Question-199774

Question Number 199774 by Mingma last updated on 09/Nov/23

Answered by ajfour last updated on 09/Nov/23

Commented by ajfour last updated on 09/Nov/23

P (perimeter of △ ABC) = 2R+2Rcos θ left part of it cut by MN = BM+BN(=AB) =Rcos θ+R =(P/2) S( Area △ABC)= R^2 sin θcos θ area cut by MN = ((BM)/2)×MN = ((Rcos θ)/2)×Rsin θ = (S/2)

$${P}\:\left({perimeter}\:{of}\:\bigtriangleup\:{ABC}\right) \\ $$$$\:\:=\:\mathrm{2}{R}+\mathrm{2}{R}\mathrm{cos}\:\theta \\ $$$${left}\:{part}\:{of}\:{it}\:{cut}\:{by}\:{MN} \\ $$$$\:=\:{BM}+{BN}\left(={AB}\right) \\ $$$$\:\:={R}\mathrm{cos}\:\theta+{R}\:\:=\frac{{P}}{\mathrm{2}} \\ $$$$\:{S}\left(\:{Area}\:\bigtriangleup{ABC}\right)=\:{R}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\:\:{area}\:{cut}\:{by}\:{MN}\:=\:\frac{{BM}}{\mathrm{2}}×{MN} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{R}\mathrm{cos}\:\theta}{\mathrm{2}}×{R}\mathrm{sin}\:\theta\:=\:\frac{{S}}{\mathrm{2}} \\ $$

Commented by Mingma last updated on 09/Nov/23

Very elegant solution!

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Question-199774

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