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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

$$\mathrm{AH}=4+2\mathrm{r},6^2={(4+\mathrm{r})}^2-{\mathrm{r}}^2$$

Commented by A5T last updated on 07/Nov/24

$$r=2.5\Rightarrow R=4.5\Rightarrow AC=7.5$$$$\Rightarrow sin\mathrm{\angle }ACI=0.8\Rightarrow cos\mathrm{\angle }ACI=0.6$$$$\Rightarrow cos\left(\mathrm{\angle }BCD\right)=cos(180-\mathrm{\angle }ACI)=-cos\mathrm{\angle }ACI$$$$\Rightarrow cos\left(\mathrm{\angle }BCD\right)=-0.6$$$$BD=\sqrt{{12}^2+2.5^2}=\frac{\sqrt{601}}{2};CD=7.5$$$$cos\left(\mathrm{\angle }BCD\right)=\frac{\frac{601}{4}-7.5^2-7^2}{-2\times 7.5\times 7}=\frac{-3}{7}$$$$Thiscontradictscos\left(\mathrm{\angle }BCD\right)=-\frac{3}{5}gottenabove.$$$$So,thediagramstillremainsimpossibleand$$$$cannotbeconstructed.$$

Answered by a.lgnaoui last updated on 08/Nov/24

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