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Question-82378




Question Number 82378 by ajfour last updated on 20/Feb/20
Commented by ajfour last updated on 20/Feb/20
If all 5 regions have equal areas,  find sides of rectangle, given  radius=1.
$$\mathrm{If}\:\mathrm{all}\:\mathrm{5}\:\mathrm{regions}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{areas}, \\ $$$$\mathrm{find}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rectangle},\:\mathrm{given} \\ $$$$\mathrm{radius}=\mathrm{1}. \\ $$
Answered by mr W last updated on 21/Feb/20
rectangle = L×B=L×(2t)=πr^2   ⇒L=((πr)/(2((t/r))))=((πr)/(2λ)) with λ=(t/r)  ⇒B=2t=2λr  segment = r^2 cos^(−1) ((t/r))−t(√(r^2 −t^2 ))=((πr^2 )/3)  ⇒cos^(−1) ((t/r))−((t/r))(√(1−((t/r))^2 ))=(π/3)  ⇒cos^(−1) λ−λ(√(1−λ^2 ))=(π/3)  ⇒λ≈0.264932    ⇒B≈0.528864r  ⇒L≈5.929055r
$${rectangle}\:=\:{L}×{B}={L}×\left(\mathrm{2}{t}\right)=\pi{r}^{\mathrm{2}} \\ $$$$\Rightarrow{L}=\frac{\pi{r}}{\mathrm{2}\left(\frac{{t}}{{r}}\right)}=\frac{\pi{r}}{\mathrm{2}\lambda}\:{with}\:\lambda=\frac{{t}}{{r}} \\ $$$$\Rightarrow{B}=\mathrm{2}{t}=\mathrm{2}\lambda{r} \\ $$$${segment}\:=\:{r}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{{t}}{{r}}\right)−{t}\sqrt{{r}^{\mathrm{2}} −{t}^{\mathrm{2}} }=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left(\frac{{t}}{{r}}\right)−\left(\frac{{t}}{{r}}\right)\sqrt{\mathrm{1}−\left(\frac{{t}}{{r}}\right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \lambda−\lambda\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{264932} \\ $$$$ \\ $$$$\Rightarrow{B}\approx\mathrm{0}.\mathrm{528864}{r} \\ $$$$\Rightarrow{L}\approx\mathrm{5}.\mathrm{929055}{r} \\ $$
Commented by ajfour last updated on 22/Feb/20
Very Nice way, Sir, thanks lots.
$$\mathrm{Very}\:\mathrm{Nice}\:\mathrm{way},\:\mathrm{Sir},\:\mathrm{thanks}\:\mathrm{lots}. \\ $$

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