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'Perfect Oval' (Posted on 2006-11-18) Difficulty: 3 of 5
A well-meaning senior citizen once erroneously contended that a "perfect oval" could only be constructed with a straight edge and a pair of compasses.

The theoretical construction that he described so very closely approximates the ellipse given by the equation: (x^2)/16 + (y^2)/9 = 1.

Required:
1. Emulate such a construction
2. Suggest the difference in area of these two entities if this construction and an ellipse representing the above equation are drawn at the same scale (let them share a common major radial length).

To my knowledge, oval and ellipse refer to the same thing, ellipse being the 'technical' term.

See The Solution Submitted by brianjn    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 5

Construction of the first quadrant portion
of the "Oval":
Label points O(0,0), A(4,0), and B(0,3).
Construct point M on line segment AB such
that |BM| = 4-3 = 1. Construct the
perpendicular bisector of line segment AM
which intersects the x-axis at point C and
the y-axis at point D. Construct circular
arc (starting at point A and ending at
point E on the perpendicular bisector) with
center C and radius |CA|. Construct
circular arc (starting at point E and
ending at point B) with center D and radius
|DB|. The rest of the "Oval" is obtained by
reflecting the first quadrant about the
x-axis, the y-axis, and the origin.
  Area of "Oval" = 4*[ Area(Sector CAE) +
                       Area(Sector DEB) -
                       Area(Triangle OCD)]
     = 4*[ (1/2)*(5/2)^2*arcsin(4/5) +
           (1/2)*(5)^2*arcsin(3/5) -
           (1/2)*(3/2)*(2)]
     ~= 37.766245665
  Area of Ellipse = (PI)*(4)*(3)
                  ~= 37.699111843
   (Area of "Oval")-(Area of Ellipse)
  ------------------------------------
             Area of Ellipse
     ~= 0.001780780

  Posted by Bractals on 2006-11-18 17:28:31
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