Question

Find the dimensions of a rectangle whose area is 221 cm2 and whose perimeter is 60 cm. (enter your answers as a comma-separated list.)

Answer 1

The dimensions of a rectangle with an area of 221 cm² and a perimeter of 60 cm are found by solving a system of equations derived from the definitions of area and perimeter. The dimensions are 13 cm by 17 cm, or equivalently, 17 cm by 13 cm.

Step-by-step explanation:

To find the dimensions of a rectangle with an area of 221 cm2 and a perimeter of 60 cm, we will let the length be x and the width be y. The area of a rectangle is found by multiplying the length and width, so we have the equation x * y = 221. The perimeter is twice the sum of the length and width, so we have 2x + 2y = 60, which simplifies to x + y = 30.

To solve these equations, divide the perimeter equation by 2 to find y = 30 - x. Substituting this into the area equation gives x(30 - x) = 221. Expanding this and bringing all terms to one side provides a quadratic equation: x2 - 30x + 221 = 0. Solving this quadratic equation by factoring or using the quadratic formula gives the dimensions of the rectangle.

The solutions to the quadratic equation are x = 13 and x = 17. Since x and y are interchangeable as length and width, the two sets of possible dimensions for the rectangle are 13 cm by 17 cm and 17 cm by 13 cm.

Answer 2

60 = 2 ( length + width )

30 = length + width

Let length be = x and width = y

x+y = 30

xy = 209

4xy = 836

( x+ y)^2 = 900

( x+y)^2 - 4xy = 900-836 = 64

( x-y)^2 = 64

x- y = 8

x+y = 30

ADD 2x =38

x = 19

y = 11

ANSWER 19cm X 11cm