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Question Number 40255 by ajfour last updated on 17/Jul/18

Commented by MJS last updated on 17/Jul/18

$$\mathrm{are}\mathrm{the}\mathrm{trapezoids}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{similar}?$$

Commented by MJS last updated on 18/Jul/18

$$...\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{we}\mathrm{have}$$$$\frac{y+\frac{1}{x}}{1}=\frac{\sqrt{\frac{1}{x^2}+1}}{\frac{x}{\sqrt{1+x^2}}}$$$$y=\sqrt{\frac{\frac{x^2+1}{x^2}}{\frac{x^2}{x^2+1}}}-\frac{1}{x}=\frac{x^2+1}{x}-\frac{1}{x}$$$$c=\frac{x^3+x^2+1}{x^2}$$$$\frac{x^3+x^2+1}{x^2}=\frac{x^3+1}{x}$$$$x^4-x^3-x^2+x-1=0$$$$x\approx 1.51288$$$$c\approx 2.94979$$

Answered by MJS last updated on 17/Jul/18

$$△(\frac{1}{x};1;\sqrt{\frac{1}{x^2}+1})$$$$△(1;x;\sqrt{1+x^2})$$$$△(\sqrt{\frac{1}{x^2}+1};\sqrt{1+x^2};\frac{1}{x}+x)$$$$△(x;x+y;\sqrt{x^2+{(x+y)}^2})$$$$c=\frac{1}{x}+x+y$$$$\frac{x+y}{x}=\frac{x}{1}$$$$y=x^2-x$$$$c=\frac{1}{x}+x^2$$$$x^3-cx+1=0$$$$\left(\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{this}\mathrm{for}\mathrm{given}c\right)$$

Commented by ajfour last updated on 17/Jul/18

$$whatmorelinestoconstruct$$$$tolocatex;isthatiwantto$$$$knowandcouldnotfind..Sir?$$

Commented by MJS last updated on 17/Jul/18

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{not}\mathrm{unique}.\mathrm{you}\mathrm{need}$$$$\mathrm{another}\mathrm{indication}$$

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