The length of a rectangle is 3 cm less than twice the width. If the area of the rectangle is 54 cm2, what is the width of the rectangle?

Answer 1

Hi there,

So to solve this you essentially need to create two algebraic equations. I will use l for length and w for width.

First since the length of the rectangle is 3 cm less than twice the width, this equates to l = 2*w - 3. Next it tells you that the area, or (l*w), is 54 cm². So, l*w = 54 and l = 2*w - 3. From here I would substitute l = 2*w - 3 into the equation l*w = 54 to get (2*w - 3)*(w) = 54. Next expand the left side of the equation to get 2*w² - 3*w = 54. Bring over 54 to the left side by subtracting it from both sides to get 2*w² - 3*w - 54 = 0. Now use the quadratic formula to get solutions to the formula as x = 6, x = -9/2. Since length can't be negative, we can throw out -9/2 as a solution, to get 6 cm as the width.

To find the length, simply plug it back into either original equation. I am going to plug it back into the equation l*w = 54 since it's much simpler. When we do this, we get 6*l = 54, and after dividing both sides by 6, l comes out to 9.

Thus, the length is 9 cm and the width is 6 cm (l = 9, w = 6).

Answer 2

w = width

l = length = 3 less than twice width = 2w - 3

area of rectangle = w* l = w * (2w - 3) = 54

2w^2 - 3w = 54 (expand)

2w^2 - 3w - 54 = 0

2w^2 - 12w + 9w - 54 = 0 (factor)

2w(w-6) + 9(w-6) = 0

(2w + 9)(w-6) = 0

2w + 9 = 0 or w- 6 = 0

w = -9/2 or 6

w = 6 is only viable solution

width = 6

Verify: length = 2w - 3 = 9

area = length * width = 6 * 9 = 54