Assuming no change in the width, yes.
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It would be 4 times greater.To find this algebraically: let L be the length and W the width originallyA = L x WWhen both are doubled, the equation becomesA = (2L) x (2W) = 4LWThe area of the rectangle is quadrupled if both the length and width are doubled.
A = lw Area of a rectangle = length times width
Let's take a look at this problem.Rectangle Perimeter = 2(l + w)Rectangle Perimeter =? 2(2l + 2w)Rectangle Perimeter =? (2)(2)(l + w)2(Rectangle Perimeter) = 2[2(l + w)]Thus, we can say that the perimeter of a rectangle is doubled when its dimensions are doubled.Rectangle Area = lwRectangle Area =? (2l)(2w)Rectangle Area =? 4lw4(Rectangle Area) = 4lwThus, we can say that the area of a rectangle is quadruplicated when its dimensions are doubled.
The length of a rectangle is twice its width. If the perimeter of the rectangle is , find its area.
Length times width gives the area of a rectangle. The rectangle with a length of 38 and a width of 28 has an area of 1064 square units.