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Question Number 196304 by cortano12 last updated on 22/Aug/23

	Answered by a.lgnaoui last updated on 24/Aug/23
	$S=shaded Area S1=Arc(AMC) S2=Arc(OBD) S=S(ABCD)−S1+S2 •Calcul de S(ABCD) OBsin 30=OAsin 45⇒ OA=5(√2) S(ABCD)=AB×AC { ((AB=OBcos 30−OAcos 45)=5((√3) −1))),((AC=2×AM=2×OAsin 45=10)) :} ⇒S(ABCD)=50((√3) −1) •calcul de S1 S1=Area[Arc(OAC)]−[area(OAC)] =π((OA^2 )/4)−OA^2 cos 45sin 45 =((25π)/2)−((50)/2)=((25π−50)/2) •calcul de S2 S2=Area[arc(OBD)]−Area(OBD) =π((OB^2 )/6)−OB^2 cos 30sin 30 =((50π)/3)−((50(√3))/2)= ((100π−150(√3))/6) alors Shaded Area=50((√3) −1)−((25π)/2)+25 + ((100π−150(√3))/6) =(((50)/3)−((25)/2))π+(50−25)(√3) −25 =((25π)/6)+25(√3) −25 Shaded Area= 31,39 cm^2$
	$S = shaded Area$ $S 1 = Arc (AMC) S 2 = Arc (OBD)$ $S = S (ABCD) - S 1 + S 2$ $∙ Calcul de S (ABCD)$ $OBsin 30 = OAsin 45 \Rightarrow OA = 5 \sqrt{2}$ $S (ABCD) = AB \times AC$ ${\begin{cases} AB = OBcos 30 - OAcos 45) = 5 (\sqrt{3} - 1) \\ AC = 2 \times AM = 2 \times OAsin 45 = 10 \end{cases}$ $\Rightarrow S (ABCD) = 50 (\sqrt{3} - 1)$ $∙ calcul de S 1$ $S 1 = Area [Arc (OAC)] - [area (OAC)]$ $= π \frac{{OA}^{2}}{4} - {OA}^{2} \cos 45 \sin 45$ $= \frac{25 π}{2} - \frac{50}{2} = \frac{25 π - 50}{2}$ $∙ calcul de S 2$ $S 2 = Area [arc (OBD)] - Area (OBD)$ $= π \frac{{OB}^{2}}{6} - {OB}^{2} \cos 30 \sin 30$ $= \frac{50 π}{3} - \frac{50 \sqrt{3}}{2} = \frac{100 π - 150 \sqrt{3}}{6}$ $alors$ $Shaded Area = 50 (\sqrt{3} - 1) - \frac{25 π}{2} + 25$ $+ \frac{100 π - 150 \sqrt{3}}{6}$ $= (\frac{50}{3} - \frac{25}{2}) π + (50 - 25) \sqrt{3} - 25$ $= \frac{25 π}{6} + 25 \sqrt{3} - 25$ $Shaded Area = 31, 39 {cm}^{2}$
	Commented by a.lgnaoui last updated on 24/Aug/23

	$look at rectification (error calcul)$
	Commented by a.lgnaoui last updated on 23/Aug/23

	Commented by cortano12 last updated on 24/Aug/23

	$no sir$
	Commented by cortano12 last updated on 24/Aug/23

	$correct sir$
	Answered by mr W last updated on 23/Aug/23

	$M E T H O D I$ $r \sin α = R \sin β$ $\Rightarrow r = \frac{R \sin β}{\sin α}$ $A B = R \cos β - r \cos α$ $A_{s h a d e} = β (R^{2} - r^{2}) + (R \cos β - r \cos α) R \sin β - (α - β) r^{2}$ $A_{s h a d e} = R^{2} [β (1 - \frac{\sin^{2} β}{\sin^{2} α}) + \sin β (\cos β - \frac{\sin β}{\tan α}) - (α - β) \frac{\sin^{2} β}{\sin^{2} α}]$ $A_{s h a d e} = 10^{2} \times \frac{6 (\sqrt{3} - 1) + π}{24} \approx 31.391$
	Commented by cortano12 last updated on 24/Aug/23

	$nice solution$
	Answered by mr W last updated on 23/Aug/23

	$M E T H O D I I$ $r = \frac{R \sin β}{\sin α}$ $A_{s h a d e} = 2 \int_{0}^{R \sin β} (\sqrt{R^{2} - x^{2}} - \sqrt{r^{2} - x^{2}}) d x$ $= {[R^{2} \sin^{- 1} \frac{x}{R} - r^{2} \sin^{- 1} \frac{x}{r} + x \sqrt{R^{2} - x^{2}} - x \sqrt{r^{2} - x^{2}}]}_{0}^{R \sin β}$ $= R^{2} [β - α \frac{\sin^{2} β}{\sin^{2} α} + \sin β \cos β - \frac{\sin^{2} β}{\tan α}]$ $= 10^{2} \times [\frac{6 (\sqrt{3} - 1) + π}{24}] \approx 31.391$
	Commented by mr W last updated on 23/Aug/23

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