Question Number 10540 by FilupS last updated on 17/Feb/17
Commented by FilupS last updated on 17/Feb/17
$$\mathrm{All}\:\mathrm{side}\:\mathrm{lenghts}\:=\:{n} \\ $$$$\angle{CAE}=\theta \\ $$$$\mathrm{0}\leqslant\theta<\frac{\pi}{\mathrm{3}} \\ $$$$\: \\ $$$$\mathrm{1}.\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{sections} \\ $$$$\mathrm{2}.\:\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{non}\:\mathrm{overlapping}\:\mathrm{area} \\ $$$$\:\:\:\:\:\:\mathrm{located}\:\mathrm{at}\:\mathrm{point}\:{C} \\ $$$$\mathrm{3}.\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bottom}−\mathrm{most} \\ $$$$\:\:\:\:\:\:\mathrm{non}\:\mathrm{overlapping}\:\mathrm{area}\:\left(\mathrm{near}\:\mathrm{points}\:{A}\:\mathrm{and}\:{B}\right) \\ $$
Commented by ajfour last updated on 17/Feb/17
$$\:{which}\:{app}\:{did}\:{u}\:{draw}\:{the}\:{triangles}\:{with}\left(\:{if}\:\:{on}\:{mobile}\right)? \\ $$
Commented by FilupS last updated on 17/Feb/17
$$\mathrm{Geogebra} \\ $$
Answered by mrW1 last updated on 18/Feb/17
![((CF)/(sin ∠CAF))=((CA)/(sin ∠CFA)) ⇒((CF)/(sin (θ/2)))=(n/(sin (π−(θ/2)−(π/3))))=(n/(sin ((θ/2)+(π/3)))) ⇒CF=((sin (θ/2))/(sin ((θ/2)+(π/3))))n similarily: ⇒CG=((sin θ)/(sin (θ+(π/3))))n ((GB)/(sin ∠BAG))=((AB)/(sin ∠AGB)) ⇒((GB)/(sin ((π/3)−θ)))=(n/(sin ((θ/2)+(π/3)))) ⇒GB=((sin ((π/3)−θ))/(sin ((θ/2)+(π/3))))n FG=CG−CF=[((sin θ)/(sin (θ+(π/3))))−((sin (θ/2))/(sin ((θ/2)+(π/3))))]n EF=CF FH=FG HD=GB A_(ΔAEF) =A_(ΔACF) =(1/2)×CF×((√3)/2)n=((√3)/4)n^2 ×((sin (θ/2))/(sin ((θ/2)+(π/3)))) A_(ΔAFH) =A_(ΔAFG) =(1/2)×FG×((√3)/2)n=((√3)/4)n^2 ×[((sin θ)/(sin (θ+(π/3))))−((sin (θ/2))/(sin ((θ/2)+(π/3))))] A_(ΔAHD) =A_(ΔAGB) =(1/2)×GB×((√3)/2)n=((√3)/4)n^2 ×((sin ((π/3)−θ))/(sin ((θ/2)+(π/3)))) (1) area of overlapping zone A_1 A_1 =2×A_(ΔAFG) =((√3)/2)n^2 ×[((sin θ)/(sin (θ+(π/3))))−((sin (θ/2))/(sin ((θ/2)+(π/3))))] (2) A_2 =A_(ΔCHF) =A_(ΔACF) −A_(ΔAFH) =((√3)/4)n^2 ×((sin (θ/2))/(sin ((θ/2)+(π/3))))−((√3)/4)n^2 ×[((sin θ)/(sin (θ+(π/3))))−((sin (θ/2))/(sin ((θ/2)+(π/3))))] =((√3)/4)n^2 ×[((2sin (θ/2))/(sin ((θ/2)+(π/3))))−((sin θ)/(sin (θ+(π/3))))] (3) A_3 =A_(ΔAGB) =((√3)/4)n^2 ×((sin ((π/3)−θ))/(sin ((θ/2)+(π/3))))](https://www.tinkutara.com/question/Q10556.png)
$$\frac{{CF}}{\mathrm{sin}\:\angle{CAF}}=\frac{{CA}}{\mathrm{sin}\:\angle{CFA}} \\ $$$$\Rightarrow\frac{{CF}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}=\frac{{n}}{\mathrm{sin}\:\left(\pi−\frac{\theta}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}\right)}=\frac{{n}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\ $$$$\Rightarrow{CF}=\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}{n} \\ $$$${similarily}: \\ $$$$\Rightarrow{CG}=\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}{n} \\ $$$$ \\ $$$$\frac{{GB}}{\mathrm{sin}\:\angle{BAG}}=\frac{{AB}}{\mathrm{sin}\:\angle{AGB}} \\ $$$$\Rightarrow\frac{{GB}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}=\frac{{n}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\ $$$$\Rightarrow{GB}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}{n} \\ $$$$ \\ $$$${FG}={CG}−{CF}=\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right]{n} \\ $$$$ \\ $$$${EF}={CF} \\ $$$${FH}={FG} \\ $$$${HD}={GB} \\ $$$$ \\ $$$${A}_{\Delta{AEF}} ={A}_{\Delta{ACF}} =\frac{\mathrm{1}}{\mathrm{2}}×{CF}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\ $$$$ \\ $$$${A}_{\Delta{AFH}} ={A}_{\Delta{AFG}} =\frac{\mathrm{1}}{\mathrm{2}}×{FG}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\ $$$${A}_{\Delta{AHD}} ={A}_{\Delta{AGB}} =\frac{\mathrm{1}}{\mathrm{2}}×{GB}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${area}\:{of}\:{overlapping}\:{zone}\:{A}_{\mathrm{1}} \\ $$$${A}_{\mathrm{1}} =\mathrm{2}×{A}_{\Delta{AFG}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${A}_{\mathrm{2}} ={A}_{\Delta{CHF}} ={A}_{\Delta{ACF}} −{A}_{\Delta{AFH}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}\right] \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\left[\frac{\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)}−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right)}\right] \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${A}_{\mathrm{3}} ={A}_{\Delta{AGB}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{n}^{\mathrm{2}} ×\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\theta\right)}{\mathrm{sin}\:\left(\frac{\theta}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)} \\ $$
Commented by FilupS last updated on 17/Feb/17
$${A}\mathrm{wesome}\:\mathrm{job}! \\ $$