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Question Number 139053 by mathocean1 last updated on 21/Apr/21

Commented by mathocean1 last updated on 21/Apr/21

Determinate R

$${Determinate}\:{R} \\ $$

Answered by mr W last updated on 21/Apr/21

x^2 +30^2 =R^2    ...(i)  (x+d)^2 +24^2 =R^2    ...(ii)  (x+2d)^2 +12^2 =R^2    ...(iii)  (iii)−(ii):  d(2x+3d)−12×36=0   ...(iv)  (ii)−(i):  d(2x+d)−6×54=0   ...(v)  (iv)−(v):  2d^2 −12×36+6×54=0  ⇒d=3(√6)  3(√6)(2x+3(√6))=6×54  ⇒x=((15(√6))/2)  (((15(√6))/2))^2 +30^2 =R^2   ⇒R=((15(√(22)))/2)≈35.178

$${x}^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$\left({x}+{d}\right)^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\left({x}+\mathrm{2}{d}\right)^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:...\left({iii}\right) \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$${d}\left(\mathrm{2}{x}+\mathrm{3}{d}\right)−\mathrm{12}×\mathrm{36}=\mathrm{0}\:\:\:...\left({iv}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${d}\left(\mathrm{2}{x}+{d}\right)−\mathrm{6}×\mathrm{54}=\mathrm{0}\:\:\:...\left({v}\right) \\ $$$$\left({iv}\right)−\left({v}\right): \\ $$$$\mathrm{2}{d}^{\mathrm{2}} −\mathrm{12}×\mathrm{36}+\mathrm{6}×\mathrm{54}=\mathrm{0} \\ $$$$\Rightarrow{d}=\mathrm{3}\sqrt{\mathrm{6}} \\ $$$$\mathrm{3}\sqrt{\mathrm{6}}\left(\mathrm{2}{x}+\mathrm{3}\sqrt{\mathrm{6}}\right)=\mathrm{6}×\mathrm{54} \\ $$$$\Rightarrow{x}=\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{15}\sqrt{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{15}\sqrt{\mathrm{22}}}{\mathrm{2}}\approx\mathrm{35}.\mathrm{178} \\ $$

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