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Question Number 77390 by Maclaurin Stickker last updated on 05/Jan/20

Commented by Maclaurin Stickker last updated on 05/Jan/20

$$findtheradiusoftheshaded$$$$circumferenceasafunctionof$$$$sideaofthesquare.$$

Answered by jagoll last updated on 06/Jan/20

$$\mathrm{r}(1+\sqrt{2})=\mathrm{a}(\sqrt{2}-1)$$$$\mathrm{r}=\mathrm{a}(3-2\sqrt{2})$$

Commented by mr W last updated on 06/Jan/20

$${it}^{\prime }snotcorrectsir!$$$$thecenterofthesmallcircle{doesn}^{\prime }t$$$$lieonthediagonalofthesquare!$$

Commented by jagoll last updated on 06/Jan/20

$$oyessiriagree$$

Answered by mr W last updated on 06/Jan/20

Commented by mr W last updated on 06/Jan/20

$$AC=a-r$$$$BC=a+r$$$$\sin \beta =\frac{DC}{AC}=\frac{r}{a-r}=\cos \alpha$$$${BC}^2={AC}^2+{AB}^2-2\times AC\times AB\times \cos \alpha$$$${(a+r)}^2={(a-r)}^2+a^2-2(a-r)a\times \frac{r}{a-r}$$$$a^2+r^2+2ar=a^2-2ar+r^2+a^2-2ar$$$$0=a^2-6ar$$$$\Rightarrow r=\frac{a}{6}$$

Commented by mr W last updated on 06/Jan/20

$$or:$$$${AD}^2={AC}^2-{DC}^2={(a-r)}^2-r^2=a^2-2ar$$$${BC}^2={AD}^2+{(AB-DC)}^2$$$${(a+r)}^2=a^2-2ar+{(a-r)}^2$$$$\Rightarrow r=\frac{a}{6}$$

Commented by jagoll last updated on 06/Jan/20

$$howAC=a-rsir?$$

Commented by mr W last updated on 06/Jan/20

Commented by mr W last updated on 06/Jan/20

$$AC=AE-CE=a-r$$

Commented by Maclaurin Stickker last updated on 06/Jan/20

$$Yes,MrW,youareright!$$

Commented by mr W last updated on 06/Jan/20

$$thanksforcomfirmingsir!$$

Commented by Tawa11 last updated on 29/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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