Wiki User
∙ 15y agoPerimeter, P = 32 m
Area, A = 60 m^2.
A = LW
60= LW
L = 60/W
P = 2L + 2W
32= 2L + 2W substitute 60/W for L;
32= 2(60/W) + 2W divide by 2 to both sides;
16 = 60/W + W multiply by W to both sides;
16W = 60 + W^2
0 = 60 + W^2 - 16W or,
W^2 - 16W + 60 = 0
W = [16 +,- square root of [16^2 - (4)(1)(60)]/2
W = [16 +,- square root of (256 -240)]/2
W = (16 +,- square root of 16)/2
W = (16 + 4)/2 or W = (16 - 4)/2
W = 20/2 or W = 12/2
W = 10 m or W = 6 m
L = 60/W = 60/10 or L = 60/6
L = 6 m or L = 10 m
Wiki User
∙ 15y agowhat are the dimensions of the rectangle with this perimeter and an area of 8000 square meters
The dimensions of the rectangle are 2 units and 15 units
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what are the dimensions of the rectangle with this perimeter and an area of 8000 square meters
The dimensions of the rectangle are 2 units and 15 units
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.
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