Question

Вопрос 8/10 Прямоугольная фотография вставлена в деревянную рамку, ширина вертикальных частей которой в три раза меньше ширины горизонтальных. Периметр фотографии - 420 мм, внешний периметр рамки – 500 мм. Чему равна ширина горизонтальных частей рамки? Ответ дайте в миллиметрах. Ответ: исключительное право на pamkan mpoenta ellmuen копирование, распростран предварительного письмен​

Answer 1

Ответ:

30

Пошаговое объяснение:

Answer 2

Calculation of the Width of the Horizontal Parts of the Frame

To find the width of the horizontal parts of the frame, we need to set up an equation using the given information.

Let's assume that the width of the horizontal parts of the frame is represented by x.

According to the problem statement, the width of the vertical parts of the frame is three times smaller than the width of the horizontal parts. Therefore, the width of the vertical parts can be represented as x/3.

The perimeter of the photograph is given as 420 mm, and the external perimeter of the frame is given as 500 mm.

The perimeter of a rectangle can be calculated by adding the lengths of all four sides. In this case, the perimeter of the photograph is equal to the sum of the lengths of the four sides of the photograph.

The perimeter of the photograph can be calculated as follows:

2 * (length + width) = 420 mm

Since the photograph is rectangular, the length and width can be represented as follows:

2 * (length + x) = 420 mm

Similarly, the external perimeter of the frame can be calculated as follows:

2 * (length + 2 * (x/3)) = 500 mm

Simplifying the equations, we get:

length + x = 210 mm (Equation 1)

length + (2/3)x = 250 mm (Equation 2)

To solve these equations, we can subtract Equation 1 from Equation 2:

[(2/3)x - x] = 250 mm - 210 mm

Simplifying further:

(-1/3)x = 40 mm

Multiplying both sides by -3:

x = -120 mm

Since the width cannot be negative, we can conclude that there is an error in the given information or in the calculations.

Unfortunately, based on the information provided, it is not possible to determine the width of the horizontal parts of the frame.