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Question Number 172412 by cherokeesay last updated on 26/Jun/22

Answered by aleks041103 last updated on 26/Jun/22

$$distancefromunknowncenterto$$$$leftsideisr.$$$$distancefromcentertotopsideisd.$$$$\Rightarrow {(6-d)}^2+{(6-r)}^2={(6+r)}^2$$$${(6-d)}^2+r^2={(6-r)}^2$$$${(6-d)}^2=(6+r-6+r)(6+r+6-r)=$$$$=24r$$$${(6-d)}^2={(6-r)}^2-r^2=36-12r$$$$\Rightarrow 36=36r$$$$\Rightarrow r=1$$

Commented by aleks041103 last updated on 26/Jun/22

$${(6-d)}^2={(6-1)}^2-1^2=24$$$$\Rightarrow 6-d=\pm 2\sqrt{6}$$$$\Rightarrow d=6\pm 2\sqrt{6}$$$$d<6$$$$\Rightarrow d=6-2\sqrt{6}$$

Commented by cherokeesay last updated on 26/Jun/22

$$Nice!thankyou.$$

Commented by Tawa11 last updated on 26/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by uochiton last updated on 27/Jun/22

$${(6+r)}^2={(6-r)}^2+{(6-d)}^2$$$$$$$${(6+r)}^2-{(6-r)}^2={(6-d)}^2$$$$[(6+r)+(6-r)]\cdot [(6+r)-(6-r)]={(6-d)}^2$$$$12\cdot 2\cdot r={(6-d)}^2$$$$24\cdot r={(6-d)}^2$$$$$$$${(6-r)}^2=r^2+{(6-d)}^2$$$${(6-r)}^2=r^2+24\cdot r$$$$36-12\cdot r+r^2=r^2+24\cdot r$$$$36\cdot r=36$$$$r=1$$

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