# Find the lengths of the sides of the rectangle if its area is 20 dm ^ 2, and the half-perimeter is 9 dm.

Let one side of the rectangle be x dm, then the second side of the rectangle is equal to (9 – x) dm (half-perimeter is the sum of the two sides (length and width) of the rectangle; therefore, from the sum of the two sides 9 dm, subtract the length of the known side x). By the condition of the problem, it is known that the area of a rectangle (the area of a rectangle is equal to the product of its sides) is equal to x (9 – x) dm ^ 2 or 20 dm ^ 2. Let’s make an equation and solve it.

x (9 – x) = 20;

9x – x ^ 2 = 20;

x ^ 2 – 9x + 20 = 0;

D = b ^ 2 – 4ac;

D = (-9) ^ 2 – 4 * 1 * 20 = 81 – 80 = 1;

x = (-b ± √D) / (2a);

x1 = (9 + 1) / 2 = 10/2 = 5 (dm) – first first side;

x2 = (9 – 1) / 2 = 8/2 = 4 (dm) – second first side;

9 – x1 = 9 – 5 = 4 (dm) – the first second side;

9 – x2 = 9 – 4 = 5 (dm (- second second side.

The sides of the rectangle are 5 inches and 4 inches, or 4 inches and 5 inches, which are the same.

Answer. 5 dm, 4 dm.