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Question Number 82499 by Maclaurin Stickket last updated on 21/Feb/20

Commented by Maclaurin Stickket last updated on 21/Feb/20

ABCD is a square with side 1.  The circles are congruent.  AC is the diagonal of square.  Find the radius r.

$${ABCD}\:{is}\:{a}\:{square}\:{with}\:{side}\:\mathrm{1}. \\ $$$${The}\:{circles}\:{are}\:{congruent}. \\ $$$${AC}\:{is}\:{the}\:{diagonal}\:{of}\:{square}. \\ $$$${Find}\:{the}\:{radius}\:{r}. \\ $$

Commented by Maclaurin Stickket last updated on 21/Feb/20

I got r=(((√(2(√2)−2))−(√(2(√2)+2))+2)/4), is this  right?

$${I}\:{got}\:{r}=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}−\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}+\mathrm{2}}{\mathrm{4}},\:{is}\:{this} \\ $$$${right}? \\ $$

Commented by mr W last updated on 26/Feb/20

(√(2(√2)+2))−(√(2(√2)−2))=2(√((√2)−1))  r=(((√(2(√2)−2))−(√(2(√2)+2))+2)/4)=((2−2(√((√2)−1)))/4)  =((1−(√((√2)−1)))/2)

$$\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}−\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}=\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$${r}=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}−\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}}+\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{2}−\mathrm{2}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 22/Feb/20

Commented by mr W last updated on 26/Feb/20

(1−((2r)/h))^2 =1−((2R)/h)  h=1  R(2+(√2))=1  R=((2−(√2))/2)  (1−2r)^2 =(√2)−1  ⇒r=((1−(√((√2)−1)))/2)≈0.178

$$\left(\mathrm{1}−\frac{\mathrm{2}{r}}{{h}}\right)^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{2}{R}}{{h}} \\ $$$${h}=\mathrm{1} \\ $$$${R}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)=\mathrm{1} \\ $$$${R}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{r}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{178} \\ $$

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