Question-222072

Question Number 222072 by fantastic last updated on 16/Jun/25

Commented by fantastic last updated on 16/Jun/25

$${If}\:{the}\:{side}\:{length}\:{of}\:{the}\:{square}\:{is}\:{a} \\ $$$$\:{what}\:{is}\:{the}\:{overlapping}\:{area}?? \\ $$

Commented by fantastic last updated on 16/Jun/25

$$\int\:{not}\:{allowed} \\ $$

Answered by mr W last updated on 16/Jun/25

Commented by mr W last updated on 16/Jun/25

a=2r α=tan^(−1) (1/2) A_1 =(((2r)^2 )/2)(2α−sin 2α)+(r^2 /2)(π−2α−sin 2α) =(3α+(π/2))r^2 −((5r^2 sin 2α)/2) =(3 tan^(−1) (1/2)+(π/2)−2)(a^2 /4)≈0.240435a^2 A_2 =(((2r)^2 )/(2 tan 2α))+r^2 tan α−(((2r)^2 )/2)((π/2)−2α)−((2αr^2 )/2) =((2/(tan 2α))+tan α−π+3α)r^2 =(2−π+3 tan^(−1) (1/2))(a^2 /4)≈0.062337a^2 A_1 +A_2 =(1/8)(12 tan^(−1) (1/2)−π)a^2 ≈0.302772a^2

$${a}=\mathrm{2}{r} \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha\right)+\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\pi−\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{3}\alpha+\frac{\pi}{\mathrm{2}}\right){r}^{\mathrm{2}} −\frac{\mathrm{5}{r}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\left(\mathrm{3}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}−\mathrm{2}\right)\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\approx\mathrm{0}.\mathrm{240435}{a}^{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} =\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}\:\mathrm{2}\alpha}+{r}^{\mathrm{2}} \mathrm{tan}\:\alpha−\frac{\left(\mathrm{2}{r}\right)^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha\right)−\frac{\mathrm{2}\alpha{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:=\left(\frac{\mathrm{2}}{\mathrm{tan}\:\mathrm{2}\alpha}+\mathrm{tan}\:\alpha−\pi+\mathrm{3}\alpha\right){r}^{\mathrm{2}} \\ $$$$\:\:=\left(\mathrm{2}−\pi+\mathrm{3}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\approx\mathrm{0}.\mathrm{062337}{a}^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{12}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\pi\right){a}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{302772}{a}^{\mathrm{2}} \\ $$

Commented by fantastic last updated on 17/Jun/25

$${I}\:{think}\:{i}\:{did}\:{it}\:{in}\:{more}\:{harder}\:{way} \\ $$

Commented by fantastic last updated on 17/Jun/25

Commented by fantastic last updated on 17/Jun/25

I got Area=[(((180^0 −θ)/(360^0 ))×π((a/2))^2 −(1/2)((a/2))^2 sin (180^0 −θ))] +[((θ/(360^0 ))π(a)^2 −(1/2)a^2 sin θ)] Where θ=tan^(−1) ((4/3))

$${I}\:{got}\: \\ $$$${Area}=\left[\left(\frac{\mathrm{180}^{\mathrm{0}} −\theta}{\mathrm{360}^{\mathrm{0}} }×\pi\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{180}^{\mathrm{0}} −\theta\right)\right)\right] \\ $$$$+\left[\left(\frac{\theta}{\mathrm{360}^{\mathrm{0}} }\pi\left({a}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} \mathrm{sin}\:\theta\right)\right] \\ $$$${Where}\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$

Commented by mr W last updated on 17/Jun/25