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Question-185387




Question Number 185387 by Rupesh123 last updated on 20/Jan/23
Answered by HeferH last updated on 21/Jan/23
Commented by HeferH last updated on 21/Jan/23
x = (√((4−r)^2 −r^2 )) ⇒   [(√((4−r)^2 −r^2 )) + (5/2)]^2 + r^2  = ((3/2) + r)^2    r = ((240)/(121))
$${x}\:=\:\sqrt{\left(\mathrm{4}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\:\left[\sqrt{\left(\mathrm{4}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}}\right]^{\mathrm{2}} +\:{r}^{\mathrm{2}} \:=\:\left(\frac{\mathrm{3}}{\mathrm{2}}\:+\:{r}\right)^{\mathrm{2}} \\ $$$$\:{r}\:=\:\frac{\mathrm{240}}{\mathrm{121}}\: \\ $$
Answered by mr W last updated on 21/Jan/23
(−(1/4)+(1/(1.5))+(2/r))^2 =2((1/4^2 )+(1/(1.5^2 ))+(2/r^2 ))  ⇒r=((240)/(121))
$$\left(−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{1}.\mathrm{5}}+\frac{\mathrm{2}}{{r}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}.\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{2}}{{r}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{r}=\frac{\mathrm{240}}{\mathrm{121}} \\ $$

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