The circumference of a circle is 10pi. Which of the followin

Step 1: the maximum side of the rectangle must be less than 2*radius (diameter). If the radius is 5 (as perimeter=2*pi*r=10*pi), the maximum side of the inscribed rectangle must be less than 10.

Step 2: let's try to find the maximum area of an inscribed rectangle: thinking a little bit, it can be found that the maximum area will be when we pick an inscribed square (see the red square in the drawing). If each black arrow (radius) measure \(5\), the sides of this red square have to be \(\sqrt{2}*5\). Why? Because \(45\) - \(45\) - \(90\) triangles have sides \(X\) - \(X\) - \(\sqrt{2}*X\).

Step 3: therefore, the maximum area is \((\sqrt{2}*5)^2=25*2=50\)

SOLUTION: \(40*\sqrt{2}=40*1.41...=56.5...\) the only possible solution is E

Posting official solution of this problem.

The important piece of this question is to know that every square is a special kind of rectangle, but not every rectangle is a square. The question would have been easier if it mentioned "quadrilateral" instead of rectangle.

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