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Question Number 41146 by ajfour last updated on 02/Aug/18

Commented by ajfour last updated on 02/Aug/18

$$Q.41138solution$$

Answered by ajfour last updated on 02/Aug/18

$$64=x^2+y^2...\left(i\right)$$$$144={(12-x)}^2+y^2....\left(ii\right)$$$${AC}^2={(12+x)}^2+y^2....\left(iii\right)$$$$\left(ii\right)+\left(iii\right)-2\left(i\right)gives$$$${AC}^2+144-128=2\times 144$$$$\Rightarrow AC=\sqrt{144+128}=\sqrt{272}$$$$orAC=4\sqrt{17}.$$

Commented by prakash jain last updated on 03/Aug/18

$$\left(ii\right)\mathrm{i}\mathrm{think}\mathrm{should}\mathrm{be}100={(12-x)}^2+y^2$$

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