The perimeter of the rectangle is 60 cm. If the length of the rectangle is increased by 10 cm and the width

The perimeter of the rectangle is 60 cm. If the length of the rectangle is increased by 10 cm and the width is reduced by 6 cm, then the area of the rectangle will decrease by 32 cm2. find the area of the rectangle.

The perimeter of a rectangle is equal to the sum of the lengths of its four sides. Half of the perimeter is the sum of the two sides, length and width. The sum of the length and width will be 60: 2 = 30 cm.

Let the length of the rectangle be x cm, then the width of the rectangle is (30 – x) cm, and its area will be x (30 – x) cm ^ 2. If the length of the rectangle is increased by 10 cm, then it will become equal to (x + 10) cm, and if the width of the rectangle is reduced by 6 cm, then it will become equal to (30 – x – 6) = (24 – x) cm, and the area will become is equal to (x + 10) (24 – x) cm ^ 2. By the condition of the problem it is known that the initial area will be larger by (x (30 – x) – (x + 10) (24 – x)) cm ^ 2 or by 32 cm ^ 2. Let’s make an equation and solve it.

x (30 – x) – (x + 10) (24 – x) = 32;

30x – x ^ 2 – (24x – x ^ 2 + 240 – 10x) = 32;

30x – x ^ 2 – 24x + x ^ 2 – 240 + 10x = 32;

16xx = 32 + 240;

16x = 272;

x = 272: 16;

x = 17 (cm) – length;

30 – 17 = 13 (cm) – width.

Find the area of ​​the rectangle. The area of ​​a rectangle is equal to the product of its sides.

S = 17 * 13 = 221 (cm ^ 2).

Answer. 221 cm ^ 2.