$\"\"$

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The triangle formed by two midpoints and an included corner of the largest square is a right triangle with side lengths of $\"\"$ , and a hypotenuse of $\"\"$ .

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(a) Find the perimeter of the square with side lengths of $\"\"$ .

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Draw the related diagram.

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$\"\"$

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Pythagorean Theorem:

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$\"\"$

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$\"\"$

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$\"\"$ .

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The perimeter of the square is $\"\"$ .

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$\"\"$

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(b) Find the sum of the perimeters of all the squares.

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The perimeter of the largest square is $\"\"$ .

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The perimeter of the second largest square is $\"\"$ .

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Consider $\"\"$ and $\"\"$ .

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The common ratio is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$ .

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$\"\"$ .

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Hence,the infinite geometric series perimeters are $\"\"$ .

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The sum of an infinite geometric series is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$

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Therefore, the sum of the squares of all the perimeters is $\"\"$ .

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$\"\"$

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(c) Find the sum of the areas of all the squares.

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The area of the largest square is $\"\"$ .

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The area of the second largest square is $\"\"$ .

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Consider $\"\"$ and $\"\"$ .

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The common ratio is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$ .

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$\"\"$ .

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Hence, the infinite geometric series areas are $\"\"$ .

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The sum of an infinite geometric series is $\"\"$ .

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Substitute $\"\"$ and $\"\"$ .

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$\"\"$

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Therefore, the sum of the squares of all the areas is $\"\"$ . \ \

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$\"\"$

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(a) The perimeter of the square is $\"\"$ . \ \

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(b) The sum of the squares of all the perimeters is $\"\"$ .

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(c) The sum of the squares of all the areas is $\"\"$ .