Question Number 221907 by alvan545 last updated on 12/Jun/25
Answered by fantastic last updated on 12/Jun/25
Commented by fantastic last updated on 12/Jun/25
![CD^2 =CN×CF ⇒18^2 =x×27 or x=((18×18)/(27))=12 So NA=12 and AF=27−12−12=3 ∴NA=12 and AF=3 now MA×AE=FA×AN so y×AE=3×12 or AE=((36)/y) Now BD^2 =BM×BE or (6(√3))^2 =y×(y+y+((36)/y)) or y(((2y^2 +36)/y))=((√(36×3)))^2 or 2y^2 +36=108 or 2y^2 =108−36=72 or y^2 =((72)/2)=36 So y=(√(36))=6[−6<0 denied]_(we can now find BE) BE=BM+MA+ME=y+y+((36)/y) =6+6+((36)/6)=12+6=18 So the final answer is ⇒BE=18](https://www.tinkutara.com/question/Q221909.png)
$${CD}^{\mathrm{2}} ={CN}×{CF} \\ $$$$\Rightarrow\mathrm{18}^{\mathrm{2}} ={x}×\mathrm{27}\: \\ $$$${or}\:{x}=\frac{\mathrm{18}×\mathrm{18}}{\mathrm{27}}=\mathrm{12} \\ $$$${So}\:{NA}=\mathrm{12}\:{and}\:{AF}=\mathrm{27}−\mathrm{12}−\mathrm{12}=\mathrm{3} \\ $$$$\therefore{NA}=\mathrm{12}\:{and}\:{AF}=\mathrm{3} \\ $$$${now}\:{MA}×{AE}={FA}×{AN} \\ $$$${so}\:{y}×{AE}=\mathrm{3}×\mathrm{12} \\ $$$${or}\:{AE}=\frac{\mathrm{36}}{{y}} \\ $$$${Now}\:{BD}^{\mathrm{2}} ={BM}×{BE} \\ $$$${or}\:\left(\mathrm{6}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={y}×\left({y}+{y}+\frac{\mathrm{36}}{{y}}\right) \\ $$$${or}\:{y}\left(\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{36}}{{y}}\right)=\left(\sqrt{\mathrm{36}×\mathrm{3}}\right)^{\mathrm{2}} \\ $$$${or}\:\mathrm{2}{y}^{\mathrm{2}} +\mathrm{36}=\mathrm{108} \\ $$$${or}\:\mathrm{2}{y}^{\mathrm{2}} =\mathrm{108}−\mathrm{36}=\mathrm{72} \\ $$$${or}\:{y}^{\mathrm{2}} =\frac{\mathrm{72}}{\mathrm{2}}=\mathrm{36} \\ $$$$\underset{{we}\:{can}\:{now}\:{find}\:{BE}} {\underbrace{\boldsymbol{{So}}\:\boldsymbol{{y}}=\sqrt{\mathrm{36}}=\mathrm{6}\left[−\mathrm{6}<\mathrm{0}\:\boldsymbol{{denied}}\right]}} \\ $$$${BE}={BM}+{MA}+{ME}={y}+{y}+\frac{\mathrm{36}}{{y}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{6}+\mathrm{6}+\frac{\mathrm{36}}{\mathrm{6}}=\mathrm{12}+\mathrm{6}=\mathrm{18} \\ $$$${So}\:{the}\:{final}\:{answer}\:{is}\:\Rightarrow{BE}=\mathrm{18} \\ $$
Answered by mr W last updated on 12/Jun/25
Commented by mr W last updated on 12/Jun/25
$${a}\left(\mathrm{2}{a}+{b}\right)=\left(\mathrm{6}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{2}} +{ab}=\mathrm{108} \\ $$$$\mathrm{2}{c}+{d}=\mathrm{27} \\ $$$${c}\left(\mathrm{2}{c}+{d}\right)=\mathrm{18}^{\mathrm{2}} \\ $$$$\Rightarrow{c}=\frac{\mathrm{18}^{\mathrm{2}} }{\mathrm{27}}=\mathrm{12} \\ $$$$\Rightarrow{d}=\mathrm{27}−\mathrm{2}×\mathrm{12}=\mathrm{3} \\ $$$${ab}={cd}=\mathrm{12}×\mathrm{3}=\mathrm{36} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{36}=\mathrm{108} \\ $$$$\Rightarrow{a}=\mathrm{6} \\ $$$$\Rightarrow{b}=\frac{\mathrm{36}}{\mathrm{6}}=\mathrm{6} \\ $$$${BE}=\mathrm{2}{a}+{b}=\mathrm{18} \\ $$