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?2x−42 x − 48x−88 x − 82x2−4x2 x 2 − 4 x4x2−8x4 x 2 − 8 x

Answer 1

If the dimensions of the rectangle were doubled, the area in square inches is: $Area = 4x^2 - 8x$

• Let the length of the rectangle be L.

• Let the width of the rectangle be W.

Given the following data:

• Length = x inches

• Width = x - 2 inches

If the dimensions of the rectangle were doubled, the value of L and W becomes;

$L = 2x$

$W = 2(x - 2)$

Mathematically, the area of a rectangle is given by the formula;

$Area = Length$ × $Width$

$Area = 2x$ × $2(x-2)$

$Area = 2x$ × $2x - 4$

$Area = 4x^2 - 8x$  square inches.

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