Question

One more rectangular-shaped piece of metal siding needs to be cut to cover the exterior of a pole barn. The area of the piece is 30 ft². The length is 1 less than 3 times the width. How wide should the metal piece be? Round to the nearest hundredth of a foot.

A.) 3.33 ft
B.) 4.3 ft
C.) 1 ft
D.) 30 ft

Answer 1

Let w be the width of the rectangle. With this representation, length becomes 3w - 1. The area of the rectangular figure is,
A = L x W
where L and W are length and width, respectively. Substituting the known values,
30 ft² = (3w - 1)(w)
The value of w from the equation is 3.33 ft.

Answer 2

Option A is correct

3.33 ft wide should the metal piece be.

Explanation:

Area of rectangle(A) is given by:

`A =lw` .....[1]

l is the length and w is the width of the rectangle respectively.

As per the statement:

One more rectangular-shaped piece of metal siding needs to be cut to cover the exterior of a pole barn. The area of the piece is 30 ft²

⇒A = 30 ft²

It is also given that the length is 1 less than 3 times the width

⇒
`l = 3w -1` ft

Substitute the given values in [1] we have;

`30 = (3w-1) \cdot w`

Using distributive property:
`a \cdot (b+c) = a\cdot b+ a\cdot c`

then;

`30 = 3w^2-w`

We can write this as:

`3w^2-w-30 = 0`

Factorize this equation:

Split the middle term we have;

`3w^2-10w+9w-30 = 0`

⇒
`w(3w-10)+3(3w-10=0)`

⇒
`(3w-10)(w+3)=0`

By zero product property we have;

`3w-10 = 0` and w+3 = 0

Since, w cannot be negative;

so,

`w = (10)/(3) = 3.3333..` ft

Therefore, to the nearest hundredth of a foot, 3.33 ft wide should the metal piece be.