Question Number 207035 by mr W last updated on 04/May/24
Answered by A5T last updated on 04/May/24
![BE=(√(6^2 +2^2 ))=2(√(10))⇒EF=FB=(√(10)) Let the line through F parallel to BC meet AB,DC at H,I resp.,then BH=3=DI; FH=1 ⇒FI=5; DG=x⇒FG=(√((3−x)^2 +5^2 )) ⇒FG^2 =x^2 −6x+34; EG=(√(4^2 +x^2 )) EG^2 =FE^2 +FG^2 −2FE×FGcos45 ⇒16+x^2 =10+FG^2 −2(√5)FG ⇒16−10−34+6x=−2(√5)FG=6x−28 ⇒6x−28=−2(√5)[(√((3−x)^2 +5^2 ))]⇒x=(1/2)](Q207036.png)
$${BE}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{EF}={FB}=\sqrt{\mathrm{10}} \\ $$$${Let}\:{the}\:{line}\:{through}\:{F}\:{parallel}\:{to}\:{BC}\:{meet}\:{AB},{DC} \\ $$$${at}\:{H},{I}\:{resp}.,{then}\:{BH}=\mathrm{3}={DI};\:{FH}=\mathrm{1} \\ $$$$\Rightarrow{FI}=\mathrm{5};\:{DG}={x}\Rightarrow{FG}=\sqrt{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$$\Rightarrow{FG}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{34};\:{EG}=\sqrt{\mathrm{4}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$${EG}^{\mathrm{2}} ={FE}^{\mathrm{2}} +{FG}^{\mathrm{2}} −\mathrm{2}{FE}×{FGcos}\mathrm{45} \\ $$$$\Rightarrow\mathrm{16}+{x}^{\mathrm{2}} =\mathrm{10}+{FG}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{5}}{FG} \\ $$$$\Rightarrow\mathrm{16}−\mathrm{10}−\mathrm{34}+\mathrm{6}{x}=−\mathrm{2}\sqrt{\mathrm{5}}{FG}=\mathrm{6}{x}−\mathrm{28} \\ $$$$\Rightarrow\mathrm{6}{x}−\mathrm{28}=−\mathrm{2}\sqrt{\mathrm{5}}\left[\sqrt{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\right]\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/May/24
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Answered by mr W last updated on 04/May/24
Commented by mr W last updated on 04/May/24
$$\mathrm{tan}\:\alpha=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$$\mathrm{tan}\:\beta=\mathrm{tan}\:\left(\alpha−\mathrm{45}°\right)=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{1}+\mathrm{3}×\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${GH}=\mathrm{5}\:\mathrm{tan}\:\beta=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$?={DH}−{GH}=\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$
Answered by HeferH24 last updated on 05/May/24
$$\:\frac{\mathrm{4}+{y}}{\mathrm{2}\sqrt{\mathrm{10}}}\:=\:\frac{\sqrt{\mathrm{10}}}{\mathrm{4}}\: \\ $$$$\:\mathrm{16}+\mathrm{4}{y}\:=\:\mathrm{20}\:\:\:;\:{y}=\mathrm{1}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{5}}\:=\:\frac{{x}}{\mathrm{3}−{x}}\:;\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by HeferH24 last updated on 05/May/24
