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Question-213492

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Question Number 213492 by a.lgnaoui last updated on 06/Nov/24
Commented by a.lgnaoui last updated on 07/Nov/24
AH=4+2r , 6^2 =(4+r)^2 −r^2
$$\mathrm{AH}=\mathrm{4}+\mathrm{2r}\:\:\:,\:\mathrm{6}^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \\ $$
Commented by A5T last updated on 07/Nov/24
r=2.5⇒R=4.5⇒AC=7.5 ⇒sin∠ACI=0.8⇒cos∠ACI=0.6 ⇒cos(∠BCD)=cos(180−∠ACI)=−cos∠ACI ⇒cos(∠BCD)=−0.6 BD=(√(12^2 +2.5^2 ))=((√(601))/2);CD=7.5 cos(∠BCD)=((((601)/4)−7.5^2 −7^2 )/(−2×7.5×7))=((−3)/7) This contradicts cos(∠BCD)=−(3/5) gotten above. So, the diagram still remains impossible and cannot be constructed.
$${r}=\mathrm{2}.\mathrm{5}\Rightarrow{R}=\mathrm{4}.\mathrm{5}\Rightarrow{AC}=\mathrm{7}.\mathrm{5} \\ $$$$\Rightarrow{sin}\angle{ACI}=\mathrm{0}.\mathrm{8}\Rightarrow{cos}\angle{ACI}=\mathrm{0}.\mathrm{6} \\ $$$$\Rightarrow{cos}\left(\angle{BCD}\right)={cos}\left(\mathrm{180}−\angle{ACI}\right)=−{cos}\angle{ACI} \\ $$$$\Rightarrow{cos}\left(\angle{BCD}\right)=−\mathrm{0}.\mathrm{6} \\ $$$${BD}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{2}.\mathrm{5}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{601}}}{\mathrm{2}};{CD}=\mathrm{7}.\mathrm{5} \\ $$$${cos}\left(\angle{BCD}\right)=\frac{\frac{\mathrm{601}}{\mathrm{4}}−\mathrm{7}.\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{−\mathrm{2}×\mathrm{7}.\mathrm{5}×\mathrm{7}}=\frac{−\mathrm{3}}{\mathrm{7}} \\ $$$${This}\:{contradicts}\:{cos}\left(\angle{BCD}\right)=−\frac{\mathrm{3}}{\mathrm{5}}\:{gotten}\:{above}. \\ $$$${So},\:{the}\:{diagram}\:{still}\:{remains}\:{impossible}\:{and} \\ $$$${cannot}\:{be}\:{constructed}. \\ $$
Answered by a.lgnaoui last updated on 08/Nov/24

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