Question Number 41146 by ajfour last updated on 02/Aug/18
Commented by ajfour last updated on 02/Aug/18
$${Q}.\mathrm{41138}\:\:\:{solution} \\ $$
Answered by ajfour last updated on 02/Aug/18
$$\mathrm{64}\:=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{144}=\:\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:….\left({ii}\right) \\ $$$${AC}^{\:\mathrm{2}} =\:\left(\mathrm{12}+{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:\:\:\:….\left({iii}\right) \\ $$$$\left({ii}\right)+\left({iii}\right)−\mathrm{2}\left({i}\right)\:\:{gives} \\ $$$${AC}^{\:\mathrm{2}} +\mathrm{144}−\mathrm{128}\:=\:\mathrm{2}×\mathrm{144} \\ $$$$\Rightarrow\:\:\:{AC}\:=\:\sqrt{\mathrm{144}+\mathrm{128}}\:=\sqrt{\mathrm{272}} \\ $$$$\:\:{or}\:\:\:{AC}=\:\mathrm{4}\sqrt{\mathrm{17}}\:.\: \\ $$
Commented by prakash jain last updated on 03/Aug/18
$$\left({ii}\right)\:\mathrm{i}\:\mathrm{think}\:\mathrm{should}\:\mathrm{be}\:\mathrm{100}=\left(\mathrm{12}−{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$