Two rectangles have the same shape. The perimeter of the larger one is twice the perimeter of the smaller one.How many times as great is the larger rectangle's area?
PerimeterLarge=2W+2L
area= WL
but the small one, is half the perimeter of the large, so permitersmall= w+L
and area = (w/2)(l/2)= 1/4 wl so the large area is 4 times the smaller.
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To solve this problem, we need to use the relationship between the perimeters and areas of rectangles.
Let's assume the length and width of the smaller rectangle are "l" and "w" respectively.
Therefore, the perimeter of the smaller rectangle would be:
Perimeter = 2 * (l + w)
According to the given information, the perimeter of the larger rectangle is twice that of the smaller rectangle. So, the perimeter of the larger rectangle would be:
Perimeter = 2 * 2 * (l + w)
Perimeter = 4 * (l + w)
Now, let's assume the length and width of the larger rectangle are "L" and "W" respectively.
Therefore, the perimeter of the larger rectangle in terms of L and W would be:
Perimeter = 2 * (L + W)
Since we know that the perimeter of the larger rectangle is twice that of the smaller rectangle, we can set up the following equation:
4 * (l + w) = 2 * (L + W)
Now, let's solve the equation to get the relationship between l, w, L, and W.
Dividing the equation by 2:
2 * (l + w) = L + W
Since the two rectangles have the same shape, the ratios of their corresponding sides (length and width) should be equal. Therefore, we can assume:
L = 2l
W = 2w
Substituting these values in the equation:
2 * (l + w) = 2l + 2w
l + w = 2l + 2w
0 = l + w - 2l - 2w
0 = l - w
From this equation, we can conclude that the length and width of the smaller rectangle are equal to the length and width of the larger rectangle. So, both rectangles are squares.
Now, let's compare the areas of the two squares.
The area of the smaller square is: Area = l * w
The area of the larger square is: Area = L * W = 2l * 2w = 4 * l * w
Comparing the area of the larger square with the smaller square, we can conclude that the larger square's area is 4 times greater than the smaller square's area.
Therefore, the larger rectangle's area is 4 times as great as the smaller rectangle's area.