Question Number 225449 by efronzo1 last updated on 26/Oct/25
Answered by fantastic last updated on 26/Oct/25
Commented by fantastic last updated on 26/Oct/25
![AG×AH=AE^2 AH=(((25−r)^2 )/2) GH=(((25−r)^2 )/2)−2=(((25−r)^2 −2^2 )/2) GH/2=(((25−r)^2 −2^z )/4) IO=(√(r^2 −((((25−r)^2 −2^2 )/4))^2 )) IO+r=25 IO=25−r IO^2 =625−50r+r^2 r^2 −((((25−r)^2 −2^2 )/4))^2 =(25−r)^2 Solving gives r=17 and r=37 r=17 [∵25−37<0 rejected]](https://www.tinkutara.com/question/Q225456.png)
$$ \\ $$$${AG}×{AH}={AE}^{\mathrm{2}} \\ $$$${AH}=\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${GH}=\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}=\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${GH}/\mathrm{2}=\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} −\mathrm{2}^{{z}} }{\mathrm{4}} \\ $$$${IO}=\sqrt{{r}^{\mathrm{2}} −\left(\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$${IO}+{r}=\mathrm{25} \\ $$$${IO}=\mathrm{25}−{r} \\ $$$${IO}^{\mathrm{2}} =\mathrm{625}−\mathrm{50}{r}+{r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} −\left(\frac{\left(\mathrm{25}−{r}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} =\left(\mathrm{25}−{r}\right)^{\mathrm{2}} \\ $$$${Solving}\:{gives}\:{r}=\mathrm{17}\:{and}\:{r}=\mathrm{37} \\ $$$${r}=\mathrm{17}\:\left[\because\mathrm{25}−\mathrm{37}<\mathrm{0}\:{rejected}\right] \\ $$
Answered by gregori last updated on 30/Oct/25
$$\:\:{R}^{\mathrm{2}} \:=\:\left(\mathrm{25}−{R}\right)^{\mathrm{2}} +\left({R}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{2}{R}−\mathrm{2}\right).\mathrm{2}\:=\:{R}^{\mathrm{2}} −\mathrm{50}{R}+\mathrm{625}\: \\ $$$$\:\:{R}^{\mathrm{2}} −\mathrm{54}{R}+\mathrm{629}\:=\:\mathrm{0} \\ $$$$\:\:\left({R}−\mathrm{37}\right)\left({R}−\mathrm{17}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\begin{array}{|c|}{{R}=\mathrm{17}\:{cm}}\\\hline\end{array} \\ $$$$\:\:\: \\ $$