Saturday, October 31, 2009

Fox 172

Simple beginnings...

3 comments:

  1. Let r=1, radius of circle.
    Notation: (S)=Area of S
    theta=angle
    (A)=(pi/2-theta)/2
    (Right triangle) = tan(theta)/2
    (C)=theta/2=(Right triangle) - (B)
    (B)=tan(theta)/2-theta/2=(A)=(pi/2-theta)/2
    tan(theta)-theta = pi/2 -theta -> tan(theta) = pi/2. -> Answer D.

    MIGUE.

    ReplyDelete
  2. An easier way:
    Call the white area C and the radius 1.
    Call the hight of the right-angled triangle h.
    Then because A(A)=A(B), is also A(A+C)=A(B+C).
    This means that the area of the quarter circle is equal to the area of the right-angled triangle.
    So: A(A+C)=pi/4.
    A(B+C)=1*h/2 = pi/4 so h = pi/2.
    tan(theta) = h/1 = h = pi/2 -> answer D.

    ReplyDelete