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?

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

Commented by MJS last updated on 18/Jul/18

...in this case we have ((y+(1/x))/1)=((√((1/x^2 )+1))/(x/( (√(1+x^2 ))))) y=(√(((x^2 +1)/x^2 )/(x^2 /(x^2 +1))))−(1/x)=((x^2 +1)/x)−(1/x) c=((x^3 +x^2 +1)/x^2 ) ((x^3 +x^2 +1)/x^2 )=((x^3 +1)/x) x^4 −x^3 −x^2 +x−1=0 x≈1.51288 c≈2.94979

$$…\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{{y}+\frac{\mathrm{1}}{{x}}}{\mathrm{1}}=\frac{\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}}}{\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}} \\ $$$${y}=\sqrt{\frac{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} }}{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}}}−\frac{\mathrm{1}}{{x}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}} \\ $$$${c}=\frac{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}\approx\mathrm{1}.\mathrm{51288} \\ $$$${c}\approx\mathrm{2}.\mathrm{94979} \\ $$

Answered by MJS last updated on 17/Jul/18

△((1/x); 1; (√((1/x^2 )+1))) △(1; x; (√(1+x^2 ))) △((√((1/x^2 )+1)); (√(1+x^2 )); (1/x)+x) △(x; x+y; (√(x^2 +(x+y)^2 ))) c=(1/x)+x+y ((x+y)/x)=(x/1) y=x^2 −x c=(1/x)+x^2 x^3 −cx+1=0 (we can solve this for given c)

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

Commented by ajfour last updated on 17/Jul/18

$${what}\:{more}\:{lines}\:{to}\:{construct} \\ $$$${to}\:{locate}\:{x}\:;\:{is}\:{that}\:{i}\:{want}\:{to} \\ $$$${know}\:{and}\:{could}\:{not}\:{find}..{Sir}\:? \\ $$

Commented by MJS last updated on 17/Jul/18

I don′t know. it′s just not unique. you need another indication

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

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