2022 AMC 12A Problems/Problem 3

The square's area is the sum of the areas of the five smaller rectangles. This is 1*6 + 2*4 + 5*6 + 2*7 + 2*3 = 64, so the square has side length 8. The square's perimeter is therefore 32. Its perimeter can also be expressed as the sum of the width and height of the four outer rectangles. Adding up the width and height of all rectangles combined gives us 1 + 6 + 2 + 4 + 5 + 6 + 2 + 7 + 2 + 3 = 38, so adding up the width and height of the inner rectangle must give us 38 - 32 = 6. Only the 2x4 rectangle has dimensions that add up to 6, so our answer is B.