Question

Периметр прямокутника дорiвнюе 120м. його довжина на 10м бiльша за ширину. Знайди площу прямокутника

Answer 1

Пусть а (ширина)- х,тогда b (длина)  10+х          Р=а+b+a+b
10+х+х+10+х+х=120
20+4 х=120
4 х=120-20
4 х=100
х=100:4=25 м ширина
25+10=35 м длина
S=a*b
S=25*35=875 м. кв.

Answer 2

Problem Analysis

We are given that the perimeter of a rectangle is 120m, and its length is 10m greater than its width. We need to find the area of the rectangle.

Solution

Let's assume the width of the rectangle is x meters. Since the length is 10m greater than the width, the length would be x + 10 meters.

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

According to the problem, the perimeter is 120m. Substituting the values into the formula, we get:

2(x + 10 + x) = 120

Simplifying the equation, we have:

2(2x + 10) = 120

4x + 20 = 120

4x = 100

x = 25

So, the width of the rectangle is 25m, and the length is 25 + 10 = 35m.

The area of a rectangle is given by the formula: length × width.

Substituting the values, we get:

Area = 35 × 25 = 875 square meters

Therefore, the area of the rectangle is 875 square meters.

Answer

The area of the rectangle is 875 square meters.