Here's my working: WLOG, if you assume half of the square side to be of unit length then area of the square = 4 = πR^2 or R= √(4/π) and the y-co-ordinate of the point where the vertical side of square intersects with the circle is=√((4-π)/π). We can now say, sin(ξ)=√(π/4)√(π/4)–(√((4-π)/π) √((4-π)/π))/(√(4/π)√(4/π))= π/4-(4-π)/4 =(2π-4)/4=(π-2)/2

I set the area of circle and square equal to each other.

(pi)r^2 = (2rcos(theta))^2

(If you draw a line from O to middle right of square, you get angle theta from that line to the next red line or the red line below the other one.)

Then you can cancel out r^2 and find that new angle theta by doing some arithmetic. Once you find that angle, you multiply it by 8 and subtract that number by 360. Once you have that number, divide it by 4 and that is the angle you are looking for. Then just sin of that angle and you get .57

Option B

ReplyDeleteDo you have the details Joe?

ReplyDeleteHere's my working: WLOG, if you assume half of the square side to be of unit length then area of the square = 4 = πR^2 or R= √(4/π) and the y-co-ordinate of the point where the vertical side of square intersects with the circle is=√((4-π)/π). We can now say, sin(ξ)=√(π/4)√(π/4)–(√((4-π)/π) √((4-π)/π))/(√(4/π)√(4/π))= π/4-(4-π)/4 =(2π-4)/4=(π-2)/2

ReplyDeleteI set the area of circle and square equal to each other.

ReplyDelete(pi)r^2 = (2rcos(theta))^2

(If you draw a line from O to middle right of square, you get angle theta from that line to the next red line or the red line below the other one.)

Then you can cancel out r^2 and find that new angle theta by doing some arithmetic. Once you find that angle, you multiply it by 8 and subtract that number by 360. Once you have that number, divide it by 4 and that is the angle you are looking for. Then just sin of that angle and you get .57

Thank you six. Your work confirms the answer.

ReplyDeleteYou can also see: "non-overlapping areas are equal", and rest will follow.

Thank you.