Question

The rectangle shown has a perimeter of 52 cm and the given area. Its length is 6 more than four times its width. Write and solve a system of equations to find the dimensions of the rectangle. The length of the rectangle is cm and the width of the rectangle is cm. (...) W L 2 A 88 cm​

Answer 1

Let's begin by defining the variables. Let "w" be the width of the rectangle and let "l" be the length of the rectangle.
We can write the equation for the perimeter as: 2w + 2l = 52.
The equation for the area is: wl = 88.
We can substitute l in the perimeter equation to yield: 2w + 2(4w + 6) = 52.
Simplifying gives us: 10w + 12 = 52.
Subtracting 12 from both sides yields: 10w = 40.
Finally, dividing by 10 gives us w = 4 cm.
Since the width of the rectangle is 4 cm, the length of the rectangle is (4 x 4) + 6 = 22 cm. Therefore, the dimensions of the rectangle are 4 cm x 22 cm.