$$\frac{{BF}}{{FA}}=\frac{{z}}{{y}}=\mathrm{2} \\ $$$$\frac{{BF}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×{BF}}=−\frac{{FA}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×{FA}} \\ $$$${BF}^{\mathrm{2}} =−\mathrm{2}×{FA}^{\mathrm{2}} +\mathrm{78} \\ $$$${FA}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow{FA}=\sqrt{\mathrm{13}}\:\Rightarrow{BA}=\mathrm{3}\sqrt{\mathrm{13}} \\ $$$${BA}^{\mathrm{2}} =\mathrm{9}×\mathrm{13}=\mathrm{117}=\mathrm{9}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow{AD}\bot{BE} \\ $$$${x}=\frac{\mathrm{3}×\mathrm{6}}{\mathrm{2}}=\mathrm{9} \\ $$$$\Delta_{{ABC}} =\frac{\mathrm{20}}{{p}}\left({y}+{z}\right)=\mathrm{12}{x}=\mathrm{12}×\mathrm{9}=\mathrm{108}\:\checkmark \\ $$