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Question-201599




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

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