Question-40255

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

Commented by MJS last updated on 17/Jul/18

are the trapezoids on the right similar ?

Commented by MJS last updated on 18/Jul/18

\dots in this case we 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} - c x + 1 = 0

(we can solve this for given c)

Commented by ajfour last updated on 17/Jul/18

w h a t m o r e l i n e s t o c o n s t r u c t

t o l o c a t e x; i s t h a t i w a n t t o

k n o w a n d c o u l d n o t f i n d . . S i r ?

Commented by MJS last updated on 17/Jul/18

I {don}^{'} t know . {it}^{'} s just not unique . you need

another indication

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