1. In the figure, $\frac{1}{8}$ of the circle and $\frac{1}{4}$ of the rectangle are shaded. The area of the square is equal to half the sum of the areas of the circle and the rectangle. If it is known that the shaded area is $40 \%$ of the square, determine the ratio in which the sum of the areas of the circle and the square relates to the area of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_06_03_5d6b13b2bd1879787abdg-3.jpg?height=437&width=348&top_left_y=1232&top_left_x=840)
1. The shaded area is $\frac{1}{8}$ of the circle and $\frac{1}{4}$ of the rectangle. The area of the square is equal to half the sum of the areas of the circle and the rectangle. If it is known that 40% of the square is shaded, determine the ratio in which the sum of the areas of the circle and the square relates to the area of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-15.jpg?height=439&width=351&top_left_y=877&top_left_x=858)

## Solution.

Let:

$p$ - area of the rectangle

$k$ - area of the square

$x$ - area of the circle

Then we have:

$\frac{1}{8} x + \frac{1}{4} p = 40\% \cdot k$

$k = \frac{x + p}{2}$

From the first equation, we get:

$5 x + 10 p = 16 k$.

Substitute $k$ from the second equation:

$5 x + 10 p = 16 \cdot \frac{x + p}{2}$

$2 p = 3 x$

$x = \frac{2}{3} p$

Further, we have:

$k = \frac{1}{2}(x + p)$

$k = \frac{1}{2}\left(\frac{2}{3} p + p\right)$

$k = \frac{5}{6} p$.

National Mathematics Competition 2021

## It follows:

$\frac{x + k}{p} = \frac{\frac{2}{3} p + \frac{5}{6} p}{p} = \frac{\frac{9}{6} p}{p} = \frac{3}{2}$.

The sum of the areas of the circle and the square is $\frac{3}{2} = 1.5$ times the area of the rectangle, or $(x + k): p = 3: 2$.