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Question Number 227957 by TonyCWX last updated on 05/Mar/26
1. Solve the question below. 2. Generalise the area of the shaded segment.
$$\mathrm{1}.\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{question}\:\mathrm{below}. \\ $$$$\mathrm{2}.\:\mathrm{Generalise}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{segment}. \\ $$
Commented by TonyCWX last updated on 05/Mar/26
Commented by TonyCWX last updated on 07/Mar/26
Post it as a new question would be a better option.
$$\mathrm{Post}\:\mathrm{it}\:\mathrm{as}\:\mathrm{a}\:\mathrm{new}\:\mathrm{question}\:\mathrm{would}\:\mathrm{be}\:\mathrm{a}\:\mathrm{better}\:\mathrm{option}. \\ $$
Answered by fantastic2 last updated on 05/Mar/26
Commented by fantastic2 last updated on 05/Mar/26
cos α=(x/(2x))=(1/2) ⇒α=60^0 ∴shaded area=((120^0 )/(360^0 ))π(2x)^2 −(1/2)×2x×2x×sin 120^0 =((4πx^2 )/3)−(√3)x^2 =x^2 (((4π)/3)−(√3)) in this case x=2 ∴area=4(((4π)/3)−(√3))≈9.826
$$\mathrm{cos}\:\alpha=\frac{{x}}{\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\mathrm{60}^{\mathrm{0}} \\ $$$$\therefore{shaded}\:{area}=\frac{\mathrm{120}^{\mathrm{0}} }{\mathrm{360}^{\mathrm{0}} }\pi\left(\mathrm{2}{x}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{x}×\mathrm{2}{x}×\mathrm{sin}\:\mathrm{120}^{\mathrm{0}} \\ $$$$=\frac{\mathrm{4}\pi{x}^{\mathrm{2}} }{\mathrm{3}}−\sqrt{\mathrm{3}}{x}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} \left(\frac{\mathrm{4}\pi}{\mathrm{3}}−\sqrt{\mathrm{3}}\right) \\ $$$${in}\:{this}\:{case} \\ $$$${x}=\mathrm{2} \\ $$$$\therefore{area}=\mathrm{4}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}−\sqrt{\mathrm{3}}\right)\approx\mathrm{9}.\mathrm{826} \\ $$
Commented by TonyCWX last updated on 05/Mar/26
Nice. Now what if the Radius = R and number of segment = n? What′s the formula of the area of the segment?
$$\mathrm{Nice}. \\ $$$$\mathrm{Now}\:\mathrm{what}\:\mathrm{if}\:\mathrm{the}\:\mathrm{Radius}\:=\:\mathrm{R}\:\mathrm{and}\:\mathrm{number}\:\mathrm{of}\:\mathrm{segment}\:=\:\mathrm{n}? \\ $$$$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}? \\ $$
Commented by fantastic2 last updated on 06/Mar/26
Commented by fantastic2 last updated on 06/Mar/26
there are n segments width of each segment=((2R)/n) ∴ length of −=R−((2R)/n) α=cos^(−1) (((R(1−(2/n)))/R))=cos^(−1) (1−(2/n)) ∴area=((2α)/(360^0 ))πR^2 −(R^2 /2)sin 2α ∴area=((2cos^(−1) (1−(2/n)))/(360^0 ))πR^2 −(R^2 /2)sin(2 cos^(−1) (1−(2/n))) putting n=4 and R=4 we get area≈9.826
$${there}\:{are}\:{n}\:{segments} \\ $$$${width}\:{of}\:{each}\:{segment}=\frac{\mathrm{2}{R}}{{n}} \\ $$$$\therefore\:{length}\:{of}\:−={R}−\frac{\mathrm{2}{R}}{{n}} \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{R}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)}{{R}}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right) \\ $$$$\therefore{area}=\frac{\mathrm{2}\alpha}{\mathrm{360}^{\mathrm{0}} }\pi{R}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\therefore{area}=\frac{\mathrm{2cos}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)}{\mathrm{360}^{\mathrm{0}} }\pi{R}^{\mathrm{2}} −\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)\right) \\ $$$${putting}\:{n}=\mathrm{4}\:{and}\:{R}=\mathrm{4} \\ $$$${we}\:{get}\:{area}\approx\mathrm{9}.\mathrm{826} \\ $$
Commented by TonyCWX last updated on 07/Mar/26
Nice!
$$\mathrm{Nice}! \\ $$

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