1. A large rectangle is divided into four congruent rectangles, one of which is colored. Then, one of the uncolored rectangles is divided into five congruent rectangles, one of which is colored again. Finally, one of those uncolored rectangles is divided into three congruent rectangles, one of which is colored, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_06_03_6c102289945b60d8f211g-3.jpg?height=228&width=517&top_left_y=1208&top_left_x=792)

Calculate the area of the uncolored part of the rectangle in the figure, if the smallest colored rectangle is a square, and the sum of the perimeters of all three colored rectangles is $19.6 \mathrm{~cm}$.
1. A large rectangle is divided into four congruent rectangles, one of which is colored. Then, one of the uncolored rectangles is divided into five congruent rectangles, one of which is colored again. Finally, one of those uncolored rectangles is divided into three congruent rectangles, one of which is colored, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_ca549eb60a6f3c3b6299g-15.jpg?height=289&width=688&top_left_y=935&top_left_x=684)

Calculate the area of the uncolored part of the rectangle in the figure, if the smallest colored rectangle is a square, and the sum of the perimeters of all three colored rectangles is $19.6 \mathrm{~cm}$.

## Solution.

Let the side length of the colored square be $x$.

Then, the side lengths of the second colored rectangle are $3x$ and $x$, and the side lengths of the largest colored rectangle are $3x$ and $5x$, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_ca549eb60a6f3c3b6299g-15.jpg?height=434&width=943&top_left_y=1596&top_left_x=251)

For the colored rectangles, we have:

$o_{1}=4x$

$o_{2}=2 \cdot (3x + x) = 8x$

$o_{3}=2 \cdot (3x + 5x) = 16x$.

The sum of the perimeters of all three colored rectangles is:

$o_{1} + o_{2} + o_{3} = 4x + 2(3x + x) + 2(3x + 5x) = 28x$.

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From the problem statement, we have $28x = 19.6$

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so $x = 0.7 \mathrm{~cm} (= 7 \mathrm{~mm})$.

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The side lengths of the original rectangle are

$12x = 12 \cdot 7 \mathrm{~mm} = 84 \mathrm{~mm}$ and $5x = 5 \cdot 7 \mathrm{~mm} = 35 \mathrm{~mm}$.

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Its area is:

$P_{4} = 84 \mathrm{~mm} \cdot 35 \mathrm{~mm} = 2940 \mathrm{~mm}^{2}$.

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The areas of the colored rectangles are:

$P_{1} = 7 \mathrm{~mm} \cdot 7 \mathrm{~mm} = 49 \mathrm{~mm}^{2}$

$P_{2} = 21 \mathrm{~mm} \cdot 7 \mathrm{~mm} = 147 \mathrm{~mm}^{2}$

$P_{3} = 21 \mathrm{~mm} \cdot 35 \mathrm{~mm} = 735 \mathrm{~mm}^{2}$

2 POINTS

The area of the uncolored part of the rectangle will be calculated as the difference between the area of the original rectangle and the sum of the areas of the colored rectangles:

$P = P_{4} - (P_{1} + P_{2} + P_{3})$

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$P = 2940 - (49 + 147 + 735)$

$P = 2940 - 931$

$P = 2009 \mathrm{~mm}^{2} \quad (= 20.09 \mathrm{~cm}^{2})$.

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TOTAL 10 POINTS