It would be 4 times greater.
To find this algebraically: let L be the length and W the width originally
A = L x W
When both are doubled, the equation becomes
A = (2L) x (2W) = 4LW
The area of the rectangle is quadrupled if both the length and width are doubled.
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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.
what are the dimensions of the rectangle with this perimeter and an area of 8000 square meters
If you know the dimensions of the missing triangle, then compute the area from those dimensions, then subtract that answer from the area of the full rectangle.
The dimensions of the rectangle are 2 units and 15 units
27:55