The perimeter of the rectangle is 62m. Find its sides if the area of the rectangle is 210m2.

1) Let x meters be the length of the rectangle, y meters – its width.

2) Then (x * y) m ^ 2 is the area of the rectangle, which, according to the condition of the problem, is 210 m ^ 2.

3) (2 * (x + y)) m – the perimeter of the rectangle, which is 62 m.

4) Let’s solve the system of equations:

x * y = 210;

2 * (x + y) = 62.

Let us express x:

x = 210 / y.

Substitute the x value in the second equation:

2 * (210 / y + y) = 62;

420 + 2y ^ 2 = 62y;

y ^ 2 – 31y + 210 = 0.

D = (-31) ^ 2 – 4 * 1 * 210 = 961 – 840 = 121.

y1 = (- (- 31) + 11) / 2;

y1 = (31 + 11) / 2;

y1 = 21;

y2 = (- (- 31) – 11) / 2;

y2 = (31 – 11) / 2;

y2 = 10.

Let’s find the values of x:

x1 = 210 / y1 = 210/21 = 10 m;

x2 = 210 / y2 = 210/10 = 21 m.

Answer: 10 meters and 21 meters; 21 meters and 10 meters.