Question

The area of a rectangle is 54yd^2 , and the length of the rectangle is 3yd more than double the width. Find the dimensions of the rectangle.

Answer 1

Given:

a.) The area of a rectangle is 54 yd^2.

b.) The length of the rectangle is 3 yd. more than double the width.

Since it's said that the length of the rectangle is 3 yd. more than double the width, the length can also be written as:

L = 2W + 3

Recall: the formula for getting the area of a rectangle.

$\text{ Area = L x W}$

Let's now determine the dimensions of the rectangle. Let's first find the width.

$\text{ Area = L x W}$$\text{ 54 = \lparen2W + 3\rparen\lparen W\rparen}$$\text{ \lparen2W + 3\rparen\lparen W\rparen = 54}$$\text{ 2W}^2\text{ + 3W = 54}$$\text{ 2W}^2\text{ + 3W - 54 = 0}$

a = 2, b = 3 and c = -54

Let's use the quadratic formula to find W.

$\text{ x = W = }\frac{-b\text{ }\pm\text{ }\sqrt{b^2-4ac}}{2a}$$\text{ = }\frac{-3\text{ }\pm\text{ }\sqrt{3^2-4(2)(-54)}}{2(2)}$$\text{ = }\frac{-3\text{ }\pm\text{ }\sqrt{9\text{ + 432 }}}{4}$$\text{ = }\frac{-3\text{ }\pm\text{ }\sqrt{441}}{4}\text{ = }\frac{-3\text{ }\pm21}{4}$$\text{ W}_1\text{ = }\frac{-3\text{ + 21}}{4}\text{ = }\frac{18}{4}\text{ = }\frac{9}{2}$$\text{ W}_2\text{ = }\frac{-3\text{ - 21}}{4}\text{ = }\frac{-24}{4}\text{ = -6}$

Therefore, the width of the rectangle is 9/2 or 4.5 yards because it could never be a negative number.

Let's now find the length.

L = 2W + 3 = 2(9/2) + 3 = 9 + 3 = 12 yards

In Summary,

Width = 4.5 yards

Length = 12 yards