Question

Периметр прямокутника 54 см. Його довжина вдвічі більша за ширину. Знайти площу.

Answer 1

P = 2A + 2B

..........................

 

 A = 2B

P = 2*2B + 2B = 4B + 2B = 6B

54 = 6B --> B = 9 (cm)

A = 2*9 = 18 (cm)

.............................

 

  S= A*B = 18 * 9 = 162 (cm^2) - Ответ

 

 

 

 

 

 

 

Answer 2

Problem Analysis

We are given that the perimeter of a rectangle is 54 cm, and its length is twice the width. We need to find the area of the rectangle.

Solution

Let's assume the width of the rectangle is x cm. Since the length is twice the width, the length would be 2x cm.

The perimeter of a rectangle is given by the formula: perimeter = 2(length + width).

Substituting the given values, we have: 54 = 2(2x + x).

Simplifying the equation, we get: 54 = 2(3x).

Dividing both sides of the equation by 2, we have: 27 = 3x.

Dividing both sides of the equation by 3, we get: 9 = x.

Therefore, the width of the rectangle is 9 cm, and the length is twice the width, which is 18 cm.

To find the area of the rectangle, we use the formula: area = length * width.

Substituting the values, we have: area = 18 * 9.

Calculating the area, we get: area = 162 square cm.

So, the area of the rectangle is 162 square cm.

Answer

The area of the rectangle is 162 square cm.