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Question Number 148532 by cherokeesay last updated on 28/Jul/21

Answered by mr W last updated on 29/Jul/21

Commented by mr W last updated on 29/Jul/21

R=6 cm DG=(√2)r+r=R ⇒r=(R/( (√2)+1))=((√2)−1)R DF=(R/( (√2)))=(((√2)R)/2) CF=DF−(√2)r=(((√2)R)/2)−(2−(√2))R=(((3(√2))/2)−2)R AF=(√(r^2 −(((3(√2))/2)−2)^2 R^2 )) AF=R(√(((√2)−1)^2 −(((3(√2))/2)−2)^2 )) AF=(1/2)R(√(16(√2)−22)) AB=2AF=(√(16(√2)−22))R≈0.792R≈4.753

$${R}=\mathrm{6}\:{cm} \\ $$$${DG}=\sqrt{\mathrm{2}}{r}+{r}={R} \\ $$$$\Rightarrow{r}=\frac{{R}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){R} \\ $$$${DF}=\frac{{R}}{\:\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}{R}}{\mathrm{2}} \\ $$$${CF}={DF}−\sqrt{\mathrm{2}}{r}=\frac{\sqrt{\mathrm{2}}{R}}{\mathrm{2}}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){R}=\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{2}\right){R} \\ $$$${AF}=\sqrt{{r}^{\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} {R}^{\mathrm{2}} } \\ $$$${AF}={R}\sqrt{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${AF}=\frac{\mathrm{1}}{\mathrm{2}}{R}\sqrt{\mathrm{16}\sqrt{\mathrm{2}}−\mathrm{22}} \\ $$$${AB}=\mathrm{2}{AF}=\sqrt{\mathrm{16}\sqrt{\mathrm{2}}−\mathrm{22}}{R}\approx\mathrm{0}.\mathrm{792}{R}\approx\mathrm{4}.\mathrm{753} \\ $$

Commented by cherokeesay last updated on 29/Jul/21

I would also like to share my solution ! thank you a lot mr W !

$${I}\:{would}\:{also}\:{like}\:{to}\:{share}\:{my}\:{solution}\:! \\ $$$${thank}\:{you}\:{a}\:{lot}\:{mr}\:{W}\:! \\ $$

Commented by Tawa11 last updated on 29/Jul/21

Nice sir. Thanks.

$$\mathrm{Nice}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$

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