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Question Number 213745 by efronzo1 last updated on 15/Nov/24

Answered by mnjuly1970 last updated on 16/Nov/24

$$\left(\mathrm{M}ediantheorem\right):7^2+R^2=2{\left(\sqrt{14}\right)}^2+\frac{{\left(2R\right)}^2}{2}$$$$49+R^2=28+2R^2$$$$21=R^2\Rightarrow R=\sqrt{21}◼$$$$$$$$$$

Answered by golsendro last updated on 15/Nov/24

$$\mathrm{\angle }\mathrm{AOC}=\alpha$$$$\left(1\right)\cos \alpha =\frac{{2\mathrm{R}}^2-14}{{2\mathrm{R}}^2}$$$$\left(2\right)\cos \alpha =\frac{{5\mathrm{R}}^2-49}{{4\mathrm{R}}^2}$$$$\Rightarrow \frac{{5\mathrm{R}}^2-49}{2}={2\mathrm{R}}^2-14$$$${\mathrm{R}}^2=21\Rightarrow \mathrm{R}=\sqrt{21}$$

Answered by mr W last updated on 15/Nov/24

$$4\times {\left(\sqrt{14}\right)}^2=2\times 7^2+2\times R^2-{\left(2R\right)}^2$$$$\Rightarrow R=\sqrt{7^2-2\times 14}=\sqrt{21}$$

Answered by A5T last updated on 16/Nov/24

$$R^2+14-2R\sqrt{14}cosACO=R^2\Rightarrow cosACO=\frac{14}{2R\sqrt{14}}$$$$R^2+14+2R\sqrt{14}cosACO=49\Rightarrow cosACO=\frac{35-R^2}{2R\sqrt{14}}$$$$\Rightarrow 35-R^2=14\Rightarrow R=\sqrt{21}$$

Answered by A5T last updated on 16/Nov/24

$$R^2\times 2R+{\left(\sqrt{14}\right)}^2\left(2R\right)=R^2\times R+7^2\times R$$$$\Rightarrow 2R^2+28=R^2+49\Rightarrow R=\sqrt{21}$$

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