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Question Number 199358 by ajfour last updated on 01/Nov/23

Commented by ajfour last updated on 01/Nov/23

$$FindR.$$

Answered by mr W last updated on 01/Nov/23

$$(R+R+1+\sqrt{{(R+1)}^2+R^2})\times 1=R(R+1)$$$$\sqrt{{(R+1)}^2+R^2}=R^2-R-1$$$$R^2-2R-3=0$$$$(R-3)(R+1)=0$$$$\Rightarrow R=3$$

Answered by Frix last updated on 01/Nov/23

$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{exactly}\mathrm{one}\mathrm{rectangular}\mathrm{triangle}$$$$\mathrm{with}\mathrm{sides}a,b=a+1\mathrm{and}\mathrm{incircle}\mathrm{of}\mathrm{radius}$$$$\mathrm{equal}\mathrm{to}1:$$$$a=R=3,b=R+1=4,c=5,r_{\mathrm{I}}=1$$

Commented by ajfour last updated on 01/Nov/23

$$yes,thanks.simpleoneforyou.$$

Answered by cortano12 last updated on 02/Nov/23

$${(2\mathrm{R}-1)}^2={(\mathrm{R}+1)}^2+{\mathrm{R}}^2$$$${4\mathrm{R}}^2-4\mathrm{R}+1={2\mathrm{R}}^2+2\mathrm{R}+1$$$${2\mathrm{R}}^2-6\mathrm{R}=0$$$$2\mathrm{R}(\mathrm{R}-3)=0$$$$\mathrm{R}=3$$

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