7.1. (14 points) Misha drew a triangle with a perimeter of 11 and cut it into parts with three straight cuts parallel to the sides, as shown in the figure. The perimeters of the three shaded figures (trapezoids) turned out to be 5, 7, and 9. Find the perimeter of the small triangle that resulted from the cutting.

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# Answer: 10.

![](https://cdn.mathpix.com/cropped/2024_05_06_38daf1cc3b5132f798e9g-3.jpg?height=325&width=434&top_left_y=89&top_left_x=1485)

Solution. Since opposite sides of parallelograms are equal, the lateral sides of each resulting trapezoid are equal to the corresponding segments on the sides of the original triangle. Therefore, the sum of the perimeters of the three trapezoids is equal to the sum of the perimeter of the original triangle (the sum of the larger bases and lateral sides of the three trapezoids) and the perimeter of the new triangle (the sum of the smaller bases of the trapezoids). Thus, the perimeter of the new triangle is equal to the sum of the perimeters of the trapezoids minus the perimeter of the original triangle: \(a + b + c - P\).