OE


To picture the smallest possible rectangle, envision one extremely short and wide, as in the figure at left, on the next page. This rectangle would have a width approaching that of the circle's diameter (12), but a height approaching o. Thus, the area would approach o. To picture the largest possible rectangle, envision a perfect square. As in the figure at right below, the diagonal PR = 12, which is also the diameter of the circle. This diameter cuts the square into two 45-45-90 triangles, which have sides in the proportion \(1:1:\sqrt{2}\) .

Therefore, the side of the square would equal \(\frac{12}{\sqrt{2}}= 6 \sqrt{2}\) , and the area would equal \((6 \sqrt{2} )^2=72\) , which is less than Quantity B (80). Therefore, the rectangle will always have an area smaller than Quantity B. Thus, Quantity B is greater.