The perimeter of the rectangle is 70 cm. If its length is reduced by 5 cm, and its width is increased by 5 cm

The perimeter of the rectangle is 70 cm. If its length is reduced by 5 cm, and its width is increased by 5 cm, then the area will increase by 50 cm2. Find the length and width of the perfect rectangle.

The perimeter of a rectangle is the doubled sum of its sides.
Since the perimeter is 70 cm, the sum of the sides will be:
70 = 2 * (a + b) = 70/2 = a + b.
35 = a + b.
Let us express the value in from the first equation.
h = 35 – a.
The area is the product of the sides of the rectangle.
In this case, we get equality:
(a – 5) * (b + 5) = a * b + 50.
Substitute the value in from the first equation into the second.
(a – 5) * (35 – a + 5) = a * (35 – a) + 50.
35 * a – a ^ 2 + 5 * a – 175 + 5 * a – 25 = 35 * a – a ^ 2 + 50.
35 * a – 35 * a – a ^ 2 + a ^ 2 + 5 * a + 5 * a = 50 + 25 + 175.
10 * a = 250.
a = 250/10 = 25 cm (initial length of the rectangle).
h = 35 – 25 = 10 cm (the initial width of the rectangle).
Answer: Length 25 cm, width 10 cm.