The largest square above has sides of length 8 and is divided into the two shaded rectangles and two smaller squares labeled I and II. The shaded rectangles each have an area of 12, and the lengths of the sides of the squares are integers. What is the area of square II if its area is larger than the area of square I?
The area of square II is 25.
Given that the largest square has a side length of 8, its total area is 64. The two shaded rectangles have a combined area of 24 (12 each), leaving an area of 40 for the two squares I and II. Since square II must have an area larger than square I, we find that the only integer area that meets this condition is 25.
If square II has an area of 9, then its side length would be 3. This would leave an area of 40 - 9 = 31 for square I. However, square I cannot have an integer area greater than 31 and still be smaller than square II, violating the condition that square II must be larger.
If square II has an area of 16, its side length would be 4. This would leave 40 - 16 = 24 for square I. The area of square I would then be 24, which is larger than 16, contradicting the requirement that square II's area must be larger than square I.
Square II having an area of 25 means its side length is 5. This allows square I to have an area of 40 - 25 = 15. Since 15 is less than 25, this satisfies the condition that square II's area is larger than that of square I.
If square II has an area of 36, its side length would be 6. This would leave only 40 - 36 = 4 for square I. However, a square with an area of 4 (side length 2) does not satisfy the condition that square II must have an area larger than square I, as both squares would be larger than 2.
The areas of the squares must adhere to the condition that square II is larger than square I. With the calculations showing that the only valid area for square II that meets all conditions is 25, we can conclude that it is the only choice consistent with the given criteria. Thus, the area of square II is confirmed as 25.
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