# What are the sides of a rectangle with an area of 12 cm and a perimeter of 26 cm?

Let “x” be the length of the rectangle and “y” the width. Let’s compose a system of equations:

(1) x * y = 12.

(2) (x + y) * 2 = 26.

(x + y) * 2 = 26;

x + y = 26: 2;

x + y = 13;

(3) x = 13 – y.

Substitute equation (3) into equation (1).

x * y = 12.

(13-y) * y = 12;

-y² + 13y – 12 = 0;

D = b² – 4ac = 132 – 4 · (−1) · (−12) = 169 – 48 = 121;

√D = √121 = 11;

y = (−b ± √D) / 2a;

y₁ = (-13 + 11) / (2 * (-1)) = (-2) / (- 2) = 1 (cm);

y₂ = (-13 – 11) / (2 * (-1)) = (-24) / (- 2) = 12 (cm).

x = 13 – y.

x₁ = 13 – 1 = 12 (cm).

x₂ = 13 – 12 = 1 (cm).

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