Question-201065

Question Number 201065 by mr W last updated on 28/Nov/23

Commented by mr W last updated on 28/Nov/23

find the side length a of the square in terms of r.

$${find}\:{the}\:{side}\:{length}\:\boldsymbol{{a}}\:{of}\:{the}\:{square} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{r}}. \\ $$

Answered by ajfour last updated on 29/Nov/23

2rcos θ=(a/2) 2rsin θ=a+p p^2 +(r−(a/2))^2 =r^2 ⇒ ar=p^2 +(a^2 /4) 4r^2 =(a^2 /4)+a^2 (1+(√((r/a)−(1/4))))^2 say (r/a)=t 4t^2 =(1/4)+(1+(√(t−(1/4))))^2 4(t−(1/4))(t+(1/4))=(1+(√(t−(1/4))))^2 say t−(1/4)=z^2 4z^2 (z^2 +(1/2))=1+z^2 +2z 4z^4 +z^2 −2z−1=0 z≈0.83813 t=(1/4)+z^2 ≈ 0.95246 a=(r/t)≈ (r/(0.95246)) ≈ 1.0499r

$$\mathrm{2}{r}\mathrm{cos}\:\theta=\frac{{a}}{\mathrm{2}} \\ $$$$\mathrm{2}{r}\mathrm{sin}\:\theta={a}+{p} \\ $$$${p}^{\mathrm{2}} +\left({r}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{ar}={p}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+{a}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{{r}}{{a}}−\frac{\mathrm{1}}{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$${say}\:\:\frac{{r}}{{a}}={t} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}+\left(\mathrm{1}+\sqrt{{t}−\frac{\mathrm{1}}{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}\left({t}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left({t}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\left(\mathrm{1}+\sqrt{{t}−\frac{\mathrm{1}}{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$${say}\:\:{t}−\frac{\mathrm{1}}{\mathrm{4}}={z}^{\mathrm{2}} \\ $$$$\mathrm{4}{z}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1}+{z}^{\mathrm{2}} +\mathrm{2}{z} \\ $$$$\mathrm{4}{z}^{\mathrm{4}} +{z}^{\mathrm{2}} −\mathrm{2}{z}−\mathrm{1}=\mathrm{0} \\ $$$${z}\approx\mathrm{0}.\mathrm{83813} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{4}}+{z}^{\mathrm{2}} \approx\:\mathrm{0}.\mathrm{95246} \\ $$$${a}=\frac{{r}}{{t}}\approx\:\frac{{r}}{\mathrm{0}.\mathrm{95246}}\:\approx\:\mathrm{1}.\mathrm{0499}{r} \\ $$

Commented by mr W last updated on 29/Nov/23

$${thanks}\:{sir}!\:{excellent}! \\ $$

Answered by Frix last updated on 29/Nov/23

r=1 (√(4−x^2 ))−(√(2x−x^2 ))=2x The usual squaring etc. leads to x^4 −(x^3 /2)−((7x^2 )/8)−(x/2)+(1/2)=0 No useable exact solution. x≈.524956679452 a=2x≈1.0499133589

$${r}=\mathrm{1} \\ $$$$\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }=\mathrm{2}{x} \\ $$$$\mathrm{The}\:\mathrm{usual}\:\mathrm{squaring}\:\mathrm{etc}.\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{4}} −\frac{{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{7}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solution}. \\ $$$${x}\approx.\mathrm{524956679452} \\ $$$${a}=\mathrm{2}{x}\approx\mathrm{1}.\mathrm{0499133589} \\ $$

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