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The length of a rectangle is 4 cm greater than twice its width. The area of the rectangle is 30 cm2. Determine the dimensions of the rectangle.

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Final answer:

The dimensions of the rectangle are width w = 3 cm and length l = 10cm.

Step-by-step explanation:

Let's denote the width of the rectangle as w and the length as l.

According to the given information:

1. The length is 4 cm greater than twice the width: l = 2w + 4.

2. The area of the rectangle is 30 cm^2: A = lw = 30.

Substitute the expression for l into the area formula:

2w + 4 x w = 30

Combine like terms:

2w^2 + 4w = 30

Rearrange into a quadratic equation:

2w^2 + 4w - 30 = 0

Now, solve for w. You can use the quadratic formula:

w = {-b pm {b^2 - 4ac}^1/2} / {2a}

In this case, a = 2, b = 4, and c = -30.

w = {-4 pm {4^2 - 4(2)(-30)}^1/2} / {2(2)}

w = {-4 pm {256}^1/2} / {4}

w = {-4 pm 16} / {4}

The two possible solutions for w are w = 3 and w = -{4} / {2}. Since the width cannot be negative, we discard the second solution.

Now, use the value of w to find l using the relationship l = 2w + 4:

l = 2(3) + 4 = 6 + 4 = 10

Therefore, the dimensions of the rectangle are: width w = 3 cm and length l = 10 cm.

User Jon Bright
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