Question

Прямоугольник разрезали на одинаковые квадраты со стороной, равной ширине прямоугольника. Затем квадраты пронумеровали подряд: 1, 2, 3, 4, и т.д. Оказалось, что сумма периметров квадратов с нечётными номерами на 24 см больше, чем сумма периметров квадратов с чётными номерами, а периметр данного прямоугольника в 3 раза больше периметра квадрата. Найдите площадь прямоугольника. Ответ объясните.

Answer 1

Обозначим квадраты буквами. Пусть каждая буква обозначает длину стороны каждого квадрата.

текст при наведении

Выразим все стороны через F

G=F+1

J=G+1=F+2

E=J+1=F+3

E+1=F+D

подставляем вместо Е F+3

F+3+1=F+D

D=4

B=E+D=F+3+4=F+7

A=B+D=F+7+4=F+11

C=G+F=2F+1

Противоположные стороны прямоугольника равны:

A+B=J+G+C

F+11+F+7=F+2+F+1+2F+1

2F+18=4F+4

2F=14

F=7

Находим стороны прямоугольника.

1-я сторона A+B

A+B = 18+2F=18+14=32

2-я сторона A+C

A+C = F+11+2F+1=3F+12=21+12=33

Стороны прямоугольника 32 и 33

Answer 2

Problem Analysis

We are given a rectangle that has been divided into equal squares, with each square numbered consecutively. The sum of the perimeters of the squares with odd numbers is 24 cm greater than the sum of the perimeters of the squares with even numbers. Additionally, the perimeter of the rectangle is three times greater than the perimeter of a single square. We need to find the area of the rectangle.

Solution

Let's assume that the width of the rectangle is "w" and the length is "l". Since the rectangle has been divided into equal squares, the side length of each square is equal to the width of the rectangle.

To find the area of the rectangle, we need to find the product of its width and length: Area = width * length.

To solve this problem, we need to find the values of "w" and "l".

Finding the Perimeter of the Rectangle

We are given that the perimeter of the rectangle is three times greater than the perimeter of a single square. The perimeter of a square is given by the formula: Perimeter = 4 * side length.

Let's denote the perimeter of the rectangle as "P_r" and the perimeter of a single square as "P_s". According to the given information, we have the equation: P_r = 3 * P_s.

Since the side length of each square is equal to the width of the rectangle, we can express the perimeters in terms of the width "w": P_r = 2 * (w + l) and P_s = 4 * w.

Substituting these values into the equation, we get: 2 * (w + l) = 3 * 4 * w.

Simplifying the equation, we have: 2w + 2l = 12w.

Rearranging the terms, we get: 2l = 12w - 2w.

Simplifying further, we have: 2l = 10w.

Dividing both sides of the equation by 2, we get: l = 5w.

Finding the Difference in Perimeters

We are given that the sum of the perimeters of the squares with odd numbers is 24 cm greater than the sum of the perimeters of the squares with even numbers. Let's denote the sum of the perimeters of the squares with odd numbers as "P_odd" and the sum of the perimeters of the squares with even numbers as "P_even".

The sum of the perimeters of the squares with odd numbers can be expressed as: P_odd = 4 * (1 + 3 + 5 + ...).

The sum of the perimeters of the squares with even numbers can be expressed as: P_even = 4 * (2 + 4 + 6 + ...).

According to the given information, we have the equation: P_odd = P_even + 24.

Substituting the formulas for P_odd and P_even, we get: 4 * (1 + 3 + 5 + ...) = 4 * (2 + 4 + 6 + ...) + 24.

Simplifying the equation, we have: 1 + 3 + 5 + ... = 2 + 4 + 6 + ... + 6.

Finding the Value of "w"

To find the value of "w", we need to consider the sum of the perimeters of the squares with odd numbers and the sum of the perimeters of the squares with even numbers.

The sum of the perimeters of the squares with odd numbers can be expressed as: P_odd = 4 * (1 + 3 + 5 + ...) = 4 * (1 + 2 + 3 + ...) = 4 * (1 + 2 + 3 + ... + n).

The sum of the perimeters of the squares with even numbers can be expressed as: P_even = 4 * (2 + 4 + 6 + ...) = 4 * (2 + 2 + 2 + ...) = 4 * (2 + 2 + 2 + ... + n).

According to the given information, we have the equation: P_odd = P_even + 24.

Substituting the formulas for P_odd and P_even, we get: 4 * (1 + 2 + 3 + ... + n) = 4 * (2 + 2 + 2 + ... + n) + 24.

Simplifying the equation, we have: 1 + 2 + 3 + ... + n = 2 + 2 + 2 + ... + n + 6.

Solving for "n"

To solve for "n", we can use the formula for the sum of an arithmetic series: Sum = (n/2) * (first term + last term).

For the sum of the odd numbers, the first term is 1 and the last term is n. For the sum of the even numbers, the first term is 2 and the last term is n.

Using the formula, we have: (n/2) * (1 + n) = (n/2) * (2 + n) + 6.

Simplifying the equation, we have: (n/2) * (1 + n) = (n/2) * (2 + n) + 6.

Expanding and rearranging the terms, we get: n^2 + n = 2n + n^2 + 4 + 6.

Simplifying further, we have: n = 12.

Finding the Value of "w" and "l"

We have found that "n" is equal to 12. Since the side length of each square is equal to the width of the rectangle, we can conclude that the width of the rectangle is 12 cm.

Using the equation we found earlier, l = 5w, we can substitute the value of "w" to find the value of "l": l = 5 * 12 = 60.

Finding the Area of the Rectangle

Now that we have the values of the width "w" and length "l", we can find the area of the rectangle using the formula: Area = width * length.

Substituting the values, we get: Area = 12 * 60 = 720.

Therefore, the area of the rectangle is 720 square cm.

Summary

To summarize, we were given a rectangle that was divided into equal squares. The sum of the perimeters of the squares with odd numbers was 24 cm greater than the sum of the perimeters of the squares with even numbers. The perimeter of the rectangle was three times greater than the perimeter of a single square. By solving the equations and finding the value of "w" and "l", we determined that the area of the rectangle is 720 square cm.