# Find the side lengths of a rectangle with an area of 51 cm2 and a perimeter of 40 cm.

July 24, 2021 | education

| Let x cm be the length of the rectangle and y be its width. Then the area of the rectangle is S = xy, and by condition 51. The perimeter of the rectangle is P = (x + y) 2, and by condition 40. Let’s compose and solve the system of equations.

{xy = 51

(x + y) 2 = 40.

We transform the second equation and express from it x

x + y = 20

x = 20 – y. Substitute in the first equation of the system.

(20-y) y = 51

20y – y ^ 2 = 51

-y ^ 2 + 20y-51 = 0

y ^ 2-20y + 51 = 0 by Vietta’s theorem

y1 + y2 = 20

y1 * y2 = 51

y1 = 3

y2 = 17.

Substitute the roots to x = 20-y and get

x1 = 17

x2 = 3.

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