Two rectangles with an equal area equal to 70 m3 are given. the length of the first rectangle is 4

Two rectangles with an equal area equal to 70 m3 are given. the length of the first rectangle is 4 more than the second, and the width is 2 less than the second. find the sides of the rectangles.

Let’s denote the length of the first rectangle through x, and the width of the first rectangle through y.

The problem statement says that the length of the 1st rectangle is 4 m longer than the 2nd, and the width of the 1st rectangle is 2 m less than the 2nd, therefore, the length of the 2nd rectangle is x – 4 m, and the width of the 2nd rectangle is y + 2 m.

It is also known that the areas of these rectangles are equal and equal to 70 m ^ 2, therefore, we can write the following ratio:

xy = 70;

xy = (x – 4) * (y + 2).

From the second equation we get:

xy = xy + 2x – 4y – 8;

xy + 2x – 4y – 8 – xy = 0;

2x – 4y – 8 = 0;

x – 2y – 4 = 0;

x = 2y + 4.

Substituting the found value x = 2y + 4 into the equation xy = 70, we get:

(2y + 4) * y = 70;

2y ^ 2 + 4y = 70;

2y ^ 2 + 4y – 70 = 0;

y ^ 2 + 2y – 35 = 0;

y = -1 ± √ (1 + 35) = -1 ± √36 = -1 ± 6;

y = -1 + 6 = 5.

Find x:

x = 2y + 4 = 2 * 5 + 4 = 14.

Find the sides of the 2nd rectangle:

x – 4 = 14 = 4 = 10 m;

y + 2 = 5 + 2 = 7 m.

Answer: the sides of the 1st rectangle are 14 m and 5 m, the sides of the 2nd rectangle are 10 m and 7 m.