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Question Number 208288 by efronzo1 last updated on 10/Jun/24

Answered by A5T last updated on 10/Jun/24

$$8(8+x)=\left(2r\right)(2r+2R)=4r^2+4rR$$$$\Rightarrow x=\frac{r^2+rR-16}{2}...\left(i\right)$$$${(8+x)}^2=R^2+{(2r+R)}^2=2R^2+4rR+4r^2$$$$\Rightarrow x=\sqrt{2R^2+4rR+4r^2}-8...\left(ii\right)$$$$BC=\sqrt{4r^2-25}$$$$\sqrt{{BC}^2+{BE}^2}=EC=\sqrt{R^2+R^2}=R\sqrt{2}$$$$\Rightarrow 2R^2=4r^2-25+{(3+x)}^2$$$$\Rightarrow x=\sqrt{2R^2-4r^2+25}-3...\left(iii\right)$$$$\frac{r^2+rR-16}{2}=\sqrt{2R^2+4rR+4r^2}-8=\sqrt{2R^2-4r^2+25}-3$$$$\Rightarrow r=\frac{\sqrt{34}}{2};R=\frac{3\sqrt{34}}{2}\Rightarrow x=9$$

Answered by mr W last updated on 10/Jun/24

Commented by mr W last updated on 12/Jun/24

$$\frac{x/2}{a}=\frac{a}{x+3+5}$$$$\Rightarrow 2a^2=x(x+8)$$$${\left(2b\right)}^2-5^2={\left(\sqrt{2}a\right)}^2-{(x+3)}^2$$$$\Rightarrow 4b^2=25+x(x+8)-{(x+3)}^2$$$$\Rightarrow 2b^2=8+x$$$$\frac{2b}{5}=\frac{x+3+5}{a+2b}$$$$\Rightarrow 5(x+8)=2ab+4b^2$$$$\Rightarrow 5(x+8)=2ab+16+2x$$$$\Rightarrow 2ab=3(x+8)$$$$\Rightarrow 4a^2{b}^2=9{(x+8)}^2$$$$\Rightarrow x(x+8)(8+x)=9{(x+8)}^2$$$$\Rightarrow x=9✓$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Nice}\mathrm{sir}.$$$$\mathrm{I}\mathrm{am}\mathrm{learning}\mathrm{questions}\mathrm{like}\mathrm{this}.$$

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