Question Number 206956 by mr W last updated on 01/May/24
Commented by mr W last updated on 01/May/24
$${find}\:{the}\:{radius}\:{of}\:{circle}. \\ $$
Answered by Frix last updated on 01/May/24
$$\mathrm{3}\:\mathrm{points}\:\mathrm{given} \\ $$$$\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:\:\:\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{6}}\end{pmatrix}\:\:\:\:\:\begin{pmatrix}{\mathrm{12}}\\{\mathrm{12}}\end{pmatrix} \\ $$$$\Rightarrow\:\mathrm{Triangle}\:\mathrm{with}\:\mathrm{sides} \\ $$$$\mathrm{6}\:\:\:\:\:\mathrm{6}\sqrt{\mathrm{5}}\:\:\:\:\:\mathrm{12}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{R}=\mathrm{3}\sqrt{\mathrm{10}} \\ $$
Commented by mr W last updated on 01/May/24
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Answered by mr W last updated on 01/May/24
Commented by mr W last updated on 01/May/24
$${BC}=\sqrt{\left(\mathrm{6}+\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{6}\sqrt{\mathrm{5}} \\ $$$${R}=\frac{{BC}}{\mathrm{2}\:\mathrm{sin}\:{A}}=\frac{\mathrm{6}\sqrt{\mathrm{5}}}{\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\mathrm{3}\sqrt{\mathrm{10}} \\ $$