1) For newcomers, the m in 7m is confusing, since it could be a dimension m = meters or a variable. (Those who have studied other problems from Math and Engineering will know that it is meters.) 2) Starting at 4:00, 7203 could simply be factored into 3 times 2401. Since 2401 = 49², we can take 49 outside the radical, so √(7203) = 49√3 (4:45). 3) As matthieudutriaux pointed out, the solution finds that B (1:30 in the video) is the midpoint of QR and A and R are the same point. Therefore, the drawing is not correct, because it shows A as a point on PR between P and R. The solution method, however, is still valid. It looks like this solution method would still give the correct answer if B were anywhere on the semicircle. Assuming that angles can only have positive measures, when B is outside the equilateral triangle to its right, a can have a positive value but must be added to 60° instead of subtracted to produce an angle
Original and impressive ! Here is another method : x^2*y^2=7203 ; then x*y=49*sqrt(3) x^2+y^2=4*r^2 QB^2=(xQ-xB)^2+(yQ-yB)^2 (in an orthonormal repair where P is the center, for example) xQ=2*r*cos(60°) then xQ=r yQ=2*r*sin(60°) then yQ=sqrt(3)*r xB=BR*cos(60°-a) with BR=x and cos(60°-a)=BR/PR=x/(2*r) then xB=x^2/(2*r) yB=BR*sin(60°-a) with BR=x and sin(60°-a)=PB/PR=y/(2*r) then yB=x*y/(2*r) QB^2=(xQ-xB)^2+(yQ-yB)^2 7^2=(r-x^2/(2*r))^2+(sqrt(3)*r-x*y/(2*r))^2 4*49=(2*r-x^2/r)^2+(2*sqrt(3)*r-x*y/r)^2 196=4*r^2-4*x^2+x^4/r^2+12*r^2-4*sqrt(3)*x*y+(x*y)^2/r^2 196+4*sqrt(3)*x*y=(16*r^2-4*x^2)+(x^4+(x*y)^2)/r^2 196+4*sqrt(3)*49*sqrt(3)=4*(4*r^2-x^2)+x^2*(x^2+y^2)/r^2 4*196=4*y^2+4*x^2 196=x^2+y^2 196=4*r^2 49=r^2 A(semicircle)=1/2*Pi*r^2 A(semicircle)=49/2*Pi m²
1) For newcomers, the m in 7m is confusing, since it could be a dimension m = meters or a variable. (Those who have studied other problems from Math and Engineering will know that it is meters.)
2) Starting at 4:00, 7203 could simply be factored into 3 times 2401. Since 2401 = 49², we can take 49 outside the radical, so √(7203) = 49√3 (4:45).
3) As matthieudutriaux pointed out, the solution finds that B (1:30 in the video) is the midpoint of QR and A and R are the same point. Therefore, the drawing is not correct, because it shows A as a point on PR between P and R. The solution method, however, is still valid.
It looks like this solution method would still give the correct answer if B were anywhere on the semicircle. Assuming that angles can only have positive measures, when B is outside the equilateral triangle to its right, a can have a positive value but must be added to 60° instead of subtracted to produce an angle
Original and impressive !
Here is another method :
x^2*y^2=7203 ; then x*y=49*sqrt(3)
x^2+y^2=4*r^2
QB^2=(xQ-xB)^2+(yQ-yB)^2 (in an orthonormal repair where P is the center, for example)
xQ=2*r*cos(60°) then xQ=r
yQ=2*r*sin(60°) then yQ=sqrt(3)*r
xB=BR*cos(60°-a) with BR=x and cos(60°-a)=BR/PR=x/(2*r) then xB=x^2/(2*r)
yB=BR*sin(60°-a) with BR=x and sin(60°-a)=PB/PR=y/(2*r) then yB=x*y/(2*r)
QB^2=(xQ-xB)^2+(yQ-yB)^2
7^2=(r-x^2/(2*r))^2+(sqrt(3)*r-x*y/(2*r))^2
4*49=(2*r-x^2/r)^2+(2*sqrt(3)*r-x*y/r)^2
196=4*r^2-4*x^2+x^4/r^2+12*r^2-4*sqrt(3)*x*y+(x*y)^2/r^2
196+4*sqrt(3)*x*y=(16*r^2-4*x^2)+(x^4+(x*y)^2)/r^2
196+4*sqrt(3)*49*sqrt(3)=4*(4*r^2-x^2)+x^2*(x^2+y^2)/r^2
4*196=4*y^2+4*x^2
196=x^2+y^2
196=4*r^2
49=r^2
A(semicircle)=1/2*Pi*r^2
A(semicircle)=49/2*Pi m²
Perfect 👍 your method is interesting and clear, thanks for sharing