Menu Close

Question-15765




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

Leave a Reply

Your email address will not be published. Required fields are marked *