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Question Number 193161 by Mingma last updated on 05/Jun/23

	Answered by a.lgnaoui last updated on 05/Jun/23
	$Triangle ABC equilaterale MN=NP=MP ∡BAH=60=∡DAH ⇒ ∡ADH=30° ∡AMN=90−30=60 MNP Triangle equilaterale MN=NP=MP Area(MNP)=Aeea(arc( MN)− area(triangleDMN) Aera(triangkeD MN)=R×Rsin (π/3)=((R^2 (√3))/2) Area(arc MN)=((πR^2 )/3) arc=((πR^2 )/3)−((R^2 (√3))/2)=(((2π−3(√3))R^2 )/6) Aeea(MNP)=Area(triangle MNP) −2Area(arcMN) Area=((R^2 (√3))/2)−2(((2π−3(√3))R^2 )/6)=((3R^2 −(4π−6(√3))R^2 )/6) Area(pink triangle)=(((6(√3) +3−4𝛑)R^2 )/6) =((√3) +((3−4π)/6))R^2 =0,1377R^2 aire (pink area)=(π/2)R^2 +((√3) +((3−4π)/6))R^2 =((√3) +((3−π)/6))R^2 ((√3) +(1/2)+(π/6))R^2 =0,795R^2 ⇒fraction=((0,1377)/(0,795))=17%$
	$Triangle ABC equilaterale$ $MN = NP = MP$ $∡ BAH = 60 = ∡ DAH$ $\Rightarrow ∡ ADH = 30 ° ∡ AMN = 90 - 30 = 60$ $MNP Triangle equilaterale$ $MN = NP = MP$ $Area (MNP) = Aeea (arc (MN) -$ $area (triangleDMN)$ $Aera (triangkeD MN) = R \times Rsin \frac{π}{3} = \frac{R^{2} \sqrt{3}}{2}$ $Area (arc MN) = \frac{π R^{2}}{3}$ $arc = \frac{π R^{2}}{3} - \frac{R^{2} \sqrt{3}}{2} = \frac{(2 π - 3 \sqrt{3}) R^{2}}{6}$ $Aeea (MNP) = Area (triangle MNP)$ $- 2 Area (arcMN)$ $Area = \frac{R^{2} \sqrt{3}}{2} - 2 \frac{(2 π - 3 \sqrt{3}) R^{2}}{6} = \frac{{3 R}^{2} - (4 π - 6 \sqrt{3}) R^{2}}{6}$ $Area (pink triangle) = \frac{(6 \sqrt{3} + 3 - 4 π) R^{2}}{6}$ $= (\sqrt{3} + \frac{3 - 4 π}{6}) R^{2}$ $= 0, {1377 R}^{2}$ $aire (pink area) = \frac{π}{2} R^{2} + (\sqrt{3} + \frac{3 - 4 π}{6}) R^{2}$ $= (\sqrt{3} + \frac{3 - π}{6}) R^{2}$ $(\sqrt{3} + \frac{1}{2} + \frac{π}{6}) R^{2} = 0, {795 R}^{2}$ $\Rightarrow fraction = \frac{0, 1377}{0, 795} = 17 %$
	Commented by a.lgnaoui last updated on 07/Jun/23

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