Solution of the Week #344 - Rectangles in a Quadrant

The simplest way to calculate the areas of the rectangles is probably to complete the circles, with three additional copies of each quadrant, then draw a diameter between two opposite points as shown:

In the first case the area of the required rectangle is 2x^2, and in the second it is 2y^2.

In the first diagram, there is a right triangle with legs 2x and 4x, and hypotenuse 2.

 

(2x)^2 + (4x)^2 = 2^2

 

4x^2 + 16x^2 = 4

 

20x^2 = 4

 

2x^2 = 4/10 = 2/5

 

Similarly in the second diagram:

 

y^2 + (5y)^2 = 2^2

 

26y^2 = 4

 

2y^2 = 4/13

 

So the areas are 2/5 and 4/13 respectively, and their difference is therefore 6/65. Or if you chose to interpret the question the other way, the first rectangle is 30% larger than the second.