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Question-70298




Question Number 70298 by naka3546 last updated on 03/Oct/19
Answered by mr W last updated on 03/Oct/19
Commented by mr W last updated on 03/Oct/19
CE=(√((x+1)^2 −x^2 ))=(√(2x+1))  ((CE)/(ED))=((CA)/(AB))  ⇒((√(2x+1))/x)=((2x+1)/(12))  ⇒(1/x)=((√(2x+1))/(12))  ⇒x(√(2x+1))=12  ⇒x^2 (2x+1)=144  ⇒2x^3 +x^2 −144=0  ⇒(x−4)(2x^2 +9x+36)=0  ⇒x=4  ⇒diameter=8
$${CE}=\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\frac{{CE}}{{ED}}=\frac{{CA}}{{AB}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{{x}}=\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{12}} \\ $$$$\Rightarrow{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}=\mathrm{12} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{144} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{144}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{4}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{36}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$$$\Rightarrow{diameter}=\mathrm{8} \\ $$

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