Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
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

	$R^{2} - {(\frac{7}{2})}^{2} = 15^{2} - {(R - \frac{7}{2})}^{2}$ $2 R^{2} - 7 R - 225 = 0$ $\Rightarrow R = \frac{25}{2} = 12.5$
	Commented by naka3546 last updated on 10/Dec/23

	$Thank you, sir .$
	Answered by mr W last updated on 10/Dec/23
	$generally (see Q166007) determinant (((R=(√((a^2 +b^2 +c^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [abc((3/(a^2 +b^2 +c^2 )))^(3/2) ]}))) R=(√((7^2 +15^2 +15^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [7×15×15((3/(7^2 +15^2 +15^2 )))^(3/2) ]} =(√((499)/3)) sin {(π/3)+(1/3)sin^(−1) [1575((3/(499)))^(3/2) ]} =12.5$
	$g e n e r a l l y (s e e Q 166007)$ $\begin{matrix} R = \sqrt{\frac{a^{2} + b^{2} + c^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [a b c {(\frac{3}{a^{2} + b^{2} + c^{2}})}^{\frac{3}{2}}]} \end{matrix}$ $R = \sqrt{\frac{7^{2} + 15^{2} + 15^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [7 \times 15 \times 15 {(\frac{3}{7^{2} + 15^{2} + 15^{2}})}^{\frac{3}{2}}]}$ $= \sqrt{\frac{499}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [1575 {(\frac{3}{499})}^{\frac{3}{2}}]}$ $= 12.5$
	Answered by cortano12 last updated on 11/Dec/23

	$\Rightarrow 7^{2} + 15^{2} + 15^{2} + \frac{7.15.15}{r} = {4 r}^{2}$ $\Rightarrow 499 + \frac{1575}{r} = {4 r}^{2}$ $\Rightarrow {4 r}^{3} - 499 r - 1575 = 0$ $\Rightarrow (r + 9) (2 r + 7) (2 r - 25) = 0$ $\Rightarrow r = \frac{25}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum