Question Number 201693 by naka3546 last updated on 10/Dec/23
Answered by mr W last updated on 10/Dec/23
Commented by mr W last updated on 10/Dec/23
$${R}^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} −\left({R}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}^{\mathrm{2}} −\mathrm{7}{R}−\mathrm{225}=\mathrm{0} \\ $$$$\Rightarrow{R}=\frac{\mathrm{25}}{\mathrm{2}}=\mathrm{12}.\mathrm{5} \\ $$
Commented by naka3546 last updated on 10/Dec/23
$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 10/Dec/23
![generally (see Q166007) determinant (((R=(√((a^2 +b^2 +c^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [abc((3/(a^2 +b^2 +c^2 )))^(3/2) ]}))) R=(√((7^2 +15^2 +15^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [7×15×15((3/(7^2 +15^2 +15^2 )))^(3/2) ]} =(√((499)/3)) sin {(π/3)+(1/3)sin^(−1) [1575((3/(499)))^(3/2) ]} =12.5](https://www.tinkutara.com/question/Q201699.png)
$${generally}\:\left({see}\:{Q}\mathrm{166007}\right) \\ $$$$\begin{array}{|c|}{{R}=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left[{abc}\left(\frac{\mathrm{3}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\}}\\\hline\end{array} \\ $$$${R}=\sqrt{\frac{\mathrm{7}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left[\mathrm{7}×\mathrm{15}×\mathrm{15}\left(\frac{\mathrm{3}}{\mathrm{7}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\} \\ $$$$\:\:\:=\sqrt{\frac{\mathrm{499}}{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \left[\mathrm{1575}\left(\frac{\mathrm{3}}{\mathrm{499}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]\right\} \\ $$$$\:\:\:=\mathrm{12}.\mathrm{5} \\ $$
Answered by cortano12 last updated on 11/Dec/23
$$\:\Rightarrow\:\mathrm{7}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\frac{\mathrm{7}.\mathrm{15}.\mathrm{15}}{\mathrm{r}}\:=\:\mathrm{4r}^{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{499}\:+\:\frac{\mathrm{1575}}{\mathrm{r}}\:=\:\mathrm{4r}^{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{4r}^{\mathrm{3}} \:−\mathrm{499r}−\mathrm{1575}\:=\:\mathrm{0} \\ $$$$\:\Rightarrow\left(\mathrm{r}+\mathrm{9}\right)\left(\mathrm{2r}+\mathrm{7}\right)\left(\mathrm{2r}−\mathrm{25}\right)=\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{r}\:=\:\frac{\mathrm{25}}{\mathrm{2}} \\ $$