Question

Решение задачи:Квадрат и прямоугольник имеют одинаковый периметр - 36см.Площадь какой фигуры больше , если длина прямоугольника 10см?

Answer 1

Рпрям=2*(а+b)
b=P:2-a
b=36:2-10=8 см ширина прямоугольника
Sпрям=a*b
Sпрям=10*8=80 см^2 площадь прямоугольника

Рквадр=4а
а=Р:4
а=36:4
а=9 см сторона квадрата
Sквадр=а^2
S=9*9=81 см^2 площадь квадрата

81>80,  следовательно, площадь квадрата больше площади прямоугольника

Answer 2

Если периметр прямоугольника 36 и одна из его сторон равна 10, а противолежащие стороны прямоугольника равны, то x(вторая сторона) +x=36-10/2
x=8. Тогда его площадь равна 8*10=80.
Если периметр квадрата равен 36, то его сторона равна: y=36/4=9. Тогда его площадь равна 9*9=81.
Получается, что площадь квадрата больше.

Answer 3

Problem Analysis

We are given that a square and a rectangle have the same perimeter of 36 cm. We need to determine which figure has a larger area, given that the length of the rectangle is 10 cm.

Solution

To solve this problem, we can start by finding the dimensions of both the square and the rectangle.

Let's assume that the side length of the square is 's' cm. Since a square has all sides equal, the perimeter of the square would be 4s cm. We are given that the perimeter is 36 cm, so we can write the equation:

4s = 36

Simplifying the equation, we find:

s = 36 / 4 = 9 cm

Therefore, the side length of the square is 9 cm.

Now, let's consider the rectangle. We are given that the length of the rectangle is 10 cm. Let's assume the width of the rectangle is 'w' cm. The perimeter of the rectangle can be calculated using the formula:

Perimeter = 2(length + width)

Since the perimeter is given as 36 cm, we can write the equation:

2(10 + w) = 36

Simplifying the equation, we find:

20 + 2w = 36 2w = 36 - 20 2w = 16 w = 16 / 2 = 8 cm

Therefore, the width of the rectangle is 8 cm.

Now that we have the dimensions of both the square and the rectangle, we can calculate their respective areas.

The area of a square is given by the formula:

Area = side length * side length

Substituting the value of the side length, we find