Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-23.jpg?height=422&width=507&top_left_y=92&top_left_x=469)
Answer: $52.5$.

Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.

Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.

![](https://cdn.mathpix.com/cropped/2024_05_06_5b323dccc142e96a9907g-23.jpg?height=417&width=495&top_left_y=821&top_left_x=479)

Fig. 2: to the solution of problem 7.8

The area of the "large" rectangle consists of the area of the "small" rectangle and the area of the "ring," which can be represented as four rectangles with diagonals $A F, F C, C H, H A$. The diagonals divide the area of each of these rectangles in half. The quadrilateral $A F C H$ consists of one half of each such rectangle and the "small" rectangle.

Let's calculate the area of $A F C H$. The area of the "ring" from the four rectangles is the difference between the areas of the "large" and "small" rectangles, which is $90-15=75$. The total area of the four triangles is $75: 2=37.5$. Therefore, the desired area is $15+37.5=52.5$.

## 8th grade