1968 AMC 12 #35 - Ratio of Areas in Semicircle
Source:
January 1, 2012
ratiogeometrytrapezoidrectangleAMC
Problem Statement
In this diagram the center of the circle is , the radius is inches, chord is parallel to chord are collinear, and is the midpoint of . Let (sq. in.) represent the area of trapezoid and let (sq. in.) represent the area of rectangle . Then, as and are translated upward so that increases toward the value , while always equals , the ratio become arbitrarily close to:
[asy]
size((270));
draw((0,0)--(10,0)..(5,5)..(0,0));
draw((5,0)--(5,5));
draw((9,3)--(1,3)--(1,1)--(9,1)--cycle);
draw((9.9,1)--(.1,1));
label("O", (5,0), S);
label("a", (7.5,0), S);
label("G", (5,1), SE);
label("J", (5,5), N);
label("H", (5,3), NE);
label("E", (1,3), NW);
label("L", (1,1), S);
label("C", (.1,1), W);
label("F", (9,3), NE);
label("M", (9,1), S);
label("D", (9.9,1), E);
[/asy]