Find the perimeter of a rectangular area of 91 m2, if one of the sides is 6 m larger than the other.

First, let’s write down the perimeter and area formulas for a rectangle:

P = 2 * (a + b) (where P is the perimeter; a is the length of the side; b is the width of the rectangle).

S = a * b (where S is the area; a is the length of the side; b is the width of the rectangle).

Let the width of this rectangle be X meters; then its length will be equal to: X + 6 (meters).

By condition, it is given that the area of ​​this rectangle is 91 m ^ 2.

Let’s compose and solve the equation:

91 = X * (X + 6);

Let’s transform:

91 = X ^ 2 + 6X;

Move the number from the left to the right with a change in signs:

0 = X ^ 2 + 6X – 91;

D1 = (6/2) ^ 2 + 91;

D1 = 9 + 91;

D1 = 100;

√D1 = √100 = 10.

X1 = -3 – 10 = -13 – does not satisfy the condition of the problem.

X2 = -3 + 10 = 7.

Thus, the width of this rectangle is 7 meters.

Then the length is equal to:

7 + 6 = 13 (meters) – the length of the rectangle.

Now let’s find the perimeter:

P = 2 * (13 + 7) = 2 * 20 = 40 (meters).

Answer: P = 40 meters.