Question Number 220353 by Tawa11 last updated on 11/May/25
Commented by fantastic last updated on 11/May/25
Commented by fantastic last updated on 11/May/25
![Here △DEC is a eqilateral triangle so ∠C =∠E =∠D =60^o so the area between the chord and the curve is( ((60^0 )/(360^0 ))×πr^2 −((√3)/4)r^2 )cm^2 ((1/6)π −((√3)/4))cm^2 [as r=1cm] =(1/2)((π/3)−((√3)/2))cm^2 Now the area of the small semicircle=(1/2)π((1/2))^2 =(π/8)cm^2 so the area of the shaded area =(π/8)−(1/2)((π/3)−((√3)/2))cm^2 =(1/2)((π/4)−(π/3)+((√3)/2))cm^2 =(1/2)(((3π−4π+6(√3))/(12)))cm^2 =(1/2)(((6(√3)−π)/(12)))cm^2 =((6(√3)−π)/(24)) cm^2 ≈0.30211300799264 cm^(2 )](https://www.tinkutara.com/question/Q220357.png)
$${Here}\:\bigtriangleup{DEC}\:{is}\:{a}\:{eqilateral}\:{triangle} \\ $$$${so}\:\angle{C}\:=\angle{E}\:=\angle{D}\:=\mathrm{60}^{{o}} \: \\ $$$${so}\:{the}\:{area}\:{between}\:{the}\:{chord}\:{and}\:{the}\:{curve}\:{is}\left(\:\frac{\mathrm{60}^{\mathrm{0}} }{\mathrm{360}^{\mathrm{0}} }×\pi{r}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{r}^{\mathrm{2}} \right){cm}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{6}}\pi\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right){cm}^{\mathrm{2}} \left[{as}\:{r}=\mathrm{1}{cm}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){cm}^{\mathrm{2}} \\ $$$${Now}\:{the}\:{area}\:{of}\:{the}\:{small}\:{semicircle}=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{8}}{cm}^{\mathrm{2}} \\ $$$${so}\:{the}\:{area}\:{of}\:{the}\:{shaded}\:{area} \\ $$$$=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){cm}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){cm}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}\pi−\mathrm{4}\pi+\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{12}}\right){cm}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{6}\sqrt{\mathrm{3}}−\pi}{\mathrm{12}}\right){cm}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}\sqrt{\mathrm{3}}−\pi}{\mathrm{24}}\:{cm}^{\mathrm{2}} \\ $$$$\approx\mathrm{0}.\mathrm{30211300799264}\:{cm}^{\mathrm{2}\:} \: \\ $$
Commented by Tawa11 last updated on 11/May/25
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 11/May/25
Commented by mr W last updated on 11/May/25
![θ=sin^(−1) (r/R) ⇒sin 2θ=2((r/R))(√(1−((r/R))^2 )) A_L =((πr^2 )/2)+((R^2 sin 2θ)/2)−θR^2 =((πr^2 )/2)+[(r/R)(√(1−((r/R))^2 ))−sin^(−1) (r/R)]R^2 =((π×0.5^2 )/2)+((1/2)(√(1−((1/2))^2 ))−sin^(−1) (1/2))×1^2 =(π/8)+((√3)/4)−(π/6)=((√3)/4)−(π/(24)) ✓](https://www.tinkutara.com/question/Q220364.png)
$$\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}}\:\Rightarrow\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\left(\frac{{r}}{{R}}\right)\sqrt{\mathrm{1}−\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} } \\ $$$${A}_{{L}} =\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{{R}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}−\theta{R}^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}+\left[\frac{{r}}{{R}}\sqrt{\mathrm{1}−\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} }−\mathrm{sin}^{−\mathrm{1}} \frac{{r}}{{R}}\right]{R}^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\pi×\mathrm{0}.\mathrm{5}^{\mathrm{2}} }{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)×\mathrm{1}^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{8}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\pi}{\mathrm{24}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 11/May/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$