
Периметр классной комнаты 28 м, а её длина 8 м. Сколько учеников по санитарным нормам должно быть в
классе,если на каждые 7 человек предусмотрено 12 квадратных метров?

Ответы на вопрос

2) 8 * 6 = 48(кв.м) - площадь классной комнаты
3) 48 : 12 = 4(раза) по 12кв.м содержится в 48кв.м
3) 7 * 4 = 28 (учеников)
Ответ: 28 учеников должно быть в классе.



6*8=48 площадь
48/12=4
4*7=28 человек



Perimeter of the Classroom
To find the perimeter of a rectangular classroom, we need to add up the lengths of all four sides. In this case, the length of the classroom is given as 8 meters. Since the classroom is rectangular, we can assume that the width is also 8 meters, as it is not specified in the question.
The formula for finding the perimeter of a rectangle is:
Perimeter = 2 * (Length + Width)
Substituting the given length and width values, we get:
Perimeter = 2 * (8 + 8) = 2 * 16 = 32 meters
Therefore, the perimeter of the classroom is 32 meters.
Number of Students According to Sanitary Norms
According to the information provided, for every 7 students, there should be 12 square meters of space. We can use this information to calculate the number of students that can be accommodated in the classroom.
Let's assume that the number of students is represented by 'n'.
The formula for calculating the number of students based on the given space requirement is:
Number of Students = (Total Area of Classroom) / (Area per Student)
The total area of the classroom can be calculated by multiplying the length and width:
Total Area of Classroom = Length * Width = 8 * 8 = 64 square meters
Substituting the given area per student value, we get:
Number of Students = 64 / 12 = 5.33
Since we cannot have a fraction of a student, we need to round up to the nearest whole number. Therefore, the number of students that should be in the classroom according to the given sanitary norms is 6.
Please note that the calculation assumes that the entire classroom is available for student use and that there are no obstructions or fixed furniture that would reduce the available space.
I hope this helps! Let me know if you have any further questions.


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