Question

A rectangle has a length of x inches and a width 2 inches less than the length. If the dimensions were doubled, what would be the area, in square inches , of the new rectangle in terms of x

Answer 1

Doubling the dimensions of the rectangle means the length and width are each multiplied by a scale factor of 2. The new area in terms of x will be 4x^2 - 8x square inches, representing an area scale factor of 4 since areas are proportional to the square of the linear dimensions.

Step-by-step explanation:

The original rectangle has a length of x inches and a width that is 2 inches less than the length, which would be x - 2 inches. The area of this rectangle is x multiplied by (x - 2) square inches.

If the dimensions of the rectangle are doubled, the new length would be 2x inches and the new width would be 2 times (x - 2), which is 2x - 4 inches. The area of the new, larger rectangle is therefore (2x) × (2x - 4) square inches, or 4x^2 - 8x square inches.

To solve using scale factors, you would recognize that doubling the dimensions corresponds to a scale factor of 2. So for a rectangle with a width of 2 inches, when doubled, it becomes 4 inches (scale factor 2), and similarly for the length x, when doubled, it becomes 2x (scale factor 2). The area scale factor will be the square of the linear scale factor (2^2 = 4).

Answer 2

Option (4)

Step-by-step explanation:

Area of a rectangle = Length × width

If the dimensions of the given rectangle is doubled,

Area of the new rectangle = 2(Length) × 2(width)

= 4(Length × width)

= 4(x)(x - 2)

= 4[x² - 2x]

= 4x² - 8x

Therefore, Option (4) will be the correct option.