Question

Периметр прямоугольника 32дм. Какова его площадь,если длина в 7 раз больше, чем его ширина? Мне надо решение и ответ.

Answer 1

Принимаем:-ширина-Х, тогда длина равна-7Х Решение: Х+Х+7Х+7Х=32      Выносим Х за скобку. Х(1+1+7+7)=32      16Х=32    Х=32:16    Х=2

Answer 2

Problem Analysis

We are given that the perimeter of a rectangle is 32 dm, and the length is 7 times greater than the width. We need to find the area of the rectangle.

Solution

Let's assume the width of the rectangle is w dm. According to the problem, the length of the rectangle is 7 times greater than the width, so the length is 7w dm.

The perimeter of a rectangle is given by the formula: perimeter = 2(length + width). Substituting the given values, we have:

32 = 2(7w + w)

Simplifying the equation, we get:

32 = 2(8w)

Dividing both sides of the equation by 2, we have:

16 = 8w

Simplifying further, we get:

w = 2

Now that we have the width of the rectangle, we can find the length by multiplying it by 7:

length = 7w = 7(2) = 14

The width of the rectangle is 2 dm and the length is 14 dm.

To find the area of the rectangle, we use the formula: area = length × width. Substituting the values, we have:

area = 14 × 2 = 28 dm²

Therefore, the area of the rectangle is 28 dm².

Answer

The area of the rectangle is 28 dm².

Verification

Let's verify the solution by calculating the perimeter using the given values:

perimeter = 2(length + width) = 2(14 + 2) = 2(16) = 32 dm

The calculated perimeter matches the given perimeter, which confirms that the solution is correct.