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Question Number 15765 by ajfour last updated on 13/Jun/17

Commented by ajfour last updated on 13/Jun/17

Q.15736 (alternate solution) Area ΔAPF =(1/2)(Area ΔABP) Maximum area of ΔAPF =(1/2)(maximum Area of ΔABP) Area of inscribed ΔABP is maximum when it is equilateral. x=rcos 30° =((r(√3))/2) y=xtan 60° = ((3r)/2) Max.Area of ΔAPF = (1/2)(((r(√3))/2))(((3r)/2)) =((3(√3)r^2 )/8) .

$${Q}.\mathrm{15736}\:\:\:\left({alternate}\:{solution}\right) \\ $$$${Area}\:\Delta{APF}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({Area}\:\Delta{ABP}\right) \\ $$$$ \\ $$$${Maximum}\:{area}\:{of}\:\Delta{APF} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\left({maximum}\:{Area}\:{of}\:\Delta{ABP}\right) \\ $$$$\:\:\:\:\:\:{Area}\:{of}\:{inscribed}\:\Delta{ABP}\:{is} \\ $$$${maximum}\:{when}\:{it}\:{is}\:{equilateral}. \\ $$$$\:\:\:\:\:\:\:\:\:{x}={r}\mathrm{cos}\:\mathrm{30}°\:=\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:=\:\frac{\mathrm{3}{r}}{\mathrm{2}} \\ $$$${Max}.{Area}\:{of}\:\Delta{APF}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{r}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}{r}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\sqrt{\mathrm{3}}{r}^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$

Commented by mrW1 last updated on 13/Jun/17

very clever!

$$\mathrm{very}\:\mathrm{clever}! \\ $$

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